Differential Equations

Level · University · builds on single-variable Calculus Chapters · 9 Feeds into · Physics · Population modeling · Engineering

✓ Official Syllabus Verified Last reviewed against official syllabus: 2025–2026 Academic Year Standards & Rigor ↗

First-Order Differential Equations, Separable Equations & Existence Theory

A differential equation expresses an intrinsic law of change, connecting an unknown function directly to its own derivatives. In nature and technology, physical laws rarely dictate exact states; instead, they dictate instantaneous rates of progression — cooling rates proportional to temperature differentials, chemical reaction kinetics determined by molecular concentrations, and gravitational acceleration dictating planetary velocities. This foundational chapter develops the analytical theory of first-order differential equations, the method of separation of variables, and the fundamental Picard–Lindelöf Existence and Uniqueness Theorem.

● Level 1 Foundation — Definition 1.1: Differential Equations, Solutions & Initial Value Problems (IVPs)

An ordinary differential equation (ODE) is an equation relating an unknown single-variable function $y = y(x)$ to one or more of its derivatives $y', y'', \dots, y^{(n)}$ . The order of the ODE is the order of the highest derivative present.

A solution on an open interval $I \subseteq \mathbb{R}$ is an $n$ -times differentiable function $y(x)$ that identically satisfies the equation for all $x \in I$ . The general solution is the complete family of all possible solutions, containing $n$ essential arbitrary constants.

An Initial Value Problem (IVP) couples an $n$ -th order ODE with $n$ prescribed conditions at a single point $x_0 \in I$ : $y(x_0) = y_0, \quad y'(x_0) = y_1, \quad \dots, \quad y^{(n-1)}(x_0) = y_{n-1}.$ A solution satisfying both the differential equation and these initial conditions is designated a particular solution.

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

Worked example 1.A — Solution Verification & Interval of Definition

Verify that $y(x) = C e^{2x}$ is a solution of $\dfrac{dy}{dx} = 2y$ for any real constant $C$ , prove that these are the only solutions on $\mathbb{R}$ , and determine the unique solution satisfying $y(0) = 5$ .

Verification: Differentiating gives $\dfrac{dy}{dx} = \dfrac{d}{dx}[C e^{2x}] = 2C e^{2x} = 2y(x)$ , satisfying the ODE for all $x \in \mathbb{R}$ .

Exhaustiveness Proof: Suppose $y(x)$ is any arbitrary solution to $y' = 2y$ . Consider the product $u(x) = e^{-2x}y(x)$ . Applying the product rule: $u'(x) = -2e^{-2x}y(x) + e^{-2x}y'(x) = e^{-2x}\big(y'(x) - 2y(x)\big) = e^{-2x}(0) = 0.$ By the Zero Derivative Theorem (a direct corollary of the Mean Value Theorem), a function whose derivative vanishes on an interval is constant: $u(x) = C$ . Hence $y(x) = C e^{2x}$ , proving no other solutions exist.

Initial Condition: Setting $x = 0$ : $y(0) = C e^0 = C = 5$ . The unique particular solution is $y(x) = 5e^{2x}$ .

● Level 1 Foundation — Definition 1.2: Autonomous & Separable Differential Equations

A first-order differential equation is separable if the derivative factors into an explicit function of $x$ multiplied by an explicit function of $y$ : $\frac{dy}{dx} = f(x)\,g(y).$

An ODE is autonomous if the independent variable does not appear explicitly: $\dfrac{dy}{dx} = g(y)$ . Any real value $y^*$ such that $g(y^*) = 0$ produces a constant solution $y(x) \equiv y^*$ , termed an equilibrium solution (or critical point). Geometrically, equilibrium solutions correspond to horizontal integral curves in the $xy$ -plane.

Analytical Procedure for Separable Equations

1. Equilibrium Extraction

Locate all roots of

g(y) = 0

. Every real root

y = y^*

is a constant equilibrium solution that must be recorded before division to avoid singular division by zero.

2. Differential Separation

On intervals where

g(y) \neq 0

, rewrite the equation in differential form:

\frac{1}{g(y)}\,dy = f(x)\,dx

3. Indefinite Integration

Integrate both differentials:

\int \frac{1}{g(y)}\,dy = \int f(x)\,dx + C

. A single arbitrary constant

C

suffices, as constants of integration combine.

4. Explicit Isolation & Domain

Invert the algebraic relation to solve explicitly for

y(x)

where possible, and determine the maximal open interval of existence

I

containing the initial point.

▲ Level 2 Analytical Theorem 1.1 — Separation on Nonzero Branches

Let $I, J \subseteq \mathbb{R}$ be open intervals. Suppose $f: I \to \mathbb{R}$ is continuous and $g: J \to \mathbb{R}$ is continuous with $g(y) \neq 0$ for all $y \in J$ . Let $F(x)$ be an antiderivative of $f(x)$ on $I$ , and let $G(y)$ be an antiderivative of $1/g(y)$ on $J$ . A differentiable function $y: I \to J$ satisfies $\frac{dy}{dx} = f(x)\,g(y)$ if and only if there exists a constant $C \in \mathbb{R}$ such that $G(y(x)) = F(x) + C.$

Proof

Sufficiency (⇒): Suppose $y(x)$ satisfies the ODE with $y(x) \in J$ . Since $g(y(x)) \neq 0$ , divide by $g(y(x))$ : $\frac{1}{g(y(x))}\,y'(x) = f(x).$ By the Chain Rule, $\dfrac{d}{dx}[G(y(x))] = G'(y(x))\,y'(x) = \dfrac{1}{g(y(x))}\,y'(x)$ . Thus: $\frac{d}{dx}[G(y(x))] = \frac{d}{dx}[F(x)].$ Integrating both sides with respect to $x$ yields $G(y(x)) - F(x) = C$ , so $G(y(x)) = F(x) + C$ .

Necessity (⇐): Conversely, suppose $G(y(x)) = F(x) + C$ . Differentiating both sides with respect to $x$ using the Chain Rule yields: $G'(y(x))\,y'(x) = F'(x) \implies \frac{1}{g(y(x))}\,y'(x) = f(x).$ Multiplying by $g(y(x))$ yields $y'(x) = f(x)\,g(y(x))$ , completing the proof.

Worked example 1.B — Initial Value Problems & Finite-Time Blowup

Solve the initial value problem $\frac{dy}{dx} = 2x y^2, \qquad y(0) = 1,$ and determine its maximal open interval of valid existence.

Step 1: Check equilibrium: $g(y) = y^2 = 0 \implies y \equiv 0$ . Since $y(0) = 1 \neq 0$ , the solution lies on the positive branch $y > 0$ .

Step 2: Separate and integrate: $\int y^{-2}\,dy = \int 2x\,dx \implies -\frac{1}{y} = x^2 + C.$ Applying $y(0) = 1$ : $-\frac{1}{1} = 0^2 + C \implies C = -1$ .

Step 3: Solve for $y(x)$ : $-\frac{1}{y} = x^2 - 1 \implies y(x) = \frac{1}{1 - x^2}.$

Step 4: Maximal interval of existence: The function $y(x) = \frac{1}{1 - x^2}$ has vertical asymptotes at $x = \pm 1$ . Because a solution to an ODE must be continuous on an interval containing the initial point $x_0 = 0$ , the maximal interval of existence is strictly: $I = (-1, 1).$ As $x \to 1^-$ or $x \to -1^+$ , $y(x) \to +\infty$ . This phenomenon is known as finite-time blowup, demonstrating that even when the ODE's right-hand side $f(x, y) = 2x y^2$ is infinitely smooth everywhere on $\mathbb{R}^2$ , solutions to non-linear equations can cease to exist in finite time.

Worked example 1.C — Analytical Application: Logistic Growth & Critical Harvesting Threshold

A biological population $P(t)$ with carrying capacity $K$ and intrinsic growth rate $r$ is subject to a constant harvesting rate $H > 0$ : $\frac{dP}{dt} = r P\left(1 - \frac{P}{K}\right) - H.$ Determine the critical harvesting threshold $H_{\text{crit}}$ above which population collapse is inevitable.

Equilibrium Analysis: Set $\frac{dP}{dt} = 0$ : $-\frac{r}{K}P^2 + r P - H = 0 \implies r P^2 - r K P + K H = 0.$ The roots are given by the quadratic formula: $P_{1, 2} = \frac{r K \pm \sqrt{r^2 K^2 - 4 r K H}}{2r} = \frac{K}{2} \left(1 \pm \sqrt{1 - \frac{4H}{rK}}\right).$

Saddle-Node Bifurcation: The discriminant $\Delta = r^2 K^2 - 4rKH$ dictates the physical dynamics:

If $ H < \frac{rK}{4} $: Two real equilibria exist: a stable carrying capacity $ P_1 $ and an unstable threshold $ P_2 $. If $ P(0) > P_2 $, the population stabilizes at $P_1$ .
If $H = H_{\text{crit}} = \frac{rK}{4}$ : A single semi-stable equilibrium exists at $P = K/2$ .
If $H > \frac{rK}{4}$ : $ \Delta < 0 $, so $ \frac{dP}{dt} < 0 $ for all $ P \ge 0 $. No equilibria exist, and the population drops to 0 in finite time.

Hence, the maximum sustainable yield is precisely $H_{\text{crit}} = \frac{rK}{4}$ .

⊙ Explorer · Slope field for

\frac{dy}{dx}

= ky

interactive

Sample points along the solution curve $y$ = y0·e^( $kx$ ), with the field's own slope at each
$x$	$y(x)$	slope $k \cdot y$

rate $k$ : 0.5

initial value $y_{0}$ : 1

\frac{dy}{dx}

(0, y_{0})

0.500

y(1)

1.649

This is the same family as Worked examples 1.A $(k = 2)$ and 1.C (radioactive decay: there, dN/ $dt = -k_{\text{decay}} \cdot N$ for a positive decay constant $k_{\text{decay}}$ ; here that's this explorer's $k = -k_{\text{decay}}$ , negative): $\frac{dy}{dx}$ = ky has exact solution $y$ = y0·e^( $kx$ ) — the amber dot marks the initial point $(0, y0)$ , and the indigo curve is that exact formula, not a numerical approximation. Every gray arrow is the field's own slope ky at that point; drag $k$ or $y_{0}$ and watch the curve stay tangent to the field along its entire length. The field itself is drawn everywhere in view, not just along this one curve — a different starting point would trace a different solution curve through that same field, which is what "the field belongs to the whole equation, not just one solution" means.

▲ Level 3 University Rigor — Theorem 1.2: Picard–Lindelöf Existence and Uniqueness Theorem

Consider the initial value problem $\frac{dy}{dx} = f(x, y), \qquad y(x_0) = y_0.$ Let $R = \{(x, y) \in \mathbb{R}^2 : |x - x_0| \le a, \, |y - y_0| \le b\}$ be a closed rectangle centered at $(x_0, y_0)$ . Suppose:

$f(x, y)$ is continuous on $R$ , so it is bounded by $ M = \max_{(x,y)\in R} |f(x, y)| < \infty $.
$f(x, y)$ satisfies a Lipschitz condition in $y$ on $R$ : there exists $L > 0$ such that $|f(x, y_1) - f(x, y_2)| \le L |y_1 - y_2| \quad \forall (x, y_1), (x, y_2) \in R.$ (A continuous partial derivative $\frac{\partial f}{\partial y}$ on $R$ guarantees this with $L = \max_R |\frac{\partial f}{\partial y}|$ ).

Then the IVP has a unique continuously differentiable solution $y(x)$ defined on the interval $[x_0 - h, x_0 + h]$ , where $h = \min\{a, \frac{b}{M}\}$ .

Picard Successive Iteration Operator: Converting the IVP into an equivalent Volterra integral equation $y(x) = y_0 + \int_{x_0}^x f(t, y(t))\,dt,$ the sequence defined by $y_0(x) = y_0$ and $y_{k+1}(x) = y_0 + \int_{x_0}^x f(t, y_k(t))\,dt$ converges uniformly on $[x_0 - h, x_0 + h]$ to the unique solution by the Banach Fixed-Point Contraction Principle.

Pathological Breakdown of Uniqueness: Consider $\frac{dy}{dx} = 3 y^{2/3}$ with $y(0) = 0$ . Here $f(x, y) = 3y^{2/3}$ is continuous on $\mathbb{R}^2$ (so Peano's Theorem guarantees existence). However, $\frac{\partial f}{\partial y} = 2y^{-1/3}$ is unbounded as $y \to 0$ , violating the Lipschitz condition. Separating variables gives $y(x) = x^3$ , but $y(x) \equiv 0$ is also a solution! In fact, for any $c > 0$ , the family $y(x) = \begin{cases} 0 & x \le c \\ (x - c)^3 & x > c \end{cases}$ is continuously differentiable and satisfies the IVP, yielding infinitely many solutions originating from $(0, 0)$ .

The geometry of solutions to first-order ODEs is fully characterized by integral curves. For the separable differential equation $\frac{dy}{dx} = -\frac{x}{y}$ , separation gives $y\,dy = -x\,dx$ , which integrates to the family of concentric circles $x^2 + y^2 = C$ , as illustrated below.

Integral curves for the separable differential equation

\frac{dy}{dx} = -\frac{x}{y}

. Separating variables gives

y\,dy = -x\,dx

, integrating to concentric circular trajectories

x^2 + y^2 = C

⚠️ Common Traps & University Pitfalls — Lost Singular Solutions & Non-Linear Intervals

Dividing by zero drops equilibrium solutions: When separating $\frac{dy}{dx} = g(y)h(x)$ by dividing by $g(y)$ , you assume $g(y) \neq 0$ . Any constant value $y = k$ where $g(k) = 0$ is a valid equilibrium solution that will not appear in the separated integral unless checked separately!

Constant of integration placement: You must add the constant $+C$ immediately after performing integration. For example, in $\ln|y| = 2x + C$ , exponentiating gives $|y| = e^{2x+C} = e^C e^{2x} = A e^{2x}$ , NOT $e^{2x} + C$ .

Finite-Time Blowup & Maximal Interval: In linear equations, solutions are guaranteed to exist wherever the coefficient functions are continuous. In non-linear equations, solutions frequently experience finite-time singularities (blowup) within bounded domains, requiring explicit verification of the maximal open interval of existence.

Worked Example

D

— First-Order Differential Equations, Separable Equations & Existence Theory (Hard Application)

Problem (Hard): What is the general solution to the initial value problem $\frac{dy}{dt} = -3y$ with $y(0) = 50$ ?

Step-by-Step Solution: Separable: $\frac{dy}{y} = -3\,dt \implies \ln|y| = -3t + C \implies y(t) = y(0)e^{-3t}$ . Using $y(0) = 50$ yields $y(t) = 50e^{-3t}$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (First-Order Differential Equations, Separable Equations & Existence Theory)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — First-Order Differential Equations, Separable Equations & Existence Theory

Quick Verification Trick (University Ordinary Differential Equations): For any indefinite integral $\int f(x)\,dx = F(x) + C$ , differentiate your answer $F'(x)$ mentally—especially the coefficient $1/k$ from chain-rule reversal $\int e^{kx}\,dx = \tfrac{1}{k}e^{kx} + C$ —before moving on.

Solve the initial value problem

\frac{dy}{dx} = 2x(y^2 + 1)

with

y(0) = 1

, and determine its maximal interval of existence.

💡 Hint 1 of 3

Separate variables:

\frac{1}{y^2 + 1}\,dy = 2x\,dx

💡 Hint 2 of 3

Integrate both sides:

\int \frac{1}{y^2 + 1}\,dy = \int 2x\,dx \implies \arctan(y) = x^2 + C

Apply the initial condition $y(0) = 1$ :

\arctan(1) = 0^2 + C \implies C = \frac{\pi}{4}

💡 Hint 3 of 3

Since

x^2 \ge 0

, the constraint is $x^2 < \frac{\pi}{4}$.

Step 1: Separate variables: $\frac{1}{y^2 + 1}\,dy = 2x\,dx.$

Step 2: Integrate both sides: $\int \frac{1}{y^2 + 1}\,dy = \int 2x\,dx \implies \arctan(y) = x^2 + C.$

Step 3: Apply the initial condition $y(0) = 1$ : $\arctan(1) = 0^2 + C \implies C = \frac{\pi}{4}.$

Step 4: Solve explicitly for $y(x)$ : $y(x) = \tan\left(x^2 + \frac{\pi}{4}\right).$

Step 5: Interval of existence: The tangent function has vertical asymptotes where its argument equals $\pm \frac{\pi}{2}$ . To contain $x_0 = 0$ : \[ -\frac{\pi}{2} < x^2 + \frac{\pi}{4} < \frac{\pi}{2} \implies -\frac{3\pi}{4} < x^2 < \frac{\pi}{4}. \] Since $ x^2 \ge 0 $, the constraint is $ x^2 < \frac{\pi}{4} \implies x \in \left(-\frac{\sqrt{\pi}}{2}, \frac{\sqrt{\pi}}{2}\right) $.

Find all equilibrium solutions to

\frac{dy}{dx} = y(4 - y) - 3

, classify their stability, and find the particular solution passing through

y(0) = 2

💡 Hint 1 of 3

Locate equilibria:

-y^2 + 4y - 3 = 0 \implies (y - 1)(3 - y) = 0 \implies y_1 = 1, \quad y_2

💡 Hint 2 of 3

Stability via derivative: Let

f(y) = -y^2 + 4y - 3

. Then

f'(y) = -2y + 4

. At

y = 1

f'(1) = 2 > 0 \implies y

is an unstable source.

💡 Hint 3 of 3

Note that as

x \to \infty

y(x) \to 3

, verifying convergence to the stable equilibrium sink.

Step 1: Locate equilibria: $-y^2 + 4y - 3 = 0 \implies (y - 1)(3 - y) = 0 \implies y_1 = 1, \quad y_2 = 3.$

Step 2: Stability via derivative: Let $f(y) = -y^2 + 4y - 3$ . Then $f'(y) = -2y + 4$ .
At $y = 1$ , $f'(1) = 2 > 0 \implies y = 1$ is an unstable source.
At $y = 3$ , $ f'(3) = -2 < 0 \implies y = 3 $ is a stable sink.

Step 3: Separate variables for IVP $y(0) = 2$ : $\frac{1}{(y - 1)(3 - y)}\,dy = dx \implies \frac{1}{2}\left(\frac{1}{y - 1} + \frac{1}{3 - y}\right)dy = dx.$ Integrating: $\frac{1}{2}\big(\ln|y - 1| - \ln|3 - y|\big) = x + C_1 \implies \ln\left|\frac{y - 1}{3 - y}\right| = 2x + C_2.$

Applying $y(0) = 2$ : $\ln|1/1| = 0 = 0 + C_2 \implies C_2 = 0$ . Since $ 1 < y < 3 $, $ \frac{y - 1}{3 - y} = e^{2x} $. Solving for $ y $: \[ y(x) = \frac{1 + 3e^{2x}}{1 + e^{2x}}. \] Note that as $ x \to \infty $, $ y(x) \to 3 $, verifying convergence to the stable equilibrium sink.

Compute the first three Picard successive approximations

y_1(x), y_2(x), y_3(x)

for the initial value problem

\frac{dy}{dx} = 2x(1 + y)

with

y(0) = 0

, and identify the Taylor series of the exact solution.

💡 Hint 1 of 3

Picard Integral Operator:

y_{k+1}(x) = y_0 + \int_0^x 2t\big(1 + y_k(t)\big)\,dt, \qquad y_0(x) = 0

💡 Hint 2 of 3

Iterate 1:

y_1(x) = 0 + \int_0^x 2t(1 + 0)\,dt = \left[t^2\right]_0^x = x^2

Iterate 2:

y_2(x) = \int_0^x 2t(1 + t^2)\,dt = \int_0^x (2t + 2t^3)\,dt = x^2 + \frac{x^4}{2}

💡 Hint 3 of 3

Verifying by separation:

\frac{dy}{1+y} = 2x\,dx \implies \ln|1+y| = x^2 \implies y(x) = e^{x^2} - 1

, confirming uniform convergence of the Picard approximations.

Picard Integral Operator: $y_{k+1}(x) = y_0 + \int_0^x 2t\big(1 + y_k(t)\big)\,dt, \qquad y_0(x) = 0.$

Iterate 1: $y_1(x) = 0 + \int_0^x 2t(1 + 0)\,dt = \left[t^2\right]_0^x = x^2.$

Iterate 2: $y_2(x) = \int_0^x 2t(1 + t^2)\,dt = \int_0^x (2t + 2t^3)\,dt = x^2 + \frac{x^4}{2}.$

Iterate 3: $y_3(x) = \int_0^x 2t\left(1 + t^2 + \frac{t^4}{2}\right)dt = \int_0^x (2t + 2t^3 + t^5)\,dt = x^2 + \frac{x^4}{2} + \frac{x^6}{6}.$

Exact Solution: By inspection, $y_k(x) = \sum_{m=1}^k \frac{(x^2)^m}{m!} \xrightarrow{k\to\infty} e^{x^2} - 1.$ Verifying by separation: $\frac{dy}{1+y} = 2x\,dx \implies \ln|1+y| = x^2 \implies y(x) = e^{x^2} - 1$ , confirming uniform convergence of the Picard approximations.

[Practice 4 · Easy] Consider the separable differential equation

\dfrac{dy}{dx} = 2x

with the initial condition

y(0) = 3

. Find the value of

y(2)

💡 Hint 1 of 3

The differential equation is

\dfrac{dy}{dx} = 2x

💡 Hint 2 of 3

This is a separable equation. We can rewrite it as

dy = 2x \, dx

💡 Hint 3 of 3

Substituting

x = 2

into the particular solution, we have

y(2) = (2)^2 + 3 = 4 + 3

The differential equation is

\dfrac{dy}{dx} = 2x

. This is a separable equation. We can rewrite it as

dy = 2x \, dx

. Integrating both sides gives

\int dy = \int 2x \, dx

, which results in

y = x^2 + C

, where C is the constant of integration. We are given the initial condition

y(0) = 3

. Substituting these values into the general solution, we get

3 = (0)^2 + C

, so

C = 3

. Therefore, the particular solution is

y = x^2 + 3

. We need to find the value of

y(2)

. Substituting

x = 2

into the particular solution, we have

y(2) = (2)^2 + 3 = 4 + 3 = 7

📝 Chapter Quiz

Questions Level

Linear First-Order Equations, Integrating Factors & Nonlinear Reductions

While Chapter 1 addressed separable equations where rates factor into independent coordinate components, physical systems routinely combine internal dissipation with external driving forces: thermal cooling under active heating, inductive-resistive electrical circuits under AC forcing, or chemical reactors subject to continuous inflow and outflow. Such systems are modeled by first-order linear differential equations. This chapter develops the exact integrating factor technique, analyzes transient versus steady-state dynamics, and presents nonlinear transformations including Bernoulli and Riccati equations.

● Level 1 Foundation — Definition 2.1: First-Order Linear Equations & Standard Normal Form

A first-order differential equation is linear if it can be expressed in the canonical standard form: $\frac{dy}{dx} + P(x)\,y = Q(x),$ where $P(x)$ and $Q(x)$ are continuous functions on an open interval $I \subseteq \mathbb{R}$ . The equation is homogeneous if $Q(x) \equiv 0$ ; otherwise it is nonhomogeneous.

The differential operator $L[y] = \frac{dy}{dx} + P(x)y$ is a linear mapping from $C^1(I) \to C(I)$ : $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]$ for any scalars $c_1, c_2 \in \mathbb{R}$ . If an equation is presented with a non-constant coefficient preceding $y'$ , e.g. $a_1(x)y' + a_0(x)y = b(x)$ , it must be normalized by dividing through by $a_1(x)$ on intervals where $a_1(x) \neq 0$ before computing integrating factors.

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

▲ Level 1 Foundation — Theorem 2.1: The Integrating Factor & Product Rule Inversion

Let $P(x)$ and $Q(x)$ be continuous on an open interval $I$ . Define the integrating factor: $\mu(x) = \exp\left(\int P(x)\,dx\right).$ Then $\mu(x) > 0$ for all $x \in I$ , and a function $y(x)$ is a solution to $y' + P(x)y = Q(x)$ if and only if $y(x) = \frac{1}{\mu(x)}\left(\int \mu(x)Q(x)\,dx + C\right)$ for some constant $C \in \mathbb{R}$ .

Proof

By the Fundamental Theorem of Calculus and Chain Rule, the derivative of $\mu(x) = e^{\int P(x)\,dx}$ is: $\mu'(x) = P(x)e^{\int P(x)\,dx} = P(x)\mu(x).$ Multiplying the standard form $y' + P(x)y = Q(x)$ by $\mu(x)$ : $\mu(x)y'(x) + \mu(x)P(x)y(x) = \mu(x)Q(x).$ Since $\mu(x)P(x) = \mu'(x)$ , the left side collapses into the exact product rule: $\frac{d}{dx}\big[\mu(x)y(x)\big] = \mu(x)Q(x).$ Integrating both sides with respect to $x$ : $\mu(x)y(x) = \int \mu(x)Q(x)\,dx + C.$ Since $\mu(x) = e^{\int P(x)\,dx} > 0$ is strictly positive everywhere on $I$ , dividing by $\mu(x)$ yields the unique general solution family.

Analytical Procedure for First-Order Linear Equations

1. Normalization

Divide through by the leading coefficient to obtain standard form

y' + P(x)y = Q(x)

. Identify

P(x)

and

Q(x)

, including any negative signs.

2. Integrating Factor

Compute

\mu(x) = e^{\int P(x)\,dx}

. Do not add a constant of integration

+C

here, as any scalar multiple of

\mu(x)

cancels.

3. Exact Product Collapse

Multiply the entire differential equation by

\mu(x)

, collapsing the left side into

\frac{d}{dx}[\mu(x)y] = \mu(x)Q(x)

4. Integration & Isolation

Integrate

\mu(x)Q(x)

with respect to

x

, add the constant

+C

, and divide the entire right side by

\mu(x)

Worked example 2.A — Integrating Factor with Variable Coefficients

Find the general solution of the differential equation $x \frac{dy}{dx} + 2y = 4x^2, \qquad x > 0.$

Step 1: Normalize: Divide by $x$ : $\frac{dy}{dx} + \frac{2}{x}y = 4x.$ Here $P(x) = \frac{2}{x}$ and $Q(x) = 4x$ .

Step 2: Integrating factor: $\mu(x) = \exp\left(\int \frac{2}{x}\,dx\right) = \exp(2\ln x) = \exp(\ln x^2) = x^2.$

Step 3: Multiply and integrate: $\frac{d}{dx}\big[x^2 y\big] = x^2(4x) = 4x^3 \implies x^2 y = \int 4x^3\,dx = x^4 + C.$

Step 4: Isolate $y(x)$ : $y(x) = x^2 + \frac{C}{x^2}.$ Notice that as $x \to 0^+$ , the homogeneous term $C/x^2$ diverges unless $C = 0$ , reflecting the singularity of $P(x)$ at $x = 0$ .

Worked example 2.B — Initial Value Problems & Singular Boundary Points

Solve the initial value problem $\frac{dy}{dx} - \frac{2}{x}y = x^3 e^x, \qquad y(1) = 0, \quad x > 0.$

Step 1: Integrating factor: $P(x) = -\frac{2}{x}$ , so $\mu(x) = \exp\left(\int -\frac{2}{x}\,dx\right) = e^{-2\ln x} = x^{-2} = \frac{1}{x^2}.$

Step 2: Collapse and integrate: $\frac{d}{dx}\left[\frac{y}{x^2}\right] = \frac{1}{x^2}\big(x^3 e^x\big) = x e^x.$ Integrating by parts ( $u = x, dv = e^x dx$ ): $\frac{y}{x^2} = \int x e^x\,dx = x e^x - e^x + C = (x - 1)e^x + C.$

Step 3: Apply initial condition $y(1) = 0$ : $\frac{0}{1^2} = (1 - 1)e^1 + C \implies C = 0.$

Step 4: Particular solution: $y(x) = x^2(x - 1)e^x.$

Worked example 2.C — Analytical Application: Dynamic Inflow-Outflow Mixing Tanks

A tank initially contains $V_0 = 100 \text{ L}$ of fresh water. Brine with salt concentration $c_{\text{in}} = 0.5 \text{ kg/L}$ flows in at $r_{\text{in}} = 4 \text{ L/min}$ . Well-mixed solution is pumped out at $r_{\text{out}} = 2 \text{ L/min}$ . Determine the total salt mass $Q(t)$ in the tank as a function of time.

Step 1: Dynamic volume: Because $r_{\text{in}} \neq r_{\text{out}}$ , volume changes with time: $V(t) = V_0 + (r_{\text{in}} - r_{\text{out}})t = 100 + 2t.$

Step 2: Differential balance equation: $\frac{dQ}{dt} = \text{Rate In} - \text{Rate Out} = r_{\text{in}} c_{\text{in}} - r_{\text{out}} \frac{Q(t)}{V(t)} = 4(0.5) - 2 \frac{Q}{100 + 2t}.$ In standard linear form: $\frac{dQ}{dt} + \frac{1}{50 + t}Q = 2, \qquad Q(0) = 0.$

Step 3: Integrating factor: $\mu(t) = \exp\left(\int \frac{1}{50 + t}\,dt\right) = 50 + t.$

Step 4: Integrate and solve: $\frac{d}{dt}\big[(50 + t)Q\big] = 2(50 + t) \implies (50 + t)Q = (50 + t)^2 + C.$ Applying $Q(0) = 0$ : $50(0) = 50^2 + C \implies C = -2500$ . Thus: $Q(t) = 50 + t - \frac{2500}{50 + t}.$

⊙ Explorer ·

\frac{dy}{dx}

+ ky =

b

(constant-coefficient linear equation)

interactive

Sample points along $y(x)$ showing decay toward the steady state $y_{\text{inf}} = b / k$
$x$	$y(x)$	$y(x) - y_{\text{inf}}$

coefficient $k$ : 1

forcing $b$ : 2

initial value $y_{0}$ : -1

steady state

y_{\text{inf}} = b / k

2.000

y(1)

0.896

For constant coefficients $P(x) = k$ and $Q(x) = b$ with $k > 0$ , the integrating factor is e^( $kx$ ) and the exact solution is $y(x) = b / k + (y0 - b/k)$ e^(- $kx$ ). The term $(y0 - b/k)$ e^(- $kx$ ) is the transient — it decays exponentially to 0 as $x$ grows — leaving only the steady state $y_{\text{inf}} = b / k$ (dashed amber line). Drag $k$ to change the decay rate (a larger $k$ pulls the curve toward the steady state faster; a smaller $k$ lets the transient linger); drag $b$ to shift the steady-state level; drag $y_{0}$ to move the starting point above or below it.

▲ Level 3 University Rigor — Theorem 2.2: Bernoulli Equations & Riccati Nonlinear Reduction

1. Bernoulli Equations: An ODE of the form $\frac{dy}{dx} + P(x)\,y = Q(x)\,y^n, \qquad n \in \mathbb{R}, \quad n \neq 0, 1$ is nonlinear. Dividing by $y^n$ yields $y^{-n}y' + P(x)y^{1-n} = Q(x)$ . Introducing the change of variable $v(x) = [y(x)]^{1-n} \implies \frac{dv}{dx} = (1 - n)y^{-n}\frac{dy}{dx}$ transforms the nonlinear ODE into an exact first-order linear ODE for $v(x)$ : $\frac{dv}{dx} + (1 - n)P(x)\,v = (1 - n)Q(x).$

2. Riccati Equations: A quadratic first-order ODE of the form $\frac{dy}{dx} = q_0(x) + q_1(x)y + q_2(x)y^2, \qquad q_2(x) \neq 0$ cannot generally be solved by quadratures. However:

If one particular solution $y_1(x)$ is known, the substitution $y(x) = y_1(x) + \frac{1}{v(x)}$ reduces the Riccati equation to a first-order linear ODE for $v(x)$ : $v' + \big(q_1(x) + 2q_2(x)y_1(x)\big)v = -q_2(x).$
Alternatively, the substitution $y(x) = -\frac{u'(x)}{q_2(x)u(x)}$ transforms the Riccati equation into a second-order linear homogeneous ODE: $u'' - \left(q_1(x) + \frac{q_2'(x)}{q_2(x)}\right)u' + q_0(x)q_2(x)u = 0.$

Linear systems under constant input display a fundamental decomposition into a transient solution (which decays to zero) and a steady-state response (which dominates as $x \to \infty$ ), as plotted below for $y(x) = 3 - 2e^{-2x}$ .

Decomposition of

y(x) = 3 - 2e^{-2x}

. The exponential term

-2e^{-2x}

is the transient component decaying rapidly toward zero, leaving the steady-state asymptote

y_{\infty} = 3

⚠️ Common Traps & University Pitfalls — Standard Form & Integrating Factors

Failure to normalize first: Never read $P(x)$ directly from $a(x)y' + b(x)y = c(x)$ ! You must divide by $a(x)$ first. If $x y' + 2y = 4$ , $P(x) = 2/x$ , NOT $2$ . Using $\mu = e^{2x}$ instead of $x^2$ is a catastrophic error.

Minus sign omissions: For $y' - 3y = e^x$ , $P(x) = -3$ and $\mu(x) = e^{-3x}$ . Dropping the negative sign results in $\mu(x) = e^{3x}$ , causing product-rule collapse to fail.

Exponent laws: $\exp(3\ln x) = \exp(\ln x^3) = x^3$ . It is NOT $3x$ or $3\ln x$ . Furthermore, remember to distribute the $1/\mu(x)$ across the constant $C$ : $y = \frac{\int \mu Q\,dx}{\mu} + \frac{C}{\mu}$ , never $+ C$ placed outside the fraction.

Worked Example

D

— Linear First-Order Equations, Integrating Factors & Nonlinear Reductions (Hard Application)

Problem (Hard): The Laplace transform of $R(m) = 48e^{20m}$ is $S(s) = \dfrac{48}{s - 20}$ for $s > 20$ . Determine the evaluated transform $S(21)$ .

Step-by-Step Solution: Substituting $s = 21$ gives $\frac{48}{21 - 20} = 48$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Linear First-Order Equations, Integrating Factors & Nonlinear Reductions)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — Linear First-Order Equations, Integrating Factors & Nonlinear Reductions

Solve the initial value problem

(x^2 + 1)\frac{dy}{dx} + 4x y = x

with

y(0) = 1

💡 Hint 1 of 3

Normalize:

\frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{x}{x^2 + 1}

💡 Hint 2 of 3

Integrating:

(x^2 + 1)^2 y = \frac{x^4}{4} + \frac{x^2}{2} + C

💡 Hint 3 of 3

Particular solution:

y(x) = \frac{x^4/4 + x^2/2 + 1}{(x^2 + 1)^2} = \frac{x^4 + 2x^2 + 4}{4(x^2 + 1)^2} = \frac{(x^2 + 1)^2 + 3}{4(x^2 + 1)^2}

Step 1: Normalize: $\frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{x}{x^2 + 1}.$

Step 2: Integrating factor: $\int P(x)\,dx = \int \frac{4x}{x^2 + 1}\,dx = 2\ln(x^2 + 1) = \ln\big((x^2 + 1)^2\big) \implies \mu(x) = (x^2 + 1)^2.$

Step 3: Multiply and integrate: $\frac{d}{dx}\Big[(x^2 + 1)^2 y\Big] = (x^2 + 1)^2 \cdot \frac{x}{x^2 + 1} = x(x^2 + 1) = x^3 + x.$ Integrating: $(x^2 + 1)^2 y = \frac{x^4}{4} + \frac{x^2}{2} + C.$

Step 4: Initial condition $y(0) = 1$ : $(0 + 1)^2 (1) = 0 + 0 + C \implies C = 1.$

Step 5: Particular solution: $y(x) = \frac{x^4/4 + x^2/2 + 1}{(x^2 + 1)^2} = \frac{x^4 + 2x^2 + 4}{4(x^2 + 1)^2} = \frac{(x^2 + 1)^2 + 3}{4(x^2 + 1)^2} = \frac{1}{4} + \frac{3}{4(x^2 + 1)^2}.$

Solve the nonlinear Bernoulli differential equation

\frac{dy}{dx} + \frac{1}{x}y = x y^2

for

x > 0

with

y(1) = -1

💡 Hint 1 of 3

Identify Bernoulli parameters:

n = 2

. Divide by

y^2

y^{-2}y' + \frac{1}{x}y^{-1} = x

💡 Hint 2 of 3

Multiply through:

\frac{d}{dx}\left[\frac{v}{x}\right] = \frac{1}{x}(-x) = -1 \implies \frac{v}{x} = -x + C \implies v(x) = C x - x^2

💡 Hint 3 of 3

Therefore:

y(x) = -\frac{1}{x^2}

Step 1: Identify Bernoulli parameters: $n = 2$ . Divide by $y^2$ : $y^{-2}y' + \frac{1}{x}y^{-1} = x.$

Step 2: Substitute $v = y^{1-2} = y^{-1}$ : $v' = -y^{-2}y' \implies y^{-2}y' = -v'$ . Substituting yields: $-v' + \frac{1}{x}v = x \implies v' - \frac{1}{x}v = -x.$

Step 3: Integrating factor for linear equation in $v$ : $\mu(x) = \exp\left(\int -\frac{1}{x}\,dx\right) = e^{-\ln x} = \frac{1}{x}.$ Multiply through: $\frac{d}{dx}\left[\frac{v}{x}\right] = \frac{1}{x}(-x) = -1 \implies \frac{v}{x} = -x + C \implies v(x) = C x - x^2.$

Step 4: Revert to $y$ : $y(x) = \frac{1}{v(x)} = \frac{1}{Cx - x^2}$ .
Apply $y(1) = -1$ : $-1 = \frac{1}{C(1) - 1^2} = \frac{1}{C - 1} \implies C - 1 = -1 \implies C = 0$ .
Therefore: $y(x) = -\frac{1}{x^2}.$

[Practice 3 · Easy] Consider the linear first-order differential equation

\frac{dy}{dx} + 2y = 6

. If the initial condition is

y(0) = 1

, find the value of

y(1)

💡 Hint 1 of 3

The given differential equation is a linear first-order ODE of the form

\frac{dy}{dx} + P(x)y = Q(x)

, where

P(x) = 2

and

Q(x) = 6

. The integrating factor

\mu(x)

is given by

\mu(x) = e^{\int P(x) dx}

💡 Hint 2 of 3

Calculating the integrating factor:

\mu(x) = e^{\int 2 dx} = e^{2x}

. Multiplying the differential equation by the integrating factor:

e^{2x}\frac{dy}{dx} + 2e^{2x}y = 6e^{2x}

💡 Hint 3 of 3

Now, we need to find

y(1)

y(1) = 3 - 2e^{-2(1)}

The given differential equation is a linear first-order ODE of the form

\frac{dy}{dx} + P(x)y = Q(x)

, where

P(x) = 2

and

Q(x) = 6

. The integrating factor

\mu(x)

is given by

\mu(x) = e^{\int P(x) dx}

. Calculating the integrating factor:

\mu(x) = e^{\int 2 dx} = e^{2x}

. Multiplying the differential equation by the integrating factor:

e^{2x}\frac{dy}{dx} + 2e^{2x}y = 6e^{2x}

. The left side is the derivative of the product

y\mu(x)

\frac{d}{dx}(ye^{2x}) = 6e^{2x}

. Integrating both sides with respect to

x

ye^{2x} = \int 6e^{2x} dx

ye^{2x} = 6 \cdot \frac{1}{2}e^{2x} + C

ye^{2x} = 3e^{2x} + C

. Solving for

y

y = 3 + Ce^{-2x}

. Using the initial condition

y(0) = 1

to find

C

1 = 3 + Ce^{-2(0)}

1 = 3 + C

C = 1 - 3 = -2

. So, the particular solution is

y(x) = 3 - 2e^{-2x}

. Now, we need to find

y(1)

y(1) = 3 - 2e^{-2(1)} = 3 - 2e^{-2}

. This matches choice C.

[Practice 4 · Medium] The separable initial value problem

\dfrac{dP}{dr} = 34r

satisfies

P(0) = 8

. Determine the solution value

P(7)

💡 Hint 1 of 3

Solve

\dfrac{dP}{dr} = 34r

over the specified interval using reference angles and quadrant symmetry.

💡 Hint 2 of 3

Integrating gives

P(r) = 17r^2 + 8

💡 Hint 3 of 3

Count the total number of distinct solutions that lie within the given domain.

Integrating gives

P(r) = 17r^2 + 8

, so

P(7) = 841

📝 Chapter Quiz

Questions Level

Second-Order Linear Homogeneous Equations, Auxiliary Roots & Wronskian Theory

While first-order equations require one initial condition to determine a trajectory, physical dynamics governed by Newton's Second Law ( $F = ma = m x''$ ) require two independent initial specifications: initial position and initial velocity. A second-order equation's solution set forms a two-dimensional vector space. This chapter establishes the complete algebraic theory of constant-coefficient linear homogeneous equations, proves the characteristic root trichotomy, and introduces Abel's Identity and the Wronskian determinant for linear independence.

● Level 1 Foundation — Definition 3.1: Second-Order Linear Homogeneous ODEs & Characteristic Polynomial

A second-order linear homogeneous differential equation with constant coefficients has the form: $a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = 0, \qquad a, b, c \in \mathbb{R}, \quad a \neq 0.$ Substituting the exponential ansatz $y(x) = e^{rx}$ (with $y' = r e^{rx}$ and $y'' = r^2 e^{rx}$ ) produces: $e^{rx}\big(a r^2 + b r + c\big) = 0.$ Because $e^{rx} \neq 0$ for all $x \in \mathbb{R}$ , $y = e^{rx}$ solves the ODE if and only if $r$ satisfies the characteristic (auxiliary) equation: $a r^2 + b r + c = 0.$

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

▲ Level 1 Foundation — Theorem 3.1: Superposition Principle & Vector Space of Solutions

Let $L[y] = a y'' + b y' + c y$ be the linear differential operator. If $y_1(x)$ and $y_2(x)$ are solutions to $L[y] = 0$ on an open interval $I$ , then any linear combination $y(x) = C_1 y_1(x) + C_2 y_2(x), \qquad C_1, C_2 \in \mathbb{R}$ is also a solution. The solution set $\ker(L)$ forms a two-dimensional vector subspace of $C^2(I)$ .

Proof

Differentiation is linear: $(C_1 y_1 + C_2 y_2)' = C_1 y_1' + C_2 y_2'$ and $(C_1 y_1 + C_2 y_2)'' = C_1 y_1'' + C_2 y_2''$ . Evaluating $L$ : $L[C_1 y_1 + C_2 y_2] = a(C_1 y_1'' + C_2 y_2'') + b(C_1 y_1' + C_2 y_2') + c(C_1 y_1 + C_2 y_2)$ $= C_1(a y_1'' + b y_1' + c y_1) + C_2(a y_2'' + b y_2' + c y_2) = C_1(0) + C_2(0) = 0.$ Thus linearity is established.

▲ Level 2 Analytical Theorem 3.2: General Solution Trichotomy (Real Distinct, Repeated Critical, Underdamped Complex)

Let $r_1, r_2$ be roots of $a r^2 + b r + c = 0$ with discriminant $\Delta = b^2 - 4ac$ . The general solution on $\mathbb{R}$ belongs to one of three mutually exclusive classes:

Case 1: $\Delta > 0$ (Overdamped, distinct real roots $r_1 \neq r_2$ ): $y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}.$
Case 2: $\Delta = 0$ (Critically Damped, repeated real root $r = -b/(2a)$ ): $y(x) = (C_1 + C_2 x)e^{rx}.$
Case 3: $ \Delta < 0 $ (Underdamped, complex conjugate roots $ r = \alpha \pm i\beta $ where $ \alpha = -b/(2a), \beta = \sqrt{4ac - b^2}/(2a) $): $y(x) = e^{\alpha x}\big(C_1 \cos(\beta x) + C_2 \sin(\beta x)\big).$

Proof (Repeated Roots & Complex Conjugate Derivations)

Repeated Root ( $\Delta = 0$ ): One solution is $y_1(x) = e^{rx}$ with $r = -b/(2a)$ . Seek a second solution using reduction of order: $y_2(x) = v(x)e^{rx}$ . Differentiating yields $y_2' = e^{rx}(v' + r v)$ and $y_2'' = e^{rx}(v'' + 2r v' + r^2 v)$ . Substituting into $a y'' + b y' + c y = 0$ : $e^{rx}\big[a v'' + (2ar + b)v' + (a r^2 + b r + c)v\big] = 0.$ Since $a r^2 + b r + c = 0$ and $2ar + b = 2a(-b/(2a)) + b = 0$ , the equation collapses to $a v'' = 0 \implies v''(x) = 0$ . Integrating twice gives $v(x) = C_1 + C_2 x$ , establishing $y(x) = (C_1 + C_2 x)e^{rx}$ .

Complex Conjugate Roots ($ \Delta < 0 $): Roots are $r = \alpha \pm i\beta$ . By Euler's formula, $e^{(\alpha \pm i\beta)x} = e^{\alpha x}(\cos\beta x \pm i\sin\beta x)$ . By Theorem 3.1, real linear combinations of solutions are solutions: $y_1 = \frac{e^{(\alpha + i\beta)x} + e^{(\alpha - i\beta)x}}{2} = e^{\alpha x}\cos(\beta x), \quad y_2 = \frac{e^{(\alpha + i\beta)x} - e^{(\alpha - i\beta)x}}{2i} = e^{\alpha x}\sin(\beta x).$ These are linearly independent since their Wronskian $W(y_1, y_2)(x) = \beta e^{2\alpha x} \neq 0$ .

Analytical Procedure for Homogeneous Second-Order Equations

1. Auxiliary Quadratic

Write down

a r^2 + b r + c = 0

directly from coefficients

a y'' + b y' + c y = 0

2. Discriminant Evaluation

Compute

\Delta = b^2 - 4ac

to classify roots as distinct real (

>0

), repeated (

=0

), or complex conjugate ($ <0 $).

3. Fundamental Basis

Construct linearly independent solutions

y_1, y_2

and assemble

y(x) = C_1 y_1(x) + C_2 y_2(x)

4. Initial Conditions System

Differentiate

y(x)

once and solve the

2 \times 2

linear algebraic system for

C_1, C_2

using

y(x_0) = y_0

and

y'(x_0) = y_1

Worked example 3.A — Overdamped Motion & Distinct Real Roots

Solve the initial value problem $y'' - 5y' + 4y = 0, \qquad y(0) = 0, \quad y'(0) = 3.$

Step 1: Auxiliary equation: $r^2 - 5r + 4 = (r - 1)(r - 4) = 0 \implies r_1 = 1, \quad r_2 = 4.$ General solution: $y(x) = C_1 e^x + C_2 e^{4x}$ .

Step 2: Derivative: $y'(x) = C_1 e^x + 4C_2 e^{4x}$ .

Step 3: Initial conditions: $y(0) = C_1 + C_2 = 0 \implies C_2 = -C_1,$ $y'(0) = C_1 + 4C_2 = 3 \implies C_1 + 4(-C_1) = -3C_1 = 3 \implies C_1 = -1, \quad C_2 = 1.$

Step 4: Particular solution: $y(x) = e^{4x} - e^x.$

Worked example 3.B — Critically Damped Motion & Repeated Roots

Solve the initial value problem $y'' + 6y' + 9y = 0, \qquad y(0) = 2, \quad y'(0) = -1.$

Step 1: Auxiliary equation: $r^2 + 6r + 9 = (r + 3)^2 = 0 \implies r = -3 \quad \text{(multiplicity 2)}.$ General solution: $y(x) = (C_1 + C_2 x)e^{-3x}$ .

Step 2: Derivative: $y'(x) = C_2 e^{-3x} - 3(C_1 + C_2 x)e^{-3x} = \big((C_2 - 3C_1) - 3C_2 x\big)e^{-3x}.$

Step 3: Initial conditions: $y(0) = C_1 = 2,$ $y'(0) = C_2 - 3C_1 = -1 \implies C_2 - 3(2) = -1 \implies C_2 = 5.$

Step 4: Particular solution: $y(x) = (2 + 5x)e^{-3x}.$

Worked example 3.C — Underdamped Harmonic Oscillator & Quasi-Frequency

A mass-spring-damper system satisfies $y'' + 2y' + 5y = 0, \qquad y(0) = 3, \quad y'(0) = 1.$ Determine its quasi-frequency $\omega_d$ and express the solution in phase-amplitude form $y(x) = A e^{-\gamma x}\cos(\omega_d x - \phi)$ .

Step 1: Auxiliary roots: $r^2 + 2r + 5 = 0 \implies r = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i.$ Damping factor $\alpha = -1$ , damped angular quasi-frequency $\omega_d = 2 \text{ rad/s}$ .
General solution: $y(x) = e^{-x}\big(C_1 \cos 2x + C_2 \sin 2x\big)$ .

Step 2: Initial conditions: $y(0) = C_1 = 3$ .
$y'(x) = -e^{-x}(C_1 \cos 2x + C_2 \sin 2x) + e^{-x}(-2C_1 \sin 2x + 2C_2 \cos 2x)$ .
$y'(0) = -C_1 + 2C_2 = -3 + 2C_2 = 1 \implies C_2 = 2$ .
Thus: $y(x) = e^{-x}\big(3\cos 2x + 2\sin 2x\big)$ .

Step 3: Amplitude-phase form: $A = \sqrt{C_1^2 + C_2^2} = \sqrt{3^2 + 2^2} = \sqrt{13} \approx 3.606, \qquad \phi = \arctan(C_2/C_1) = \arctan(2/3) \approx 0.588 \text{ rad}.$ $y(x) = \sqrt{13} e^{-x}\cos(2x - 0.588).$

⊙ Explorer · The auxiliary equation's three cases,

y'' + by' + cy = 0

interactive

damping $b$ : 1

stiffness $c$ : 4

discriminant

b^{2} - 4c

-15.000

RegimeUnderdamped

Drag $b$ (damping) and $c$ (stiffness) to shift the characteristic roots $r^{2}$ + br + $c = 0$ across the three dynamical regimes: when $b^{2} > 4c$ , roots are real and distinct (overdamped, no oscillations); when $b^{2} = 4c$ , the root is repeated (critically damped, fastest return to equilibrium without overshoot); when $b^{2} < 4c$, roots are complex conjugate (underdamped, oscillatory decay within an exponential envelope).

▲ Level 3 University Rigor — Theorem 3.3: Abel's Identity, Wronskian Determinant & Reduction of Order

1. Abel's Identity: Let $y_1(x)$ and $y_2(x)$ be solutions to the normalized second-order linear homogeneous ODE $y'' + P(x)y' + Q(x)y = 0$ on an open interval $I$ where $P, Q \in C(I)$ . The Wronskian determinant is defined by $W(y_1, y_2)(x) = \det\begin{pmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{pmatrix} = y_1(x)y_2'(x) - y_1'(x)y_2(x).$ Differentiating $W$ with respect to $x$ : $W'(x) = y_1 y_2'' - y_1'' y_2 = y_1\big(-P y_2' - Q y_2\big) - \big(-P y_1' - Q y_1\big)y_2 = -P(x)\big(y_1 y_2' - y_1' y_2\big) = -P(x)W(x).$ Integrating yields Abel's Formula: $W(y_1, y_2)(x) = W(y_1, y_2)(x_0) \exp\left(-\int_{x_0}^x P(t)\,dt\right).$ Consequently, $W(x)$ is either identically zero everywhere on $I$ (in which case $y_1, y_2$ are linearly dependent) or never zero anywhere on $I$ (in which case $\{y_1, y_2\}$ forms a fundamental solution set).

2. Reduction of Order Formula: If one non-zero solution $y_1(x)$ is known, setting $y_2(x) = v(x)y_1(x)$ produces $y_1 y_2' - y_1' y_2 = y_1^2 v' = W(x)$ . Thus: $v'(x) = \frac{W(x)}{y_1(x)^2} = \frac{C e^{-\int P(x)\,dx}}{y_1(x)^2} \implies y_2(x) = y_1(x) \int \frac{e^{-\int P(x)\,dx}}{y_1(x)^2}\,dx.$ This formula systematically generates the second linearly independent solution for any linear ODE, including Cauchy-Euler equations $a x^2 y'' + b x y' + c y = 0$ .

In underdamped oscillators, the oscillatory sinusoidal trajectory is strictly bounded within an exponential decay envelope $\pm A e^{-\gamma x}$ , touching the envelope tangentially at quasi-extrema, as graphed below.

Underdamped harmonic oscillator

y(x) = 2e^{-0.7x}\cos(3x)

. The solution oscillates with quasi-frequency

\omega_d = 3 \text{ rad/s}

strictly bounded within the exponential envelope

\pm 2e^{-0.7x}

⚠️ Common Traps & University Pitfalls — Repeated and Complex Characteristic Roots

Missing secular factor in repeated roots: When $b^2 - 4ac = 0$ , the roots are $r_1 = r_2 = r$ . The general solution is $(C_1 + C_2 x)e^{rx}$ , NOT $C_1 e^{rx} + C_2 e^{rx}$ . Writing $C_1 e^{rx} + C_2 e^{rx} = (C_1 + C_2)e^{rx} = C e^{rx}$ collapses two degrees of freedom into one, making it impossible to satisfy arbitrary initial conditions.

Product rule errors on repeated roots: When differentiating $y = (C_1 + C_2 x)e^{rx}$ , students frequently write $y' = r(C_1 + C_2 x)e^{rx}$ , forgetting the derivative of $C_2 x$ . The correct derivative is $y' = C_2 e^{rx} + r(C_1 + C_2 x)e^{rx}$ .

Complex numbers in real solutions: Real physical ODEs produce real solutions. For $r = \alpha \pm i\beta$ , the basis is $\{e^{\alpha x}\cos\beta x, e^{\alpha x}\sin\beta x\}$ . Do not leave imaginary units $i$ in final real-variable solutions.

Worked Example

D

— Second-Order Linear Homogeneous Equations, Auxiliary Roots & Wronskian Theory (Easy Application)

Problem (Easy): Consider the second-order linear homogeneous differential equation $y'' - 5y' + 6y = 0$ . If the auxiliary equation has roots $r_1$ and $r_2$ , find the value of $r_1 + r_2$ .

Step-by-Step Solution: The given differential equation is $y'' - 5y' + 6y = 0$ . The auxiliary equation for a second-order linear homogeneous differential equation of the form $ay'' + by' + cy = 0$ is $ar^2 + br + c = 0$ . In this case, $a=1$ , $b=-5$ , and $c=6$ . So the auxiliary equation is $r^2 - 5r + 6 = 0$ . Let the roots of this quadratic equation be $r_1$ and $r_2$ . According to Vieta's formulas, for a quadratic equation $ar^2 + br + c = 0$ , the sum of the roots $r_1 + r_2$ is equal to $-\frac{b}{a}$ . In our auxiliary equation $r^2 - 5r + 6 = 0$ , $a=1$ and $b=-5$ . Therefore, the sum of the roots is $r_1 + r_2 = - \frac{-5}{1} = 5$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Second-Order Linear Homogeneous Equations, Auxiliary Roots & Wronskian Theory)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — Second-Order Linear Homogeneous Equations, Auxiliary Roots & Wronskian Theory

Solve the initial value problem

y'' + 4y' + 13y = 0

with

y(0) = 2

and

y'(0) = -1

💡 Hint 1 of 3

Characteristic roots:

r^2 + 4r + 13 = 0 \implies r = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i

💡 Hint 2 of 3

Thus

\alpha = -2

\beta = 3

. General solution:

y(x) = e^{-2x}\big(C_1 \cos 3x + C_2 \sin 3x\big)

Apply $y(0) = 2$ :

y(0) = e^0(C_1 \cos 0 + C_2 \sin 0) = C_1

💡 Hint 3 of 3

Particular solution:

y(x) = e^{-2x}\big(2\cos 3x + \sin 3x\big)

Step 1: Characteristic roots: $r^2 + 4r + 13 = 0 \implies r = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i.$ Thus $\alpha = -2$ , $\beta = 3$ . General solution: $y(x) = e^{-2x}\big(C_1 \cos 3x + C_2 \sin 3x\big).$

Step 2: Apply $y(0) = 2$ : $y(0) = e^0(C_1 \cos 0 + C_2 \sin 0) = C_1 = 2.$

Step 3: Differentiate: $y'(x) = -2e^{-2x}(C_1 \cos 3x + C_2 \sin 3x) + e^{-2x}(-3C_1 \sin 3x + 3C_2 \cos 3x)$ $y'(0) = -2C_1 + 3C_2 = -2(2) + 3C_2 = -4 + 3C_2.$ Setting $y'(0) = -1$ : $-4 + 3C_2 = -1 \implies 3C_2 = 3 \implies C_2 = 1$ .

Step 4: Particular solution: $y(x) = e^{-2x}\big(2\cos 3x + \sin 3x\big).$

Solve the Cauchy-Euler differential equation

x^2 y'' - 3x y' + 4y = 0

for

x > 0

with initial conditions

y(1) = 3

and

y'(1) = 8

💡 Hint 1 of 3

Substitute Cauchy-Euler ansatz $y = x^m$ :

y' = m x^{m-1}

y'' = m(m-1)x^{m-2}

💡 Hint 2 of 3

Rewrite and simplify to obtain:

x^2\big(m(m-1)x^{m-2}\big) - 3x\big(m x^{m-1}\big) + 4x^m = 0 \implies x^m\big[m(m-1) - 3m + 4\big] = 0

Auxiliary equation:

m^2 - 4m + 4 = (m - 2)^2 = 0 \implies m

💡 Hint 3 of 3

Particular solution:

y(x) = x^2(3 + 2\ln x)

Step 1: Substitute Cauchy-Euler ansatz $y = x^m$ : $y' = m x^{m-1}$ , $y'' = m(m-1)x^{m-2}$ . $x^2\big(m(m-1)x^{m-2}\big) - 3x\big(m x^{m-1}\big) + 4x^m = 0 \implies x^m\big[m(m-1) - 3m + 4\big] = 0.$ Auxiliary equation: $m^2 - 4m + 4 = (m - 2)^2 = 0 \implies m = 2 \quad \text{(repeated root)}.$

Step 2: Second solution via reduction of order or variable change $x = e^t$ : For repeated root $m$ in Cauchy-Euler equations, the fundamental basis is $\{x^m, x^m \ln x\}$ . Thus: $y(x) = x^2\big(C_1 + C_2 \ln x\big).$

Step 3: Initial conditions at $x = 1$ : $y(1) = 1^2(C_1 + C_2 \ln 1) = C_1 = 3.$ Differentiating: $y'(x) = 2x(C_1 + C_2 \ln x) + x^2\left(\frac{C_2}{x}\right) = 2x(C_1 + C_2 \ln x) + C_2 x.$ $y'(1) = 2(C_1 + 0) + C_2 = 2(3) + C_2 = 6 + C_2.$ Setting $y'(1) = 8 \implies 6 + C_2 = 8 \implies C_2 = 2$ .

Step 4: Particular solution: $y(x) = x^2(3 + 2\ln x).$

[Practice 3 · Medium] Consider the second-order linear homogeneous differential equation

y'' - 2y' + y = 0

. \\ If one of the fundamental solutions is

y_1(x) = e^x

, and the Wronskian of two linearly independent solutions

y_1(x)

and

y_2(x)

is given by

W(y_1, y_2)(x) = 5e^{2x}

, find the unknown parameter

k

y_2(x) = kxe^x

💡 Hint 1 of 3

The given differential equation is

y'' - 2y' + y = 0

. The characteristic equation is

r^2 - 2r + 1 = 0

, which factors as

(r-1)^2 = 0

. Thus, the roots are repeated:

r_1 = r_2

💡 Hint 2 of 3

Rewrite and simplify to obtain:

y_2'(x) = \frac{d}{dx}(kxe^x) = k(1 \cdot e^x + x \cdot e^x) = ke^x(1+x)

💡 Hint 3 of 3

We equate this to the given Wronskian to determine

ke^{2x}

The given differential equation is

y'' - 2y' + y = 0

. The characteristic equation is

r^2 - 2r + 1 = 0

, which factors as

(r-1)^2 = 0

. Thus, the roots are repeated:

r_1 = r_2 = 1

. \\ One fundamental solution is given as

y_1(x) = e^x

. For a second-order linear homogeneous equation

y'' + p(x)y' + q(x)y = 0

with a known solution

y_1(x)

, the Wronskian is given by

W(y_1, y_2)(x) = y_1(x)y_2'(x) - y_1'(x)y_2(x) = C e^{-\int p(x) dx}

. \\ In our equation,

p(x) = -2

. So,

-\int p(x) dx = -\int (-2) dx = 2x

. Therefore,

W(y_1, y_2)(x) = C e^{2x}

. \\ We are given that

W(y_1, y_2)(x) = 5e^{2x}

, which means

C = 5

. \\ We are also given

y_1(x) = e^x

and

y_2(x) = kxe^x

. \\ Let's compute the Wronskian using these forms: \\

y_1'(x) = e^x

y_2'(x) = \frac{d}{dx}(kxe^x) = k(1 \cdot e^x + x \cdot e^x) = ke^x(1+x)

\\ Now, substitute into the Wronskian formula: \\

W(y_1, y_2)(x) = y_1(x)y_2'(x) - y_1'(x)y_2(x)

W(y_1, y_2)(x) = e^x \cdot [ke^x(1+x)] - e^x \cdot [kxe^x]

W(y_1, y_2)(x) = ke^{2x}(1+x) - kxe^{2x}

W(y_1, y_2)(x) = ke^{2x} + kxe^{2x} - kxe^{2x}

W(y_1, y_2)(x) = ke^{2x}

\\ We equate this to the given Wronskian: \\

ke^{2x} = 5e^{2x}

\\ Therefore,

k = 5

[Practice 4 · Hard] Consider the second-order linear homogeneous differential equation

y'' - 3y' + 2y = 0.

Find the general solution to this equation.

💡 Hint 1 of 3

The characteristic equation for the differential equation

y'' - 3y' + 2y = 0

r^2 - 3r + 2 = 0

💡 Hint 2 of 3

Factoring this quadratic equation gives

(r-1)(r-2) = 0

. The roots are

r_1 = 1

and

r_2 = 2

. Since the roots are real and distinct, the general solution is of the form

y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}

💡 Hint 3 of 3

Substituting the roots, we get

y(x) = c_1 e^x + c_2 e^{2x}

The characteristic equation for the differential equation

y'' - 3y' + 2y = 0

r^2 - 3r + 2 = 0

. Factoring this quadratic equation gives

(r-1)(r-2) = 0

. The roots are

r_1 = 1

and

r_2 = 2

. Since the roots are real and distinct, the general solution is of the form

y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}

. Substituting the roots, we get

y(x) = c_1 e^x + c_2 e^{2x}

. Choice B is incorrect because it uses negative roots. Choice C is incorrect because it corresponds to complex roots. Choice D is incorrect because it corresponds to a repeated root.

📝 Chapter Quiz

Questions Level

Nonhomogeneous Equations, Undetermined Coefficients & Resonance

When physical systems are subjected to time-varying external inputs — such as aerodynamic gusts on an aircraft wing, periodic magnetic fields driving an electric motor, or oscillating alternating currents driving an RLC filter — the governing equation becomes nonhomogeneous: $a y'' + b y' + c y = g(x)$ . The total response decomposes into internal natural dynamics (the complementary function) plus forced response (the particular solution). This chapter develops the method of undetermined coefficients, formalizes the Annihilator Method, and rigorously analyzes physical resonance and secular amplitude growth.

● Level 1 Foundation — Definition 4.1: Linear Nonhomogeneous Equations, Complementary & Particular Solutions

A second-order linear nonhomogeneous ODE with constant coefficients is written: $a y'' + b y' + c y = g(x), \qquad a \neq 0.$ The associated homogeneous equation $a y'' + b y' + c y = 0$ is called the reduced (complementary) equation, and its general solution is the complementary function $y_c(x) = C_1 y_1(x) + C_2 y_2(x)$ . Any single specific function $y_p(x)$ satisfying $a y_p'' + b y_p' + c y_p = g(x)$ is termed a particular solution.

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

⊙ Explorer · Undetermined Coefficients: Beats vs. Pure Resonance

(\omega \to \omega_{0})

interactive

Live numerical table for Undetermined Coefficients: Beats vs. Pure Resonance $(\omega \to \omega_{0})$
Parameter / Metric	Value

Driving Frequency $\omega =$ 2.4

Natural Frequency $\omega_{0} =$ 2

Detuning

|\omega - \omega_{0}|

|\omega -\omega_{0}| = 0.40

Amplitude Gain

1 / |\omega_{0}^{2} - \omega^{2}|

Gain = 0.57

Regime (Beats vs. Resonance)Periodic Beats Envelope

For y'' + $\omega_{0}^{2}$ y = $\cos(\omega t)$ , when $\omega \ne \omega_{0}$ the response exhibits periodic beats with envelope $\pm 1 / |\omega_{0}^{2} - \omega^{2}|$ . When $\omega = \omega_{0}$ , undetermined coefficients requires the secular term $t \cdot \sin(\omega_{0}t)$ (pure resonance!).

▲ Level 1 Foundation — Theorem 4.1: General Solution Structure & Affine Subspaces

Let $L[y] = a y'' + b y' + c y$ . If $y_p(x)$ is any fixed particular solution of $L[y] = g(x)$ , then the complete general solution is $y(x) = y_c(x) + y_p(x) = C_1 y_1(x) + C_2 y_2(x) + y_p(x),$ where $\{y_1, y_2\}$ is a fundamental solution set of $L[y] = 0$ . Geometrically, the solution set is an affine subspace of $C^2(I)$ formed by shifting $\ker(L)$ by $y_p$ .

Proof

Let $y(x)$ be any arbitrary solution to $L[y] = g(x)$ . Consider the difference $u(x) = y(x) - y_p(x)$ . By linearity of $L$ : $L[u] = L[y - y_p] = L[y] - L[y_p] = g(x) - g(x) = 0.$ Thus $u(x) \in \ker(L)$ , so $u(x) = C_1 y_1(x) + C_2 y_2(x) = y_c(x)$ for some unique constants $C_1, C_2$ . Therefore, $y(x) = u(x) + y_p(x) = y_c(x) + y_p(x)$ .

Standard Trial Forms for Undetermined Coefficients

Forcing Term g(x)	Basic Trial Form y_p(x)
Polynomial $P_n(x) = a_n x^n + \dots + a_0$	$A_n x^n + A_{n-1} x^{n-1} + \dots + A_1 x + A_0$
Exponential $e^{\alpha x}$	$A e^{\alpha x}$
Sinusoidal $\cos(\beta x)$ or $\sin(\beta x)$	$A \cos(\beta x) + B \sin(\beta x)$
Product $P_n(x)e^{\alpha x}\cos(\beta x)$	$(A_n x^n + \dots + A_0)e^{\alpha x}\cos(\beta x) + (B_n x^n + \dots + B_0)e^{\alpha x}\sin(\beta x)$

● Level 2 Analytical Rule — The Resonance Multiplication Rule & Duplication Elimination

If any term in the standard trial form $y_p(x)$ replicates a term in the complementary function $y_c(x)$ (meaning its associated characteristic root is a root of the auxiliary equation $a r^2 + b r + c = 0$ of multiplicity $s \ge 1$ ), then the basic trial form must be multiplied by $x^s$ :

y_p(x) = x^s \cdot \big[\text{Basic Trial Form}\big],

where $s = 1$ if the root is simple, and $s = 2$ if the root is repeated. This multiplication prevents the differential operator $L$ from annihilating the trial form to zero, allowing the undetermined coefficients to be matched uniquely.

Analytical Procedure for Undetermined Coefficients

1. Solve Homogeneous Equation

Find characteristic roots of

a r^2 + b r + c = 0

and write out the complementary function

y_c(x) = C_1 y_1(x) + C_2 y_2(x)

2. Select Basic Trial Form

From the shape of

g(x)

, select the general family containing

g(x)

and all of its derivatives.

3. Check Duplication (Multiply by x^s)

Compare with

y_c(x)

. If any term is duplicated, multiply the entire trial candidate by

x

(or

x^2

) until no duplication remains.

4. Substitute & Match Coefficients

Compute

y_p', y_p''

, substitute into

a y_p'' + b y_p' + c y_p = g(x)

, group like functions, and solve the linear system for coefficients.

Worked example 4.A — Polynomial & Exponential Forcing without Resonance

Find the general solution of $y'' - 3y' + 2y = 4x^2.$

Step 1: Complementary function: $r^2 - 3r + 2 = (r - 1)(r - 2) = 0 \implies y_c(x) = C_1 e^x + C_2 e^{2x}.$

Step 2: Trial form: For $g(x) = 4x^2$ , trial is $y_p(x) = A x^2 + B x + C$ . No term is in $y_c(x)$ , so $s = 0$ .

Step 3: Differentiate and substitute: $y_p' = 2Ax + B$ , $y_p'' = 2A$ . $(2A) - 3(2Ax + B) + 2(Ax^2 + Bx + C) = 4x^2$ $(2A)x^2 + (-6A + 2B)x + (2A - 3B + 2C) = 4x^2.$

Step 4: Match coefficients: $2A = 4 \implies A = 2,$ $-6(2) + 2B = 0 \implies 2B = 12 \implies B = 6,$ $2(2) - 3(6) + 2C = 0 \implies 4 - 18 + 2C = 0 \implies 2C = 14 \implies C = 7.$

Thus $y_p(x) = 2x^2 + 6x + 7$ , and general solution is: $y(x) = C_1 e^x + C_2 e^{2x} + 2x^2 + 6x + 7.$

Worked example 4.B — Exponential Forcing with Resonance

Solve the initial value problem $y'' - 4y' + 4y = 6e^{2x}, \qquad y(0) = 0, \quad y'(0) = 1.$

Step 1: Complementary function: $r^2 - 4r + 4 = (r - 2)^2 = 0 \implies y_c(x) = (C_1 + C_2 x)e^{2x}.$

Step 2: Trial form with resonance: The standard trial $A e^{2x}$ and $A x e^{2x}$ are both already present in $y_c$ because $r = 2$ is a double root (multiplicity $s = 2$ ). Therefore: $y_p(x) = A x^2 e^{2x}.$

Step 3: Derivatives: $y_p'(x) = A(2x e^{2x} + 2x^2 e^{2x}) = 2A(x + x^2)e^{2x},$ $y_p''(x) = 2A(1 + 2x)e^{2x} + 4A(x + x^2)e^{2x} = 2A(1 + 4x + 2x^2)e^{2x}.$

Step 4: Substitute: $e^{2x}\Big[2A(1 + 4x + 2x^2) - 4\cdot 2A(x + x^2) + 4\cdot A x^2\Big] = 6e^{2x}$ $e^{2x}\Big[2A + (8A - 8A)x + (4A - 8A + 4A)x^2\Big] = 2A e^{2x} = 6e^{2x} \implies 2A = 6 \implies A = 3.$ Thus $y_p(x) = 3x^2 e^{2x}$ , and general solution is $y(x) = (C_1 + C_2 x + 3x^2)e^{2x}$ .

Step 5: Initial conditions: $y(0) = C_1 = 0$ .
$y'(x) = (C_2 + 6x)e^{2x} + 2(C_2 x + 3x^2)e^{2x} \implies y'(0) = C_2 = 1$ .
Particular solution: $y(x) = (x + 3x^2)e^{2x}.$

Worked example 4.C — Analytical Application: Driven RLC Circuit & Steady-State Gain

A series RLC circuit with charge $Q(t)$ is driven by alternating voltage: $L \frac{d^2Q}{dt^2} + R \frac{dQ}{dt} + \frac{1}{C}Q = V_0 \cos(\omega t).$ Compute the steady-state current $I_{\text{ss}}(t) = \frac{dQ_{\text{ss}}}{dt}$ and the resonant driving frequency $\omega_{\text{res}}$ that maximizes current amplitude.

Step 1: Complex trial method: Consider the complex ODE $L Q'' + R Q' + \frac{1}{C}Q = V_0 e^{i\omega t}$ . Trial: $Q_p(t) = Q_0 e^{i\omega t}$ . $Q_0\left(-L\omega^2 + i R \omega + \frac{1}{C}\right)e^{i\omega t} = V_0 e^{i\omega t} \implies Q_0 = \frac{V_0}{(1/C - L\omega^2) + i R\omega}.$

Step 2: Steady-state current: $I_p(t) = Q_p'(t) = i\omega Q_0 e^{i\omega t} = \frac{V_0 e^{i\omega t}}{R + i(L\omega - \frac{1}{C\omega})} = \frac{V_0}{Z(\omega)} e^{i(\omega t - \phi)}$ , where impedance magnitude is: $|Z(\omega)| = \sqrt{R^2 + \left(L\omega - \frac{1}{\omega C}\right)^2}.$

Step 3: Resonant frequency: Current amplitude $I_0 = \frac{V_0}{|Z(\omega)|}$ is maximized when impedance $|Z|$ is minimized: $L\omega - \frac{1}{\omega C} = 0 \implies \omega_{\text{res}} = \frac{1}{\sqrt{LC}}.$ At electrical resonance, impedance is purely resistive $Z = R$ , and current reaches its absolute physical peak $I_{\max} = V_0/R$ .

▲ Level 3 University Rigor — Theorem 4.2: Annihilator Operator Theory, Pure Resonance & Secular Amplitude Catastrophe

1. Annihilator Operator Theory: Let $D = \frac{d}{dx}$ be the differential operator. A polynomial operator $A(D)$ is an annihilator of $g(x)$ if $A(D)[g(x)] \equiv 0$ . Fundamental annihilators include:

$D^{k+1}$ annihilates $x^k$ .
$(D - \alpha)^{k+1}$ annihilates $x^k e^{\alpha x}$ .
$(D^2 - 2\alpha D + \alpha^2 + \beta^2)^{k+1}$ annihilates $x^k e^{\alpha x}\cos(\beta x)$ and $x^k e^{\alpha x}\sin(\beta x)$ .

Applying $A(D)$ to $P(D)y = g(x)$ yields the higher-order homogeneous ODE $A(D)P(D)y = 0.$ Any solution to $P(D)y = g(x)$ must be a solution to this homogeneous equation. The roots of $A(r)$ that are distinct from $P(r)$ generate the basic trial form; roots of $A(r)$ that match roots of $P(r)$ with multiplicity $s$ require multiplying by $x^s$ , rigorously establishing the resonance rule.

2. Pure Resonance & Secular Amplitude Catastrophe: Consider the undamped oscillator $y'' + \omega_0^2 y = F_0 \cos(\omega t)$ with $y(0) = 0, y'(0) = 0$ . For $\omega \neq \omega_0$ , the exact solution is the beat waveform: $y(t) = \frac{F_0}{\omega_0^2 - \omega^2}\big(\cos(\omega t) - \cos(\omega_0 t)\big).$ Taking the pure resonance limit as $\omega \to \omega_0$ via L'Hôpital's Rule: $\lim_{\omega \to \omega_0} y(t) = \lim_{\omega \to \omega_0} \frac{\frac{d}{d\omega}[F_0(\cos\omega t - \cos\omega_0 t)]}{\frac{d}{d\omega}[\omega_0^2 - \omega^2]} = \lim_{\omega \to \omega_0} \frac{-F_0 t \sin(\omega t)}{-2\omega} = \frac{F_0}{2\omega_0} t \sin(\omega_0 t).$ The secular factor $t$ causes the oscillation amplitude to grow linearly toward infinity: $|y(t)| \propto t$ , leading inevitably to physical destruction (secular catastrophe) in the absence of damping.

When an undamped mechanical system is driven at its exact natural frequency, energy accumulates continuously without dissipation, generating linear secular amplitude growth, as plotted below.

Pure resonance in

y'' + 4y = 4\cos(2t)

. The particular solution

y_p(t) = t\sin(2t)

displays an amplitude expanding proportionally to

t

, bounded by the linear envelopes

\pm t

⚠️ Common Traps & University Pitfalls — Duplication with Homogeneous Solutions

Incomplete trigonometric pairs: If $g(x) = \cos(\beta x)$ , the trial form MUST be $A\cos(\beta x) + B\sin(\beta x)$ . Omitting the sine term will cause the linear system to be inconsistent whenever $b \neq 0$ , because the first derivative $y'$ produces sines from cosines.

Resonance scaling: When multiplying by $x$ due to root duplication, the ENTIRE trial pair must be multiplied: $x(A\cos\beta x + B\sin\beta x)$ , NOT just $A x\cos\beta x + B\sin\beta x$ .

Method limitations: Undetermined coefficients strictly applies ONLY when $g(x)$ is a linear combination of products of polynomials, exponentials, and sines/cosines. For forcing terms like $\tan x, \sec x, \ln x, 1/x$ , or $e^{x^2}$ , you MUST use Variation of Parameters (Chapter 5).

Worked Example

D

— Nonhomogeneous Equations, Undetermined Coefficients & Resonance (Hard Application)

Problem (Hard): Consider the second-order linear nonhomogeneous differential equation: $y''(x) - 3y'(x) + 2y(x) = 4$ . Find the particular solution, $y_p(x)$ , using the method of Undetermined Coefficients.

Step-by-Step Solution: The given differential equation is $y''(x) - 3y'(x) + 2y(x) = 4$ . Since the right-hand side is a constant, we guess a particular solution of the form $y_p(x) = A$ , where A is a constant. Then, $y_p'(x) = 0$ and $y_p''(x) = 0$ . Substituting these into the differential equation: $0 - 3(0) + 2(A) = 4$ $2A = 4$ $A = 2$ Therefore, the particular solution is $y_p(x) = 2$ .

▲ Exam Pacing & Verification Strategy — Nonhomogeneous Equations, Undetermined Coefficients & Resonance

Find the general solution of the differential equation

y'' + y = 4\sin x + 2e^{-x}

💡 Hint 1 of 3

Complementary function:

r^2 + 1 = 0 \implies r = \pm i \implies y_c(x) = C_1 \cos x + C_2 \sin x

💡 Hint 2 of 3

Particular solution trial form: For

4\sin x

: roots

\pm i

match

y_c

(multiplicity

s=1

), so trial is

y_{p1} = x(A\cos x + B\sin x)

. For

2e^{-x}

r = -1

is not a root of

r^2 + 1 = 0

, so trial is

y_{p2} = C e^{-x}

💡 Hint 3 of 3

General solution:

y(x) = C_1 \cos x + C_2 \sin x - 2x\cos x + e^{-x}

Step 1: Complementary function: $r^2 + 1 = 0 \implies r = \pm i \implies y_c(x) = C_1 \cos x + C_2 \sin x.$

Step 2: Particular solution trial form:
For $4\sin x$ : roots $\pm i$ match $y_c$ (multiplicity $s=1$ ), so trial is $y_{p1} = x(A\cos x + B\sin x)$ .
For $2e^{-x}$ : $r = -1$ is not a root of $r^2 + 1 = 0$ , so trial is $y_{p2} = C e^{-x}$ .
Total trial: $y_p = x(A\cos x + B\sin x) + C e^{-x}$ .

Step 3: Solve for $y_{p1}$ : $y_{p1}'' + y_{p1} = -2A\sin x + 2B\cos x = 4\sin x \implies -2A = 4 \implies A = -2$ , and $2B = 0 \implies B = 0$ . Thus $y_{p1} = -2x\cos x$ .

Step 4: Solve for $y_{p2}$ : $y_{p2}'' + y_{p2} = C e^{-x} + C e^{-x} = 2C e^{-x} = 2e^{-x} \implies C = 1$ . Thus $y_{p2} = e^{-x}$ .

Step 5: General solution: $y(x) = C_1 \cos x + C_2 \sin x - 2x\cos x + e^{-x}.$

Solve the initial value problem

y'' - 2y' + y = (x + 1)e^x

with

y(0) = 1

and

y'(0) = 0

💡 Hint 1 of 3

Complementary function:

r^2 - 2r + 1 = (r - 1)^2 = 0 \implies y_c(x) = (C_1 + C_2 x)e^x

💡 Hint 2 of 3

Thus

y_p(x) = \left(\frac{x^3}{6} + \frac{x^2}{2}\right)e^x

💡 Hint 3 of 3

Particular solution:

y(x) = \left(1 - x + \frac{x^2}{2} + \frac{x^3}{6}\right)e^x

Step 1: Complementary function: $r^2 - 2r + 1 = (r - 1)^2 = 0 \implies y_c(x) = (C_1 + C_2 x)e^x.$

Step 2: Particular trial form: Basic trial for $(x + 1)e^x$ is $(A x + B)e^x$ . Both $x e^x$ and $e^x$ are in $y_c$ (multiplicity $s = 2$ ). Multiply by $x^2$ : $y_p(x) = x^2(Ax + B)e^x = (A x^3 + B x^2)e^x.$

Step 3: Derivatives and substitution: Using operator $(D - 1)^2[u e^x] = e^x D^2[u]$ : $(D - 1)^2\big[(Ax^3 + Bx^2)e^x\big] = e^x \frac{d^2}{dx^2}[Ax^3 + Bx^2] = e^x(6Ax + 2B).$ Match with $(x + 1)e^x$ : $6Ax + 2B = x + 1 \implies 6A = 1 \implies A = 1/6, \quad 2B = 1 \implies B = 1/2.$ Thus $y_p(x) = \left(\frac{x^3}{6} + \frac{x^2}{2}\right)e^x$ .

Step 4: General solution and initial conditions: $y(x) = \left(C_1 + C_2 x + \frac{x^2}{2} + \frac{x^3}{6}\right)e^x.$ $y(0) = C_1 = 1$ .
$y'(x) = \left(C_2 + x + \frac{x^2}{2}\right)e^x + \left(1 + C_2 x + \frac{x^2}{2} + \frac{x^3}{6}\right)e^x \implies y'(0) = C_2 + 1 = 0 \implies C_2 = -1$ .
Particular solution: $y(x) = \left(1 - x + \frac{x^2}{2} + \frac{x^3}{6}\right)e^x.$

[Practice 3 · Easy] Consider the nonhomogeneous second-order linear differential equation

y'' + 4y = 8

. What is the particular solution,

y_p(x)

, to this equation?

💡 Hint 1 of 3

The given differential equation is

y'' + 4y = 8

💡 Hint 2 of 3

This is a nonhomogeneous second-order linear differential equation. We use the method of undetermined coefficients to find a particular solution,

y_p(x)

💡 Hint 3 of 3

Therefore, the particular solution is

y_p(x)

The given differential equation is

y'' + 4y = 8

. This is a nonhomogeneous second-order linear differential equation. We use the method of undetermined coefficients to find a particular solution,

y_p(x)

. The nonhomogeneous term is a constant, 8. We guess a particular solution of the form

y_p(x) = A

, where A is a constant. Then,

y_p'(x) = 0

and

y_p''(x) = 0

. Substituting these into the differential equation:

0 + 4(A) = 8

. Solving for A, we get

4A = 8

, which means

A = 2

. Therefore, the particular solution is

y_p(x) = 2

[Practice 4 · Medium] Consider the nonhomogeneous second-order linear differential equation

y'' - 5y' + 6y = 12x + 8

y_h(x) = c_1e^{2x} + c_2e^{3x}

is the general solution to the corresponding homogeneous equation. Find the particular solution

y_p(x)

using the method of undetermined coefficients, and then determine the value of the constant

A

such that

y_p(x) = Ax + B

for some constant

B

💡 Hint 1 of 3

The given nonhomogeneous differential equation is

y'' - 5y' + 6y = 12x + 8

💡 Hint 2 of 3

The form of the nonhomogeneous term is

g(x) = 12x + 8

, which is a polynomial of degree 1. Thus, we assume a particular solution of the form

y_p(x) = Ax + B

. We need to find the first and second derivatives of

y_p(x)

y_p'(x) = A

💡 Hint 3 of 3

Therefore, the particular solution is

y_p(x) = 2x + 3

. The value of the constant

A

is 2.

The given nonhomogeneous differential equation is

y'' - 5y' + 6y = 12x + 8

. The form of the nonhomogeneous term is

g(x) = 12x + 8

, which is a polynomial of degree 1. Thus, we assume a particular solution of the form

y_p(x) = Ax + B

. We need to find the first and second derivatives of

y_p(x)

y_p'(x) = A

y_p''(x) = 0

Substitute these into the differential equation:

0 - 5(A) + 6(Ax + B) = 12x + 8

-5A + 6Ax + 6B = 12x + 8

6Ax + (6B - 5A) = 12x + 8

Equating the coefficients of like powers of

x

: For the

x

term:

6A = 12

\Rightarrow

A = 2

. For the constant term:

6B - 5A = 8

. Substitute

A = 2

into the second equation:

6B - 5(2) = 8

6B - 10 = 8

6B = 18

B = 3

Therefore, the particular solution is

y_p(x) = 2x + 3

. The value of the constant

A

is 2.

📝 Chapter Quiz

Questions Level

Variation of Parameters, Wronskian Systems & Green's Kernel

While undetermined coefficients requires forcing terms to belong to finite-dimensional derivative-invariant families (polynomials, exponentials, sines, and cosines), modern applied mathematics frequently encounters non-elementary inputs: aerodynamic drag terms, logarithmic potentials, tangent/secant waveforms in nonlinear oscillators, and Dirac impulse distributions. The method of Variation of Parameters (originated by Joseph-Louis Lagrange) provides a universally applicable analytical technique to construct particular solutions for any continuous forcing function, formalizing the transition to Green's functions and linear response theory.

● Level 1 Foundation — Definition 5.1: The Fundamental Matrix & Wronskian Determinant

Consider the second-order linear differential equation in standard normalized form: $y'' + P(x)y' + Q(x)y = g(x),$ where $P, Q, g \in C(I)$ . Let $\{y_1(x), y_2(x)\}$ be a fundamental set of solutions to the homogeneous equation $y'' + P y' + Q y = 0$ . Their Wronskian determinant is $W(y_1, y_2)(x) = y_1(x)y_2'(x) - y_1'(x)y_2(x) \neq 0 \quad \forall x \in I.$

Instead of treating arbitrary constants $C_1, C_2$ as fixed numbers, Lagrange replaced them with unknown variable parameter functions $u_1(x), u_2(x)$ : $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x).$

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

⊙ Explorer · Variation of Parameters: Wronskian

W(y_{1}, y_{2})

& Causal Green's Kernel

G(t, s)

interactive

Live numerical table for Variation of Parameters: Wronskian $W(y_{1}, y_{2})$ & Causal Green's Kernel $G(t, s)$
Parameter / Metric	Value

Impulse Time $s =$ 1.5

Frequency $\omega =$ 1.5

Wronskian

W(\cos \omega t, \sin \omega t) = \omega

W = 1.50

Green's Kernel Max

1 / \omega

1/\omega = 0.667

Causal Response

G(4.0, s)

G(4, s) = -0.381

Plot fundamental solutions $y_{1}(t) = \cos(\omega t)$ , $y_{2}(t) = \sin(\omega t)$ (dashed) and the causal Green's kernel $G(t, s) = (y_{1}(s)y_{2}(t) - y_{2}(s)y_{1}(t)) / W(s) = \sin(\omega (t - s)) / \omega$ for $t \ge s$ (solid amber).

▲ Level 1 Foundation — Theorem 5.1: Variation of Parameters Formula & Lagrange Substitution

Let $\{y_1, y_2\}$ be a fundamental solution set of $y'' + P(x)y' + Q(x)y = 0$ with Wronskian $W(x)$ . A particular solution to the nonhomogeneous equation $y'' + P(x)y' + Q(x)y = g(x)$ is given by: $y_p(x) = -y_1(x)\int \frac{y_2(x)g(x)}{W(x)}\,dx + y_2(x)\int \frac{y_1(x)g(x)}{W(x)}\,dx.$

Proof

Differentiating $y_p = u_1 y_1 + u_2 y_2$ via the product rule: $y_p' = (u_1 y_1' + u_2 y_2') + (u_1' y_1 + u_2' y_2).$ Because we have two unknown functions $u_1, u_2$ but only one governing differential equation, we are free to impose one auxiliary constraint. To avoid second derivatives of $u_1, u_2$ , Lagrange imposed the condition: $u_1'(x)y_1(x) + u_2'(x)y_2(x) = 0.$ With this constraint, the derivative simplifies to $y_p' = u_1 y_1' + u_2 y_2'$ . Differentiating again: $y_p'' = u_1 y_1'' + u_2 y_2'' + u_1' y_1' + u_2' y_2'.$ Substituting $y_p, y_p', y_p''$ into $y'' + Py' + Qy = g(x)$ : $u_1(y_1'' + Py_1' + Qy_1) + u_2(y_2'' + Py_2' + Qy_2) + u_1' y_1' + u_2' y_2' = g(x).$ Since $y_1, y_2$ solve the homogeneous equation, the first two grouped expressions vanish, leaving: $u_1'(x)y_1'(x) + u_2'(x)y_2'(x) = g(x).$ We now have a simultaneous linear algebraic system for the derivatives $u_1', u_2'$ : $\begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix} \begin{pmatrix} u_1' \\ u_2' \end{pmatrix} = \begin{pmatrix} 0 \\ g(x) \end{pmatrix}.$ Solving this system via Cramer's Rule (determinant is $W(x) \neq 0$ ): $u_1'(x) = \frac{\det\begin{pmatrix} 0 & y_2 \\ g & y_2' \end{pmatrix}}{W(x)} = -\frac{y_2(x)g(x)}{W(x)}, \qquad u_2'(x) = \frac{\det\begin{pmatrix} y_1 & 0 \\ y_1' & g \end{pmatrix}}{W(x)} = \frac{y_1(x)g(x)}{W(x)}.$ Integrating $u_1'$ and $u_2'$ yields the stated particular solution formula.

Analytical Procedure for Variation of Parameters

1. Standard Normalization

Verify the equation is in standard form

y'' + P(x)y' + Q(x)y = g(x)

. If the leading coefficient is

a(x) \neq 1

, divide through by

a(x)

so that

g(x) = f(x)/a(x)

2. Homogeneous Basis & Wronskian

Determine

\{y_1, y_2\}

from

a r^2 + b r + c = 0

and evaluate

W(x) = y_1 y_2' - y_1' y_2

3. Parameter Derivative Integrals

Compute

u_1(x) = -\int \frac{y_2(x)g(x)}{W(x)}\,dx

and

u_2(x) = \int \frac{y_1(x)g(x)}{W(x)}\,dx

. Constants of integration may be taken as zero.

4. Assemble General Solution

Combine to write

y(x) = C_1 y_1(x) + C_2 y_2(x) + u_1(x)y_1(x) + u_2(x)y_2(x)

Worked example 5.A — Non-Elementary Forcing: Secant Function

Find the general solution of $y'' + 4y = \sec(2x), \qquad x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right).$

Step 1: Complementary solutions: Auxiliary equation $r^2 + 4 = 0 \implies r = \pm 2i$ . Basis: $y_1(x) = \cos(2x), \qquad y_2(x) = \sin(2x).$

Step 2: Wronskian: $W(x) = \det\begin{pmatrix} \cos 2x & \sin 2x \\ -2\sin 2x & 2\cos 2x \end{pmatrix} = 2\cos^2(2x) + 2\sin^2(2x) = 2.$

Step 3: Integrate parameter derivatives: $u_1'(x) = -\frac{y_2(x)g(x)}{W(x)} = -\frac{\sin(2x)\sec(2x)}{2} = -\frac{1}{2}\tan(2x).$ $u_1(x) = -\frac{1}{2}\int \tan(2x)\,dx = \frac{1}{4}\ln|\cos(2x)|.$ $u_2'(x) = \frac{y_1(x)g(x)}{W(x)} = \frac{\cos(2x)\sec(2x)}{2} = \frac{1}{2} \implies u_2(x) = \frac{x}{2}.$

Step 4: Particular solution: $y_p(x) = u_1 y_1 + u_2 y_2 = \frac{1}{4}\ln|\cos(2x)|\cos(2x) + \frac{x}{2}\sin(2x).$ General solution: $y(x) = C_1 \cos(2x) + C_2 \sin(2x) + \frac{1}{4}\cos(2x)\ln|\cos(2x)| + \frac{x}{2}\sin(2x).$

Worked example 5.B — A Repeated Root with Logarithmic Forcing

Find a particular solution of $y'' + 2y' + y = e^{-x}\ln x, \qquad x > 0.$

Step 1: Complementary basis: $r^2 + 2r + 1 = 0 \implies r = -1$ (repeated). Basis: $y_1 = e^{-x}, y_2 = x e^{-x}$ .

Step 2: Wronskian: $W(x) = e^{-x}\big(e^{-x} - x e^{-x}\big) - \big(-e^{-x}\big)x e^{-x} = e^{-2x}(1 - x + x) = e^{-2x}.$

Step 3: Parameter functions: $u_1'(x) = -\frac{x e^{-x}(e^{-x}\ln x)}{e^{-2x}} = -x\ln x \implies u_1(x) = -\int x\ln x\,dx = -\frac{x^2}{2}\ln x + \frac{x^2}{4}.$ $u_2'(x) = \frac{e^{-x}(e^{-x}\ln x)}{e^{-2x}} = \ln x \implies u_2(x) = \int \ln x\,dx = x\ln x - x.$

Step 4: Particular solution: $y_p(x) = e^{-x}\left(-\frac{x^2}{2}\ln x + \frac{x^2}{4}\right) + x e^{-x}(x\ln x - x) = e^{-x}\left(\frac{x^2}{2}\ln x - \frac{3x^2}{4}\right).$

Worked example 5.C — Comparing Efficiency with Undetermined Coefficients

Solve $y'' - y = e^{2x}$ using Variation of Parameters, and compare with Undetermined Coefficients.

Step 1: Basis: $r^2 - 1 = 0 \implies y_1 = e^x, y_2 = e^{-x}$ . Wronskian: $W = e^x(-e^{-x}) - e^x e^{-x} = -2$ .

Step 2: Variation of Parameters integrals: $u_1' = -\frac{e^{-x}e^{2x}}{-2} = \frac{1}{2}e^x \implies u_1(x) = \frac{1}{2}e^x,$ $u_2' = \frac{e^x e^{2x}}{-2} = -\frac{1}{2}e^{3x} \implies u_2(x) = -\frac{1}{6}e^{3x}.$ $y_p = u_1 y_1 + u_2 y_2 = \left(\frac{1}{2}e^x\right)e^x + \left(-\frac{1}{6}e^{3x}\right)e^{-x} = \frac{1}{2}e^{2x} - \frac{1}{6}e^{2x} = \frac{1}{3}e^{2x}.$

Undetermined Coefficients check: Guess $y_p = A e^{2x}$ . $4A e^{2x} - A e^{2x} = 3A e^{2x} = e^{2x} \implies A = 1/3$ . While both methods succeed for elementary functions, Variation of Parameters remains fully operational when Undetermined Coefficients fails completely (e.g. $\sec x$ or $\ln x$ ).

▲ Level 3 University Rigor — Theorem 5.2: Generalization to

n

-th Order Linear ODEs & Green's Kernel Representation

1. Generalization to $n$ -th Order Equations: Consider the normalized $n$ -th order linear ODE $y^{(n)} + p_{n-1}(x)y^{(n-1)} + \dots + p_1(x)y' + p_0(x)y = g(x).$ Let $\{y_1, y_2, \dots, y_n\}$ be a fundamental solution set. Then $y_p(x) = \sum_{k=1}^n u_k(x)y_k(x)$ , where the derivatives $u_k'(x)$ satisfy the algebraic system: $\begin{pmatrix} y_1 & y_2 & \dots & y_n \\ y_1' & y_2' & \dots & y_n' \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \dots & y_n^{(n-1)} \end{pmatrix} \begin{pmatrix} u_1' \\ u_2' \\ \vdots \\ u_n' \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ g(x) \end{pmatrix}.$ By Cramer's Rule: $u_k'(x) = \frac{W_k(x)g(x)}{W(x)},$ where $W_k(x)$ is the determinant of the matrix obtained by replacing the $k$ -th column of the Wronskian matrix with $(0, 0, \dots, 0, 1)^T$ .

2. One-Sided Green's Function (Cauchy Kernel): The particular solution satisfying zero initial conditions $y(x_0) = 0, y'(x_0) = 0$ can be written as an integral operator: $y_p(x) = \int_{x_0}^x G(x, t) g(t)\,dt,$ where the Green's kernel is: $G(x, t) = \frac{\det\begin{pmatrix} y_1(t) & y_2(t) \\ y_1(x) & y_2(x) \end{pmatrix}}{W(t)} = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)}.$ $G(x, t)$ represents the physical impulse response of the linear system to a unit Dirac delta impulse applied at time $t$ .

The linear independence of fundamental solutions $y_1 = \cos(2x), y_2 = \sin(2x)$ ensures that the Wronskian determinant $W = 2$ remains strictly non-zero everywhere, guaranteeing that the parameter integrals never encounter singular denominators.

Fundamental basis

\{\cos 2x, \sin 2x\}

and constant non-vanishing Wronskian

W(x) = 2

. Abel's Theorem guarantees that

W(x)

never crosses zero on

\mathbb{R}

⚠️ Common Traps & University Pitfalls — Standard Form & Signs in Parameter Derivatives

Non-standard leading coefficients: If the differential equation is $2y'' + 8y = \sec(2x)$ , you MUST divide by 2 first so $g(x) = \frac{1}{2}\sec(2x)$ . Using the unnormalized right-hand side is the #1 student error in Variation of Parameters exams.

Sign error in $u_1'$ : Remember that $u_1'(x) = -\frac{y_2(x)g(x)}{W(x)}$ has a mandatory NEGATIVE sign from Cramer's Rule, while $u_2'(x) = +\frac{y_1(x)g(x)}{W(x)}$ is positive.

Arbitrary constants of integration: When integrating $u_1'$ and $u_2'$ , constants $+C_1, +C_2$ are unnecessary because adding them merely regenerates the complementary function $C_1 y_1 + C_2 y_2$ .

Worked Example

D

— Variation of Parameters, Wronskian Systems & Green's Kernel (Medium Application)

Problem (Medium): Consider the second-order linear nonhomogeneous differential equation $y'' - y = \sin x$ . If $y_p(x) = A \cos x + B \sin x$ is a particular solution, find the values of the constants $A$ and $B$ .

Step-by-Step Solution: We are given the differential equation $y'' - y = \sin x$ and a proposed particular solution $y_p(x) = A \cos x + B \sin x$ . We need to find the derivatives of $y_p(x)$ and substitute them into the differential equation. First derivative: $y_p'(x) = -A \sin x + B \cos x$ . Second derivative: $y_p''(x) = -A \cos x - B \sin x$ . Substitute these into the differential equation: $(-A \cos x - B \sin x) - (A \cos x + B \sin x) = \sin x$ Combine like terms: $(-A - A) \cos x + (-B - B) \sin x = \sin x$ $-2A \cos x - 2B \sin x = \sin x$ For this equation to hold for all $x$ , the coefficients of $\cos x$ and $\sin x$ on both sides must be equal. Comparing coefficients of $\cos x$ : $-2A = 0$ , which implies $A = 0$ . Comparing coefficients of $\sin x$ : $-2B = 1$ , which implies $B = -\frac{1}{2}$ . Therefore, the values of the constants are $A = 0$ and $B = -\frac{1}{2}$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Variation of Parameters, Wronskian Systems & Green's Kernel)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — Variation of Parameters, Wronskian Systems & Green's Kernel

Find the general solution of

y'' + y = \tan x

for

x \in (-\pi/2, \pi/2)

💡 Hint 1 of 3

Complementary basis & Wronskian:

y_1 = \cos x, y_2 = \sin x

💡 Hint 2 of 3

Rewrite and simplify to obtain:

W(x) = \cos x(\cos x) - (-\sin x)(\sin x) = 1

Parameter integrals:

u_1'(x) = -\frac{\sin x \tan x}{1} = -\frac{\sin^2 x}{\cos x} = -\frac{1 - \cos^2 x}{\cos x} = \cos x - \sec x

💡 Hint 3 of 3

General solution:

y(x) = C_1 \cos x + C_2 \sin x - \cos x \ln|\sec x + \tan x|

Step 1: Complementary basis & Wronskian: $y_1 = \cos x, y_2 = \sin x$ . $W(x) = \cos x(\cos x) - (-\sin x)(\sin x) = 1.$

Step 2: Parameter integrals: $u_1'(x) = -\frac{\sin x \tan x}{1} = -\frac{\sin^2 x}{\cos x} = -\frac{1 - \cos^2 x}{\cos x} = \cos x - \sec x.$ $u_1(x) = \int (\cos x - \sec x)\,dx = \sin x - \ln|\sec x + \tan x|.$ $u_2'(x) = \frac{\cos x \tan x}{1} = \sin x \implies u_2(x) = -\cos x.$

Step 3: Particular solution: $y_p(x) = u_1 y_1 + u_2 y_2 = \big(\sin x - \ln|\sec x + \tan x|\big)\cos x + (-\cos x)\sin x = -\cos x \ln|\sec x + \tan x|.$

Step 4: General solution: $y(x) = C_1 \cos x + C_2 \sin x - \cos x \ln|\sec x + \tan x|.$

Use the Green's function representation to solve the initial value problem

y'' - 2y' + y = \frac{e^x}{x^2 + 1}

with

y(0) = 0

and

y'(0) = 0

💡 Hint 1 of 3

Fundamental basis:

r^2 - 2r + 1 = 0 \implies y_1(x) = e^x, y_2(x) = x e^x

💡 Hint 2 of 3

Wronskian:

W(t) = e^t(e^t + t e^t) - e^t(t e^t) = e^{2t}

. Construct Green's function:

G(x, t) = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)} = \frac{e^t (x e^x) - e^x (t e^t)}{e^{2t}} = \frac{e^{x+t}(x - t)}{e^{2t}} = (x - t)e^{x - t}

💡 Hint 3 of 3

Thus:

y(x) = e^x\left(x \arctan x - \frac{1}{2}\ln(x^2 + 1)\right)

Step 1: Fundamental basis: $r^2 - 2r + 1 = 0 \implies y_1(x) = e^x, y_2(x) = x e^x$ .
Wronskian: $W(t) = e^t(e^t + t e^t) - e^t(t e^t) = e^{2t}$ .

Step 2: Construct Green's function: $G(x, t) = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)} = \frac{e^t (x e^x) - e^x (t e^t)}{e^{2t}} = \frac{e^{x+t}(x - t)}{e^{2t}} = (x - t)e^{x - t}.$

Step 3: Evaluate integral with zero initial conditions: $y(x) = \int_0^x G(x, t) g(t)\,dt = \int_0^x (x - t)e^{x - t} \left(\frac{e^t}{t^2 + 1}\right)dt = e^x \int_0^x \frac{x - t}{t^2 + 1}\,dt$ $= e^x \left(x \int_0^x \frac{1}{t^2 + 1}\,dt - \int_0^x \frac{t}{t^2 + 1}\,dt\right) = e^x \left(x \arctan x - \frac{1}{2}\ln(x^2 + 1)\right).$

Thus: $y(x) = e^x\left(x \arctan x - \frac{1}{2}\ln(x^2 + 1)\right).$

[Practice 3 · Hard] Consider the second-order linear non-homogeneous differential equation:

\frac{d^2y}{dx^2} + 4y = \sec(2x)

. Suppose a particular solution is found using the method of variation of parameters. Let the fundamental set of solutions for the associated homogeneous equation be

y_1(x) = \cos(2x)

and

y_2(x) = \sin(2x)

. If the particular solution is expressed in the form

y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)

, determine the value of

u_2'(x)

💡 Hint 1 of 3

The given differential equation is

\frac{d^2y}{dx^2} + 4y = \sec(2x)

. The associated homogeneous equation is

\frac{d^2y}{dx^2} + 4y = 0

. The characteristic equation is

r^2 + 4 = 0

, which has roots

r = \pm 2i

. Thus, the fundamental set of solutions for the homogeneous equation is

y_1(x) = \cos(2x)

and

y_2(x) = \sin(2x)

💡 Hint 2 of 3

We use the method of variation of parameters to find a particular solution of the form

y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)

. The Wronskian of

y_1

and

y_2

is:

W(y_1, y_2)(x) = y_1y_2' - y_1'y_2 = \cos(2x)(2\cos(2x)) - (-\sin(2x))(2\sin(2x))

💡 Hint 3 of 3

The value of

u_2'(x)

\frac{1}{2}

The given differential equation is

\frac{d^2y}{dx^2} + 4y = \sec(2x)

. The associated homogeneous equation is

\frac{d^2y}{dx^2} + 4y = 0

. The characteristic equation is

r^2 + 4 = 0

, which has roots

r = \pm 2i

. Thus, the fundamental set of solutions for the homogeneous equation is

y_1(x) = \cos(2x)

and

y_2(x) = \sin(2x)

. We use the method of variation of parameters to find a particular solution of the form

y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)

. The Wronskian of

y_1

and

y_2

is:

W(y_1, y_2)(x) = y_1y_2' - y_1'y_2 = \cos(2x)(2\cos(2x)) - (-\sin(2x))(2\sin(2x))

W(y_1, y_2)(x) = 2\cos^2(2x) + 2\sin^2(2x) = 2(\cos^2(2x) + \sin^2(2x)) = 2

. The non-homogeneous term is

g(x) = \sec(2x)

. The formulas for

u_1'(x)

and

u_2'(x)

are:

u_1'(x) = -\frac{y_2(x)g(x)}{W(x)}

u_2'(x) = \frac{y_1(x)g(x)}{W(x)}

Substituting the given functions:

u_2'(x) = \frac{\cos(2x) \cdot \sec(2x)}{2}

u_2'(x) = \frac{\cos(2x)}{2\cos(2x)}

u_2'(x) = \frac{1}{2}

. The value of

u_2'(x)

\frac{1}{2}

[Practice 4 · Easy] Find a particular solution of

y''-y=2

using variation of parameters.

💡 Hint 1 of 3

Set up the initial relation:

y_1=e^{x}

y_2=e^{-x}

W=-2

💡 Hint 2 of 3

Rewrite and simplify to obtain:

u_1'=-\dfrac{y_2(2)}{W}=e^{-x}\Rightarrow u_1=-e^{-x}

💡 Hint 3 of 3

Use the resulting form to finish:

u_2'=\dfrac{y_1(2)}{W}=-e^{x} \implies u_2=-e^{x}

y_1=e^{x}

y_2=e^{-x}

W=-2

u_1'=-\dfrac{y_2(2)}{W}=e^{-x}\Rightarrow u_1=-e^{-x}

;

u_2'=\dfrac{y_1(2)}{W}=-e^{x}\Rightarrow u_2=-e^{x}

. So

y_p=u_1y_1+u_2y_2=-e^{-x}e^{x}-e^{x}e^{-x}=-2

📝 Chapter Quiz

Questions Level

The Laplace Transform, Discontinuous Forcing & Convolution Theory

In electrical engineering, structural dynamics, and control theory, physical systems are often driven by inputs that are discontinuous, piecewise-defined, or instantaneous — such as switching on a power supply, applying a hammer blow to a structural beam, or tracking digital pulse sequences. Traditional calculus techniques struggle with non-smooth forcing terms. The Laplace transform acts as an operational calculus, converting differential operators into algebraic polynomials, incorporating initial conditions automatically, and handling Heaviside step functions, Dirac delta distributions, and convolution integrals with mathematical precision.

● Level 1 Foundation — Definition 6.1: The One-Sided Laplace Transform & Exponential Order

Let $f: [0, \infty) \to \mathbb{R}$ be a piecewise continuous function. The one-sided Laplace transform of $f(t)$ is the function $F(s) = \mathcal{L}\{f(t)\}$ defined by the improper integral: $F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\,dt,$ for all complex or real numbers $s$ for which the integral converges.

A function $f(t)$ is of exponential order $c$ if there exist constants $M > 0$ and $T \ge 0$ such that $|f(t)| \le M e^{c t}$ for all $t \ge T$ . If $f(t)$ is piecewise continuous on $[0, \infty)$ and of exponential order $c$ , the integral converges absolutely for all $\operatorname{Re}(s) > c$ , and $\lim_{s\to\infty} F(s) = 0$ .

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

⊙ Explorer · Laplace Transform: Heaviside Step Forcing

u(t - a)

& Delayed Response

interactive

Live numerical table for Laplace Transform: Heaviside Step Forcing $u(t - a)$ & Delayed Response
Parameter / Metric	Value

Step Switch-On Time $a =$ 2

Damping Ratio $\zeta =$ 0.25

Laplace Factor e^(−as)/

s

e^{-2.00s}/s

Peak Response Overshoot

y_{\text{max}} = 1.44

Steady-State Limit

y(\infty )

y(\infty ) = 1.00

By the Second Shifting Theorem ℒ⁻¹{e^(−as) $F(s)$ } = $u(t - a)f(t - a)$ , the system remains quiescent $(y = 0)$ until forcing $u(t - a)$ activates at $t = a$ .

▲ Level 1 Foundation — Theorem 6.1: Linearity, First Shift Theorem & Fundamental Transform Pairs

1. Linearity: For any constants $c_1, c_2$ : $\mathcal{L}\{c_1 f(t) + c_2 g(t)\} = c_1 \mathcal{L}\{f(t)\} + c_2 \mathcal{L}\{g(t)\}.$

2. First Shift Theorem (Frequency Shift): If $\mathcal{L}\{f(t)\} = F(s)$ , then for any constant $a \in \mathbb{R}$ : $\mathcal{L}\{e^{at}f(t)\} = F(s - a), \qquad \operatorname{Re}(s) > c + a.$

3. Elementary Transform Pairs:

$\mathcal{L}\{1\} = \frac{1}{s} \quad (s > 0)$
$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \quad (n \in \mathbb{N}, s > 0)$
$\mathcal{L}\{e^{at}\} = \frac{1}{s - a} \quad (s > a)$
$\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}, \quad \mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2} \quad (s > 0)$
$\mathcal{L}\{e^{at}\cos(\omega t)\} = \frac{s - a}{(s - a)^2 + \omega^2}, \quad \mathcal{L}\{e^{at}\sin(\omega t)\} = \frac{\omega}{(s - a)^2 + \omega^2}$

Proof (Linearity & First Shift Theorem)

Linearity follows directly from the linearity of integration: $\int_0^\infty e^{-st}\big(c_1 f(t) + c_2 g(t)\big)dt = c_1 \int_0^\infty e^{-st}f(t)\,dt + c_2 \int_0^\infty e^{-st}g(t)\,dt = c_1 F(s) + c_2 G(s).$

For the First Shift Theorem: $\mathcal{L}\{e^{at}f(t)\} = \int_0^\infty e^{-st}\big(e^{at}f(t)\big)dt = \int_0^\infty e^{-(s - a)t}f(t)\,dt = F(s - a).$

▲ Level 1 Foundation — Theorem 6.2: Operational Calculus for Derivatives & IVPs

Let $f(t)$ be continuously differentiable on $[0, \infty)$ with $f'(t)$ of exponential order. Then: $\mathcal{L}\{f'(t)\} = s \mathcal{L}\{f(t)\} - f(0) = s F(s) - f(0).$ Applying this result inductively to higher derivatives: $\mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0) - f'(0),$ $\mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \dots - f^{(n-1)}(0).$ Initial conditions at $t = 0$ are thus baked directly into the algebraic domain.

Proof (Derivative Transform via Integration by Parts)

Evaluating by integration by parts with $u = e^{-st}$ and $dv = f'(t)dt$ (so $du = -s e^{-st}dt$ and $v = f(t)$ ): $\mathcal{L}\{f'(t)\} = \int_0^\infty e^{-st}f'(t)\,dt = \left[e^{-st}f(t)\right]_0^\infty - \int_0^\infty (-s e^{-st})f(t)\,dt.$ Because $f(t)$ is of exponential order $c$ , for $\operatorname{Re}(s) > c$ , $\lim_{t\to\infty} e^{-st}f(t) = 0$ . At the lower limit $t = 0$ , $e^0 f(0) = f(0)$ . Hence: $\mathcal{L}\{f'(t)\} = 0 - f(0) + s \int_0^\infty e^{-st}f(t)\,dt = s F(s) - f(0).$ Replacing $f$ by $f'$ immediately yields $\mathcal{L}\{f''\} = s(s F(s) - f(0)) - f'(0) = s^2 F(s) - s f(0) - f'(0)$ .

Worked example 6.A — Solving a 2nd-Order IVP via Partial Fractions

Solve the initial value problem $y'' - 3y' + 2y = 4e^{3t}, \qquad y(0) = 1, \quad y'(0) = 0.$

Step 1: Apply Laplace transform: $\mathcal{L}\{y''\} - 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = 4\mathcal{L}\{e^{3t}\}$ $\big(s^2 Y(s) - s y(0) - y'(0)\big) - 3\big(s Y(s) - y(0)\big) + 2Y(s) = \frac{4}{s - 3}.$ Substituting $y(0) = 1, y'(0) = 0$ : $(s^2 - 3s + 2)Y(s) - s + 3 = \frac{4}{s - 3} \implies (s - 1)(s - 2)Y(s) = s - 3 + \frac{4}{s - 3}.$ $(s - 1)(s - 2)Y(s) = \frac{(s - 3)^2 + 4}{s - 3} = \frac{s^2 - 6s + 13}{s - 3} \implies Y(s) = \frac{s^2 - 6s + 13}{(s - 1)(s - 2)(s - 3)}.$

Step 2: Partial fraction decomposition: $\frac{s^2 - 6s + 13}{(s - 1)(s - 2)(s - 3)} = \frac{A}{s - 1} + \frac{B}{s - 2} + \frac{C}{s - 3}.$ Using Heaviside cover-up method:
$A = \frac{1^2 - 6(1) + 13}{(1 - 2)(1 - 3)} = \frac{8}{(-1)(-2)} = 4$ .
$B = \frac{2^2 - 6(2) + 13}{(2 - 1)(2 - 3)} = \frac{5}{(1)(-1)} = -5$ .
$C = \frac{3^2 - 6(3) + 13}{(3 - 1)(3 - 2)} = \frac{4}{(2)(1)} = 2$ .

Step 3: Inverse transform: $y(t) = \mathcal{L}^{-1}\left\{\frac{4}{s - 1} - \frac{5}{s - 2} + \frac{2}{s - 3}\right\} = 4e^t - 5e^{2t} + 2e^{3t}.$

● Level 2 Analytical Tools — Definition 6.2: Heaviside Step Function, Shift Theorems & Dirac Delta

1. Heaviside Unit Step Function: Defined by \[ u_a(t) = u(t - a) = \begin{cases} 0 & t < a \\ 1 & t \ge a \end{cases} \qquad (a \ge 0). \] Its Laplace transform is $ \mathcal{L}\{u(t - a)\} = \int_a^\infty e^{-st}\,dt = \frac{e^{-as}}{s} $.

2. Second Shift Theorem (Time Shift): If $\mathcal{L}\{f(t)\} = F(s)$ , then: $\mathcal{L}\{u(t - a)f(t - a)\} = e^{-as}F(s), \qquad \mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t - a)f(t - a).$

3. Dirac Delta Distribution (Unit Impulse): The idealized impulse $\delta(t - a)$ represents a pulse of infinite intensity and zero duration concentrated at $t = a$ with total area 1: $\int_0^\infty f(t)\delta(t - a)\,dt = f(a) \implies \mathcal{L}\{\delta(t - a)\} = e^{-as}, \quad \mathcal{L}\{\delta(t)\} = 1.$

Operational Properties of the Laplace Transform

First Shift: e^(at) f(t)

\mathcal{L}\{e^{at}f(t)\} = F(s - a)

. Replaces every

s

F(s)

with

s - a

Second Shift: u(t - a) f(t - a)

\mathcal{L}\{u(t - a)f(t - a)\} = e^{-as}F(s)

. Incorporates time delays and piecewise switching.

Derivative of Transform: t^n f(t)

\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}F(s)

. Multiplication by

t

corresponds to frequency differentiation.

Integration: int_0^t f(tau) dtau

\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\} = \frac{F(s)}{s}

. Integration in time corresponds to division by

s

Worked example 6.B — Discontinuous Pulse & Impact Forcing

Solve the oscillator IVP driven by a finite rectangular pulse of duration 2: $y'' + 4y = u(t - 1) - u(t - 3), \qquad y(0) = 0, \quad y'(0) = 0.$

Step 1: Transform equation: $(s^2 + 4)Y(s) = \frac{e^{-s} - e^{-3s}}{s} \implies Y(s) = (e^{-s} - e^{-3s}) \cdot \frac{1}{s(s^2 + 4)}.$

Step 2: Partial fractions on transfer factor: $\frac{1}{s(s^2 + 4)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 4} = \frac{1/4}{s} - \frac{s/4}{s^2 + 4}.$ Inverse transform: $h(t) = \mathcal{L}^{-1}\left\{\frac{1}{4}\left(\frac{1}{s} - \frac{s}{s^2 + 4}\right)\right\} = \frac{1}{4}\big(1 - \cos(2t)\big).$

Step 3: Apply Second Shift Theorem: $y(t) = u(t - 1)h(t - 1) - u(t - 3)h(t - 3)$ $= \frac{1}{4}u(t - 1)\big(1 - \cos(2(t - 1))\big) - \frac{1}{4}u(t - 3)\big(1 - \cos(2(t - 3))\big).$

▲ Level 3 University Rigor — Theorem 6.3: The Convolution Theorem & Volterra Integral Equations

1. Convolution Definition: The convolution of two functions $f, g: [0, \infty) \to \mathbb{R}$ is defined by: $(f * g)(t) = \int_0^t f(\tau)g(t - \tau)\,d\tau = \int_0^t f(t - \tau)g(\tau)\,d\tau = (g * f)(t).$

2. Convolution Theorem: If $F(s) = \mathcal{L}\{f(t)\}$ and $G(s) = \mathcal{L}\{g(t)\}$ exist for $s > c$ , then: $\mathcal{L}\{(f * g)(t)\} = F(s)\,G(s) \iff \mathcal{L}^{-1}\{F(s)G(s)\} = (f * g)(t).$

Proof: By definition of the double integral: $\mathcal{L}\{(f * g)(t)\} = \int_0^\infty e^{-st}\left(\int_0^t f(\tau)g(t - \tau)\,d\tau\right)dt = \int_0^\infty \int_0^t e^{-st}f(\tau)g(t - \tau)\,d\tau\,dt.$ The region of integration is the triangular wedge $ \{0 \le \tau \le t < \infty\} $. Reversing the order of integration via Fubini's Theorem: \[ \int_0^\infty f(\tau) \left(\int_\tau^\infty e^{-st}g(t - \tau)\,dt\right)d\tau. \] Substitute $ u = t - \tau $ (so $ t = u + \tau $ and $ dt = du $): \[ \int_0^\infty f(\tau) \left(\int_0^\infty e^{-s(u + \tau)}g(u)\,du\right)d\tau = \left(\int_0^\infty e^{-s\tau}f(\tau)\,d\tau\right)\left(\int_0^\infty e^{-su}g(u)\,du\right) = F(s)G(s). \]

3. Volterra Integral Equations: An equation of the form $y(t) = f(t) + \int_0^t K(t - \tau)y(\tau)\,d\tau$ is recognized as $y = f + K * y$ . Applying the Laplace transform yields $Y(s) = F(s) + K(s)Y(s) \implies Y(s) = \frac{F(s)}{1 - K(s)}$ , solving the integral equation algebraically.

The kernel $e^{-st}$ acts as an exponential damping filter that suppresses functions of exponential order, ensuring convergence of the improper integral over $[0, \infty)$ , as illustrated below for varying parameter values $s$ .

Exponential decay of the Laplace kernel

e^{-st}

for

s = 1, 2, 3

. Higher values of

s

induce faster damping, guaranteeing absolute convergence for all functions of exponential order $ c < s $.

⚠️ Common Traps & University Pitfalls — Derivative Transform Initial Conditions

Sign of initial values in derivative transforms: Remember that $\mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0)$ . Both initial condition terms are SUBTRACTED. When $y(0) = -3$ , $-s(-3) = +3s$ . A sign error here ruins the entire partial fraction decomposition.

Second Shift unshifted arguments: $\mathcal{L}\{u(t - a)g(t)\} \neq e^{-as}\mathcal{L}\{g(t)\}$ ! You must write $g(t) = f(t - a)$ before transforming: $\mathcal{L}\{u(t - a)g(t)\} = e^{-as}\mathcal{L}\{g(t + a)\}$ . For example, $\mathcal{L}\{u(t - 2)t\} = e^{-2s}\mathcal{L}\{t + 2\} = e^{-2s}\left(\frac{1}{s^2} + \frac{2}{s}\right)$ , NOT $\frac{e^{-2s}}{s^2}$ .

Pointwise multiplication fallacy: $\mathcal{L}^{-1}\{F(s)G(s)\} \neq f(t)g(t)$ . The inverse of a product of transforms is always the convolution $(f * g)(t) = \int_0^t f(\tau)g(t - \tau)\,d\tau$ .

Worked Example

C

— The Laplace Transform, Discontinuous Forcing & Convolution Theory (Hard Application)

Problem (Hard): Consider the differential equation $y''(t) + 4y(t) = u(t-2)$ , where $u(t)$ is the Heaviside step function, with initial conditions $y(0) = 0$ and $y'(0) = 0$ . Apply the Laplace Transform to both sides of the equation. If $Y(s) = \mathcal{L}\{y(t)\}$ and $\mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s}$ , which of the following expressions represents $Y(s)$ ?

Step-by-Step Solution: Applying the Laplace Transform to the given differential equation: $\mathcal{L}\{y''(t)\} + 4\mathcal{L}\{y(t)\} = \mathcal{L}\{u(t-2)\}$ . Using the properties of the Laplace Transform and the given initial conditions $y(0) = 0$ and $y'(0) = 0$ , we have $\mathcal{L}\{y''(t)\} = s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s)$ . The Laplace Transform of $y(t)$ is $Y(s)$ . The Laplace Transform of the Heaviside step function $u(t-2)$ is $\frac{e^{-2s}}{s}$ . Substituting these into the transformed equation gives: $s^2 Y(s) + 4 Y(s) = \frac{e^{-2s}}{s}$ . Factoring out $Y(s)$ , we get $Y(s)(s^2 + 4) = \frac{e^{-2s}}{s}$ . Solving for $Y(s)$ , we obtain $Y(s) = \frac{e^{-2s}}{s(s^2 + 4)}$ .

Worked Example

D

— The Laplace Transform, Discontinuous Forcing & Convolution Theory (Easy Application)

Problem (Easy): The second-order linear homogeneous ODE $f''(x) - 39f'(x) + 350f(x) = 0$ has characteristic roots $\lambda_1, \lambda_2$ . Determine the trace sum $\lambda_1 + \lambda_2$ .

Step-by-Step Solution: From $\lambda^2 - 39\lambda + 350 = 0$ , the sum of roots is $39$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (The Laplace Transform, Discontinuous Forcing & Convolution Theory)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — The Laplace Transform, Discontinuous Forcing & Convolution Theory

Solve the initial value problem with Dirac delta forcing:

y'' + 4y = \delta(t - \pi), \qquad y(0) = 1, \quad y'(0) = 0.

💡 Hint 1 of 3

Transform equation:

(s^2 + 4)Y(s) - s(1) - 0 = \mathcal{L}\{\delta(t - \pi)\} = e^{-\pi s} \implies (s^2 + 4)Y(s) = s + e^{-\pi s}

💡 Hint 2 of 3

Rewrite and simplify to obtain:

Y(s) = \frac{s}{s^2 + 4} + e^{-\pi s} \cdot \frac{1}{s^2 + 4} = \frac{s}{s^2 + 4} + \frac{1}{2} e^{-\pi s} \cdot \frac{2}{s^2 + 4}

Inverse transform:

y(t) = \cos(2t) + \frac{1}{2}u(t - \pi)\sin\big(2(t - \pi)\big)

💡 Hint 3 of 3

Physical interpretation: For $0 \le t < \pi$, the system oscillates as $\cos(2t)$. At $t = \pi$, the delta impulse imparts an instantaneous velocity jump of $\Delta y' = 1$, altering the amplitude and phase for $t \ge \pi$.

Step 1: Transform equation: $(s^2 + 4)Y(s) - s(1) - 0 = \mathcal{L}\{\delta(t - \pi)\} = e^{-\pi s} \implies (s^2 + 4)Y(s) = s + e^{-\pi s}.$ $Y(s) = \frac{s}{s^2 + 4} + e^{-\pi s} \cdot \frac{1}{s^2 + 4} = \frac{s}{s^2 + 4} + \frac{1}{2} e^{-\pi s} \cdot \frac{2}{s^2 + 4}.$

Step 2: Inverse transform: $y(t) = \cos(2t) + \frac{1}{2}u(t - \pi)\sin\big(2(t - \pi)\big).$ Using $\sin(2t - 2\pi) = \sin(2t)$ : $y(t) = \cos(2t) + \frac{1}{2}u(t - \pi)\sin(2t).$

Physical interpretation: For $ 0 \le t < \pi $, the system oscillates as $ \cos(2t) $. At $ t = \pi $, the delta impulse imparts an instantaneous velocity jump of $ \Delta y' = 1 $, altering the amplitude and phase for $ t \ge \pi $.

Use the Laplace transform and the Convolution Theorem to solve the Volterra integral equation

y(t) = t + \int_0^t \sin(t - \tau)y(\tau)\,d\tau.

💡 Hint 1 of 3

Identify convolution: The equation is

y(t) = t + \sin(t) * y(t)

💡 Hint 2 of 3

Solve for $Y(s)$ :

Y(s)\left(1 - \frac{1}{s^2 + 1}\right) = \frac{1}{s^2} \implies Y(s)\left(\frac{s^2}{s^2 + 1}\right) = \frac{1}{s^2} \implies Y(s) = \frac{s^2 + 1}{s^4} = \frac{1}{s^2} + \frac{1}{s^4}

💡 Hint 3 of 3

Inverse transform:

y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2} + \frac{1}{6} \cdot \frac{3!}{s^4}\right\}

Step 1: Identify convolution: The equation is $y(t) = t + \sin(t) * y(t)$ .

Step 2: Transform both sides: $Y(s) = \frac{1}{s^2} + \left(\frac{1}{s^2 + 1}\right)Y(s).$

Step 3: Solve for $Y(s)$ : $Y(s)\left(1 - \frac{1}{s^2 + 1}\right) = \frac{1}{s^2} \implies Y(s)\left(\frac{s^2}{s^2 + 1}\right) = \frac{1}{s^2} \implies Y(s) = \frac{s^2 + 1}{s^4} = \frac{1}{s^2} + \frac{1}{s^4}.$

Step 4: Inverse transform: $y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2} + \frac{1}{6} \cdot \frac{3!}{s^4}\right\} = t + \frac{1}{6}t^3.$

[Practice 3 · Medium] Consider the initial value problem

y'' + 4y = \delta(t- \pi)

with

y(0) = 0

and

y'(0) = 0

. Which of the following represents the Laplace Transform of the solution,

Y(s) = \mathcal{L}\{y(t)\}

💡 Hint 1 of 3

Taking the Laplace transform of the differential equation, we have

s^2 Y(s) - sy(0) - y'(0) + 4Y(s) = \mathcal{L}\{\delta(t-\pi)\}

💡 Hint 2 of 3

Given the initial conditions

y(0)=0

and

y'(0)=0

, this simplifies to

s^2 Y(s) + 4Y(s) = e^{-\pi s}

. Factoring out

Y(s)

, we get

Y(s)(s^2 + 4) = e^{-\pi s}

💡 Hint 3 of 3

Therefore, the Laplace transform of the solution is

Y(s) = \frac{e^{-\pi s}}{s^2 + 4}

Taking the Laplace transform of the differential equation, we have

s^2 Y(s) - sy(0) - y'(0) + 4Y(s) = \mathcal{L}\{\delta(t-\pi)\}

. Given the initial conditions

y(0)=0

and

y'(0)=0

, this simplifies to

s^2 Y(s) + 4Y(s) = e^{-\pi s}

. Factoring out

Y(s)

, we get

Y(s)(s^2 + 4) = e^{-\pi s}

. Therefore, the Laplace transform of the solution is

Y(s) = \frac{e^{-\pi s}}{s^2 + 4}

[Practice 4 · Hard] Solve

y''-12y'+35y=0

y(0)=-1

y'(0)=-13

. Find

y(2)

💡 Hint 1 of 3

Auxiliary equation

r^2-12r+35=0

has roots

r=5,7

, so

y=C_1e^{5x}+C_2e^{7x}

💡 Hint 2 of 3

Solving

y(0)=C_1+C_2=-1

and

y'(0)=5C_1+7C_2=-13

gives

C_1=3

C_2=-4

💡 Hint 3 of 3

y=3e^{5x}-4e^{7x}

, and

y(2)

follows by substitution.

Auxiliary equation

r^2-12r+35=0

has roots

r=5,7

, so

y=C_1e^{5x}+C_2e^{7x}

. Solving

y(0)=C_1+C_2=-1

and

y'(0)=5C_1+7C_2=-13

gives

C_1=3

C_2=-4

. So

y=3e^{5x}-4e^{7x}

, and

y(2)

follows by substitution.

📝 Chapter Quiz

Questions Level

Systems of Linear Differential Equations, Phase Plane Topology & Matrix Exponentials

Most realistic mechanical, ecological, and chemical systems consist of interconnected components whose states evolve simultaneously — such as predator-prey populations, multi-loop electrical networks, coupled harmonic oscillators, or satellite orbital attitudes. These multi-dimensional dynamics are modeled by systems of linear differential equations $\mathbf{x}' = A\mathbf{x}$ . This chapter bridges linear algebra and dynamical systems, exploring the eigenvalue-eigenvector method, geometric phase portraits, the trace-determinant classification, and the matrix exponential $e^{At}$ .

● Level 1 Foundation — Definition 7.1: First-Order Linear Systems & Phase Space

A system of two first-order linear homogeneous differential equations with constant coefficients is written in matrix-vector form: $\frac{d\mathbf{x}}{dt} = A\mathbf{x}, \qquad \mathbf{x}(t) = \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix}, \quad A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{R}^{2 \times 2}.$ The $x_1 x_2$ -plane is called the phase plane. At each point $\mathbf{x}$ , the matrix product $A\mathbf{x}$ defines a velocity vector $\mathbf{x}'$ , establishing a continuous direction field. A curve $\mathbf{x}(t)$ tangent to this field at every instant is a trajectory (orbit), and the complete geometric collection of trajectories is the phase portrait.

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

▲ Level 1 Foundation — Theorem 7.1: Eigenpairs & Fundamental Solution Sets

Substituting the exponential ansatz $\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$ into $\mathbf{x}' = A\mathbf{x}$ yields: $\lambda e^{\lambda t}\mathbf{v} = A(e^{\lambda t}\mathbf{v}) = e^{\lambda t} A\mathbf{v} \iff A\mathbf{v} = \lambda \mathbf{v}.$ Thus $\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$ is a non-trivial solution if and only if $\lambda$ is an eigenvalue of $A$ with associated non-zero eigenvector $\mathbf{v}$ .

If $A$ has two distinct real eigenvalues $\lambda_1 \neq \lambda_2$ with linearly independent eigenvectors $\mathbf{v}_1, \mathbf{v}_2$ , the general solution is: $\mathbf{x}(t) = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2.$

Proof

Since $A\mathbf{v}_1 = \lambda_1 \mathbf{v}_1$ , $\frac{d}{dt}[e^{\lambda_1 t}\mathbf{v}_1] = \lambda_1 e^{\lambda_1 t}\mathbf{v}_1 = e^{\lambda_1 t} A\mathbf{v}_1 = A(e^{\lambda_1 t}\mathbf{v}_1)$ , satisfying the system. Linear combinations satisfy the system by linearity of matrix multiplication: $\frac{d}{dt}[C_1 \mathbf{x}_1 + C_2 \mathbf{x}_2] = C_1 \mathbf{x}_1' + C_2 \mathbf{x}_2' = C_1 A\mathbf{x}_1 + C_2 A\mathbf{x}_2 = A(C_1 \mathbf{x}_1 + C_2 \mathbf{x}_2).$ Because $\lambda_1 \neq \lambda_2$ , their eigenvectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent in $\mathbb{R}^2$ . The Wronskian $W(t) = \det[e^{\lambda_1 t}\mathbf{v}_1 \quad e^{\lambda_2 t}\mathbf{v}_2] = e^{(\lambda_1 + \lambda_2)t}\det[\mathbf{v}_1 \quad \mathbf{v}_2] \neq 0$ , proving that every initial vector $\mathbf{x}(0) = \mathbf{x}_0$ can be uniquely represented.

▲ Level 2 Analytical Tools — Theorem 7.2: Trace-Determinant Plane & Phase Portrait Topology

The eigenvalues of $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ are the roots of the characteristic equation $\det(A - \lambda I) = \lambda^2 - \tau \lambda + \Delta = 0,$ where $\tau = \operatorname{tr}(A) = a + d$ and $\Delta = \det(A) = ad - bc$ . The discriminant is $D = \tau^2 - 4\Delta$ . The topological classification of the equilibrium point at the origin $(0, 0)$ is completely determined by $(\tau, \Delta)$ :

Saddle Point ($ \Delta < 0 $): Eigenvalues are real with opposite signs ( $\lambda_1 > 0 > \lambda_2$ ). Always unstable. Trajectories approach along the eigenvector direction of $ \lambda_2 < 0 $ and escape along the direction of $ \lambda_1 > 0 $.
Nodal Sink / Stable Node ($ \Delta > 0, D > 0, \tau < 0 $): Real negative eigenvalues ($ \lambda_2 < \lambda_1 < 0 $). All trajectories converge to origin tangentially to $ \mathbf{v}_1 $. Asymptotically stable.
Nodal Source / Unstable Node ( $\Delta > 0, D > 0, \tau > 0$ ): Real positive eigenvalues ( $\lambda_1 > \lambda_2 > 0$ ). All trajectories diverge to infinity. Unstable.
Spiral Sink ($ \Delta > 0, D < 0, \tau < 0 $): Complex roots $\lambda = \alpha \pm i\beta$ with $ \alpha < 0 $. Trajectories spiral inward to origin. Asymptotically stable.
Spiral Source ($ \Delta > 0, D < 0, \tau > 0 $): Complex roots with $\alpha > 0$ . Trajectories spiral outward. Unstable.
Center ($ \Delta > 0, D < 0, \tau = 0 $): Purely imaginary roots $\lambda = \pm i\beta$ . Trajectories form closed concentric ellipses. Neutrally stable.

Worked example 7.A — Saddle Point & Stable/Unstable Manifolds

Find the general solution and classify the phase portrait for $\mathbf{x}' = \begin{pmatrix} 1 & 1 \\ 4 & -2 \end{pmatrix}\mathbf{x}.$

Step 1: Characteristic polynomial: $\det(A - \lambda I) = \det\begin{pmatrix} 1 - \lambda & 1 \\ 4 & -2 - \lambda \end{pmatrix} = (1 - \lambda)(-2 - \lambda) - 4 = \lambda^2 + \lambda - 6 = (\lambda - 2)(\lambda + 3) = 0.$ Eigenvalues: $\lambda_1 = 2$ (positive), $\lambda_2 = -3$ (negative). Since $ \Delta = -6 < 0 $, the origin is a saddle point.

Step 2: Eigenvectors:
For $\lambda_1 = 2$ : $(A - 2I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 4 & -4 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0} \implies -v_1 + v_2 = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$
For $\lambda_2 = -3$ : $(A + 3I)\mathbf{v} = \begin{pmatrix} 4 & 1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0} \implies 4v_1 + v_2 = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -4 \end{pmatrix}.$

Step 3: General solution: $\mathbf{x}(t) = C_1 e^{2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-3t}\begin{pmatrix} 1 \\ -4 \end{pmatrix}.$ The line spanned by $(1, -4)^T$ is the stable manifold (inflowing as $t \to \infty$ ); the line spanned by $(1, 1)^T$ is the unstable manifold (outflowing as $t \to \infty$ ).

Worked example 7.B — Complex Conjugate Eigenvalues & Spiral Sink

Solve $\mathbf{x}' = \begin{pmatrix} -1 & -4 \\ 1 & -1 \end{pmatrix}\mathbf{x}$ , and verify the rotational direction of trajectories.

Step 1: Characteristic polynomial: $\det\begin{pmatrix} -1 - \lambda & -4 \\ 1 & -1 - \lambda \end{pmatrix} = (\lambda + 1)^2 + 4 = 0 \implies \lambda = -1 \pm 2i.$ Damping $ \alpha = -1 < 0 \implies $ asymptotically stable spiral sink.

Step 2: Complex eigenvector for $\lambda = -1 + 2i$ : $(A - \lambda I)\mathbf{v} = \begin{pmatrix} -2i & -4 \\ 1 & -2i \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0} \implies v_1 - 2i v_2 = 0 \implies \mathbf{v} = \begin{pmatrix} 2i \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} + i \begin{pmatrix} 2 \\ 0 \end{pmatrix}.$

Step 3: Decompose into real and imaginary parts: $e^{\lambda t}\mathbf{v} = e^{-t}(\cos 2t + i\sin 2t)\left[\begin{pmatrix} 0 \\ 1 \end{pmatrix} + i\begin{pmatrix} 2 \\ 0 \end{pmatrix}\right]$ $= e^{-t}\left[\cos 2t \begin{pmatrix} 0 \\ 1 \end{pmatrix} - \sin 2t \begin{pmatrix} 2 \\ 0 \end{pmatrix}\right] + i e^{-t}\left[\sin 2t \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \cos 2t \begin{pmatrix} 2 \\ 0 \end{pmatrix}\right].$ Fundamental real solutions: $\mathbf{x}_1(t) = e^{-t}\begin{pmatrix} -2\sin 2t \\ \cos 2t \end{pmatrix}, \qquad \mathbf{x}_2(t) = e^{-t}\begin{pmatrix} 2\cos 2t \\ \sin 2t \end{pmatrix}.$ $\mathbf{x}(t) = C_1 e^{-t}\begin{pmatrix} -2\sin 2t \\ \cos 2t \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} 2\cos 2t \\ \sin 2t \end{pmatrix}.$

Step 4: Rotational sense: Test the velocity vector at $(1, 0)$ : $\mathbf{x}' = A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}.$ Since $x_2' = +1 > 0$ , motion is directed upward into the second quadrant, establishing counterclockwise inward spiraling.

⊙ Explorer · Phase portrait of

x'

= Ax

interactive

$a$ : 1

$b$ : 1

$c$ : 4

$d$ : -2

eigenvalues

2.000, -3.000

classificationSaddle

Default $a = 1$ , $b = 1$ , $c = 4$ , $d = -2$ is Worked example 7.A's saddle. The two amber curves are trajectories launched from $(1.5, 0.5)$ and $(-1.5, -0.5)$ , integrated forward with a small Runge-Kutta step; the grey arrows are the direction field ẋ = Ax itself. Try $a = -1$ , $b = 1$ , $c = -1$ , $d = -1$ (Worked example 7.B's spiral) or $a = -1$ , $b = 0$ , $c = 0$ , $d = -1$ (a node — both eigenvalues -1) to see the classification table's other rows.

▲ Level 3 University Rigor — Theorem 7.3: The Matrix Exponential e^(At), Defective Systems & Generalized Eigenvectors

1. The Matrix Exponential: For any $n \times n$ matrix $A$ , the matrix exponential is defined by the absolutely convergent power series: $e^{At} = \sum_{k=0}^\infty \frac{t^k}{k!} A^k = I + t A + \frac{t^2}{2!} A^2 + \dots$ Differentiating term-by-term yields $\frac{d}{dt} e^{At} = A e^{At}$ . Consequently, the unique solution to the IVP $\mathbf{x}' = A\mathbf{x}, \mathbf{x}(0) = \mathbf{x}_0$ is: $\mathbf{x}(t) = e^{At}\mathbf{x}_0.$

2. Cayley-Hamilton Construction: By the Cayley-Hamilton Theorem, $A$ satisfies its own characteristic equation: $A^2 - \tau A + \Delta I = 0$ . Thus higher powers $A^k$ reduce linearly to $A$ and $I$ , enabling exact evaluation of $e^{At} = a_0(t)I + a_1(t)A$ .

3. Defective Systems & Generalized Eigenvectors: When $A$ has a repeated eigenvalue $\lambda$ with only one linearly independent eigenvector $\mathbf{v}_1$ (algebraic multiplicity 2, geometric multiplicity 1), $A$ is defective. A second linearly independent solution requires a generalized eigenvector $\mathbf{v}_2$ satisfying: $(A - \lambda I)\mathbf{v}_2 = \mathbf{v}_1.$ The general solution is then given exactly by: $\mathbf{x}(t) = C_1 e^{\lambda t}\mathbf{v}_1 + C_2 e^{\lambda t}\big(\mathbf{v}_2 + t\mathbf{v}_1\big).$

In a saddle equilibrium, trajectories approach along the stable eigenspace and deflect hyperbolically away along the unstable eigenspace, forming distinct topological sectors separated by straight-line separatrices, as shown below.

Hyperbolic trajectories of a saddle point. The stable manifold pulls state trajectories inward toward the origin, while the unstable manifold drives trajectories outward to infinity.

⚠️ Common Traps & University Pitfalls — Defective Eigenvalues & Spiral Direction

Incorrect defective ansatz: For a repeated eigenvalue, students often mistakenly write $C_1 e^{\lambda t}\mathbf{v}_1 + C_2 t e^{\lambda t}\mathbf{v}_1$ . Substituting $t e^{\lambda t}\mathbf{v}_1$ gives $\mathbf{x}' = e^{\lambda t}\mathbf{v}_1 + \lambda t e^{\lambda t}\mathbf{v}_1$ , but $A\mathbf{x} = \lambda t e^{\lambda t}\mathbf{v}_1$ — leaving an unbalanced $e^{\lambda t}\mathbf{v}_1$ ! The correct second term is $e^{\lambda t}(\mathbf{v}_2 + t\mathbf{v}_1)$ where $(A - \lambda I)\mathbf{v}_2 = \mathbf{v}_1$ .

Spiral orientation test: Eigenvalues $\alpha \pm i\beta$ tell you stability and frequency, but CANNOT tell you whether the spiral turns clockwise or counterclockwise. Always test the velocity vector at $(1, 0)$ : compute $A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ c \end{pmatrix}$ . If $c > 0$ , rotation is counterclockwise; if $ c < 0 $, rotation is clockwise.

Worked Example

C

— Systems of Linear Differential Equations, Phase Plane Topology & Matrix Exponentials (Medium Application)

Problem (Medium): Consider the system of linear differential equations given by: $\qquad \begin{cases} \dfrac{dx}{dt} = -x + 2y \\ \dfrac{dy}{dt} = 3x - 2y \end{cases}$ Let the general solution be of the form $\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t}$ , where $ \lambda_1 < \lambda_2 $. Determine the value of $ \lambda_2 $.

Step-by-Step Solution: The given system can be written in matrix form as $\dfrac{d\mathbf{x}}{dt} = A\mathbf{x}$ , where $A = \begin{pmatrix} -1 & 2 \\ 3 & -2 \end{pmatrix}$ . To find the eigenvalues $\lambda$ , we solve the characteristic equation $\det(A - \lambda I) = 0$ . $\det \begin{pmatrix} -1-\lambda & 2 \\ 3 & -2-\lambda \end{pmatrix} = 0$ $(-1-\lambda)(-2-\lambda) - (2)(3) = 0$ $(1+\lambda)(2+\lambda) - 6 = 0$ $2 + \lambda + 2\lambda + \lambda^2 - 6 = 0$ $\lambda^2 + 3\lambda - 4 = 0$ Factoring the quadratic equation: $(\lambda + 4)(\lambda - 1) = 0$ The eigenvalues are $\lambda_1 = -4$ and $\lambda_2 = 1$ . Since the problem states $ \lambda_1 < \lambda_2 $, we have $ \lambda_1 = -4 $ and $ \lambda_2 = 1 $. The question asks for $ \lambda_2 $.

Worked Example

D

— Systems of Linear Differential Equations, Phase Plane Topology & Matrix Exponentials (Hard Application)

Problem (Hard): Taking the Laplace transform of $y'+2y=2 e^{t}$ , $y(0)=-3$ , gives $Y(s)=$

Step-by-Step Solution: $sY-(-3)+2Y=\frac{2}{s - 1}$ , so $Y=\frac{5 - 3 s}{\left(s - 1\right) \left(s + 2\right)}$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Systems of Linear Differential Equations, Phase Plane Topology & Matrix Exponentials)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — Systems of Linear Differential Equations, Phase Plane Topology & Matrix Exponentials

Solve the initial value problem

\mathbf{x}' = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}\mathbf{x}, \qquad \mathbf{x}(0) = \begin{pmatrix} 3 \\ 1 \end{pmatrix}.

💡 Hint 1 of 3

Eigenvalues:

\det\begin{pmatrix} 3 - \lambda & -2 \\ 2 & -2 - \lambda \end{pmatrix} = (3 - \lambda)(-2 - \lambda) + 4 = \lambda^2 - \lambda - 2 = (\lambda - 2)(\lambda + 1)

💡 Hint 2 of 3

Rewrite and simplify to obtain:

\lambda_1 = 2, \qquad \lambda_2 = -1

Eigenvectors: For

\lambda_1 = 2

\begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_1 - 2v_2 = 0 \implies \mathbf{v}_1

💡 Hint 3 of 3

Particular solution:

x(t) = \frac{5}{3}e^{2t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} - \frac{7}{3}e^{-t}\begin{pmatrix} 1 \\ 2 \end{pmatrix}

Step 1: Eigenvalues: $\det\begin{pmatrix} 3 - \lambda & -2 \\ 2 & -2 - \lambda \end{pmatrix} = (3 - \lambda)(-2 - \lambda) + 4 = \lambda^2 - \lambda - 2 = (\lambda - 2)(\lambda + 1) = 0.$ $\lambda_1 = 2, \qquad \lambda_2 = -1.$

Step 2: Eigenvectors:
For $\lambda_1 = 2$ : $\begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_1 - 2v_2 = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ .
For $\lambda_2 = -1$ : $\begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies 2v_1 - v_2 = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ .

Step 3: Apply initial condition $\mathbf{x}(0) = (3, 1)^T$ : $C_1 \begin{pmatrix} 2 \\ 1 \end{pmatrix} + C_2 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} \implies \begin{cases} 2C_1 + C_2 = 3 \\ C_1 + 2C_2 = 1 \end{cases}$ Multiplying second equation by 2 and subtracting: $3C_1 = 5 \implies C_1 = 5/3$ , $C_2 = 1 - 2(5/3) = -7/3$ .

Step 4: Particular solution: $\mathbf{x}(t) = \frac{5}{3}e^{2t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} - \frac{7}{3}e^{-t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{10}{3}e^{2t} - \frac{7}{3}e^{-t} \\ \frac{5}{3}e^{2t} - \frac{14}{3}e^{-t} \end{pmatrix}.$

Solve the defective linear system

\mathbf{x}' = \begin{pmatrix} 4 & 1 \\ -1 & 2 \end{pmatrix}\mathbf{x}.

💡 Hint 1 of 3

Eigenvalues:

\det\begin{pmatrix} 4 - \lambda & 1 \\ -1 & 2 - \lambda \end{pmatrix} = (4 - \lambda)(2 - \lambda) + 1 = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 = 0 \implies \lambda

💡 Hint 2 of 3

Generalized eigenvector: Solve

(A - 3I)\mathbf{v}_2 = \mathbf{v}_1

\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}\begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \implies v_{21} + v_{22}

💡 Hint 3 of 3

General solution:

x(t) = C_1 e^{3t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 e^{3t}\left[\begin{pmatrix} 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \end{pmatrix}\right]

Step 1: Eigenvalues: $\det\begin{pmatrix} 4 - \lambda & 1 \\ -1 & 2 - \lambda \end{pmatrix} = (4 - \lambda)(2 - \lambda) + 1 = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 = 0 \implies \lambda = 3 \quad \text{(multiplicity 2)}.$

Step 2: Eigenvector: $(A - 3I)\mathbf{v}_1 = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}\begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \mathbf{0} \implies v_{11} + v_{12} = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.$ Only one independent eigenvector exists (geometric multiplicity 1).

Step 3: Generalized eigenvector: Solve $(A - 3I)\mathbf{v}_2 = \mathbf{v}_1$ : $\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}\begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \implies v_{21} + v_{22} = 1.$ Choosing $v_{22} = 0 \implies v_{21} = 1$ , so $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ .

Step 4: General solution: $\mathbf{x}(t) = C_1 e^{3t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 e^{3t}\left[\begin{pmatrix} 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \end{pmatrix}\right] = e^{3t}\begin{pmatrix} C_1 + C_2(1 + t) \\ -C_1 - C_2 t \end{pmatrix}.$

[Practice 3 · Easy] Consider the system of linear differential equations:

\begin{aligned} \frac{dx}{dt} &= ax + by \\ \frac{dy}{dt} &= cx + dy \end{aligned}

where

a, b, c, d

are constants. The characteristic equation for the system is

\lambda^2 - (a+d)\lambda + (ad-bc) = 0

. We are given that the eigenvalues of the system are

\lambda_1 = 2

and

\lambda_2 = -1

. The trace of the coefficient matrix,

T = a+d

, is

1

. Determine the value of the determinant of the coefficient matrix,

D = ad-bc

💡 Hint 1 of 3

The characteristic equation of a 2x2 system of linear differential equations

\frac{dx}{dt} = ax + by

💡 Hint 2 of 3

Rewrite and simplify to obtain:

\frac{dy}{dt} = cx + dy

is given by

\lambda^2 - (a+d)\lambda + (ad-bc) = 0

. This can be written in terms of the trace (T = a+d) and determinant (D = ad-bc) of the coefficient matrix as

\lambda^2 - T\lambda + D = 0

💡 Hint 3 of 3

Using the product of the eigenvalues:

D = \lambda_1 \lambda_2 = (2)(-1)

The characteristic equation of a 2x2 system of linear differential equations

\begin{aligned} \frac{dx}{dt} &= ax + by \\ \frac{dy}{dt} &= cx + dy \end{aligned}

is given by

\lambda^2 - (a+d)\lambda + (ad-bc) = 0

. This can be written in terms of the trace (T = a+d) and determinant (D = ad-bc) of the coefficient matrix as

\lambda^2 - T\lambda + D = 0

. We are given that the eigenvalues are

\lambda_1 = 2

and

\lambda_2 = -1

. The sum of the roots of a quadratic equation

p\lambda^2 + q\lambda + r = 0

-q/p

, and the product of the roots is

r/p

. For the characteristic equation

\lambda^2 - T\lambda + D = 0

, the sum of the eigenvalues is

\lambda_1 + \lambda_2 = -(-T)/1 = T

, and the product of the eigenvalues is

\lambda_1 \lambda_2 = D/1 = D

. We are given that the trace

T = a+d = 1

. Let's verify this with the given eigenvalues:

\lambda_1 + \lambda_2 = 2 + (-1) = 1

, which matches the given trace. We need to find the determinant

D = ad-bc

. Using the product of the eigenvalues:

D = \lambda_1 \lambda_2 = (2)(-1) = -2

[Practice 4 · Medium] Consider the system of linear differential equations given by:

\begin{aligned} \frac{dx}{dt} &= x + 2y \\ \frac{dy}{dt} &= 3x + 2y \end{aligned}

If the solution to this system is of the form

x(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}

and

y(t) = k_1 c_1 e^{\lambda_1 t} + k_2 c_2 e^{\lambda_2 t}

, what is the value of the larger eigenvalue

\lambda

💡 Hint 1 of 3

The given system can be written in matrix form as

\mathbf{x}'(t) = A \mathbf{x}(t)

, where

A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}

and

\mathbf{x}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}

💡 Hint 2 of 3

To find the eigenvalues

\lambda

, we solve the characteristic equation

\det(A - \lambda I) = 0

, where

I

is the identity matrix.

\det \begin{pmatrix} 1-\lambda & 2 \\ 3 & 2-\lambda \end{pmatrix} = 0

💡 Hint 3 of 3

The eigenvalues are

\lambda_1 = 4

and

\lambda_2

. The larger eigenvalue is 4.

The given system can be written in matrix form as

\mathbf{x}'(t) = A \mathbf{x}(t)

, where

A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}

and

\mathbf{x}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}

. To find the eigenvalues

\lambda

, we solve the characteristic equation

\det(A - \lambda I) = 0

, where

I

is the identity matrix.

\det \begin{pmatrix} 1-\lambda & 2 \\ 3 & 2-\lambda \end{pmatrix} = 0

(1-\lambda)(2-\lambda) - (2)(3) = 0

2 - \lambda - 2\lambda + \lambda^2 - 6 = 0

\lambda^2 - 3\lambda - 4 = 0

Factoring the quadratic equation:

(\lambda - 4)(\lambda + 1) = 0

The eigenvalues are

\lambda_1 = 4

and

\lambda_2 = -1

. The larger eigenvalue is 4.

📝 Chapter Quiz

Questions Level

Power Series Solutions, Ordinary Points & the Method of Frobenius

Many of the most important differential equations in theoretical physics and applied mathematics — including Airy's equation (quantum tunneling and diffraction optics), Legendre's equation (spherical harmonics and gravitational multipoles), Bessel's equation (cylindrical heat flow and drumhead vibrations), and Hermite's equation (quantum harmonic oscillator) — have variable coefficients and cannot be integrated using elementary functions. When closed-form formulas fail, power series provide an exact representation. This chapter explores series solutions about ordinary points, index-shifting algebra, and the Method of Frobenius for regular singular points.

● Level 1 Foundation — Definition 8.1: Ordinary vs Singular Points & Analytic Coefficients

Consider the second-order homogeneous linear equation in standard form: $y'' + P(x)y' + Q(x)y = 0.$ A point $x_0 \in \mathbb{R}$ is called an ordinary point if both coefficient functions $P(x)$ and $Q(x)$ are analytic at $x_0$ — that is, they possess convergent Taylor series expansions in a neighborhood of $x_0$ . If either $P(x)$ or $Q(x)$ fails to be analytic at $x_0$ , the point $x_0$ is called a singular point.

For rational coefficient equations $A(x)y'' + B(x)y' + C(x)y = 0$ , a point $x_0$ is ordinary if and only if $A(x_0) \neq 0$ . Real or complex roots of $A(x) = 0$ constitute singular points.

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

⊙ Explorer · Power Series ODE Solution y'' +

x \cdot y = 0

(Airy Series Truncation)

interactive

Truncation Degree $N =$ 5

Initial Slope y'(0) = 0

Recurrence Ratio c_{n+2}/c_{n-1}

-1 / ((k+2)(k+1))

Series Value

y_N(2.0)

Degree

N = 5

Approximation Bandwidth

|x| \approx 1.9

Compare the exact numerical Airy solution of y'' + $x \cdot y = 0$ (indigo) against its power series ∑_{n=0}^N $c_n x^{n}$ (amber) generated by the 3-term recurrence c_{k+2} = −c_{k-1}/( $(k+2)(k+1)$ ).

▲ Level 1 Foundation — Theorem 8.1: Existence of Analytic Power Series Solutions & Radius of Convergence

If $x_0$ is an ordinary point of $y'' + P(x)y' + Q(x)y = 0$ , then there exist two linearly independent solutions analytic at $x_0$ , represented by power series: $y(x) = \sum_{n=0}^\infty a_n (x - x_0)^n = a_0 y_1(x) + a_1 y_2(x),$ where $a_0 = y(x_0)$ and $a_1 = y'(x_0)$ serve as the two essential arbitrary constants.

Radius of Convergence Guarantee: The radius of convergence $R$ for both series solutions is at least as large as the distance in the complex plane from $x_0$ to the nearest singular point of $P(x)$ or $Q(x)$ : $R \ge \min\{|z - x_0| : z \in \mathbb{C} \text{ is a singularity of } P(z) \text{ or } Q(z)\}.$

Analytical Mechanics of Power Series Solutions

1. Power Series Expansions

Write

y = \sum_{n=0}^\infty a_n x^n

y' = \sum_{n=1}^\infty n a_n x^{n-1}

, and

y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}

2. Index Shifting

Shift summation indices so that every sum involves the identical power

x^k

. Adjust summation limits accordingly.

3. Recurrence Relation

Combine sums and set the coefficient of each power

x^k

to zero (Identity Theorem for power series) to extract the recurrence relation.

4. Basis Decomposition

Express higher coefficients in terms of

a_0

and

a_1

to form

y(x) = a_0 y_1(x) + a_1 y_2(x)

◆ Why It Works — First-Principles Derivation (Power Series Solutions, Ordinary Points & the Method of Frobenius)

Why Integration by Parts & Integrating Factors Work: Differentiating a product $u(x)v(x)$ gives $(uv)' = u'v + uv'$ . Integrating both sides and rearranging yields the Integration by Parts formula:

\int u\,\frac{dv}{dx}\,dx = u v - \int v\,\frac{du}{dx}\,dx

Similarly, for a linear first-order ODE $y' + P(x)y = Q(x)$ , multiplying by the integrating factor $\mu(x) = e^{\int P(x)\,dx}$ (which satisfies $\mu'(x) = P(x)\mu(x)$ ) collapses the left-hand side into an exact product derivative: $\dfrac{d}{dx}\bigl[\mu(x)y\bigr] = \mu(x)Q(x)$ .

Worked example 8.A — Airy's Differential Equation

Find the power series solution about the ordinary point $x_0 = 0$ for Airy's equation: $y'' - x y = 0.$

Step 1: Series substitution: $y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}, \qquad x y = \sum_{n=0}^\infty a_n x^{n+1}.$ $\sum_{n=2}^\infty n(n-1)a_n x^{n-2} - \sum_{n=0}^\infty a_n x^{n+1} = 0.$

Step 2: Shift indices to power $x^k$ :
In first sum, let $k = n - 2 \implies n = k + 2$ : $\sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k$ .
In second sum, let $k = n + 1 \implies n = k - 1$ : $\sum_{k=1}^\infty a_{k-1} x^k$ .
Extract $k = 0$ term from first sum: $2(1)a_2 + \sum_{k=1}^\infty \Big[(k+2)(k+1)a_{k+2} - a_{k-1}\Big]x^k = 0.$

Step 3: Recurrence relation: $2a_2 = 0 \implies a_2 = 0.$ $a_{k+2} = \frac{a_{k-1}}{(k+2)(k+1)} \quad \text{for } k \ge 1.$

Step 4: Compute coefficients:
$k=1$ : $a_3 = \frac{a_0}{3 \cdot 2} = \frac{a_0}{6}$ .
$k=2$ : $a_4 = \frac{a_1}{4 \cdot 3} = \frac{a_1}{12}$ .
$k=3$ : $a_5 = \frac{a_2}{5 \cdot 4} = 0$ .
$k=4$ : $a_6 = \frac{a_3}{6 \cdot 5} = \frac{a_0}{180}$ .
$k=5$ : $a_7 = \frac{a_4}{7 \cdot 6} = \frac{a_1}{504}$ .

Step 5: Fundamental solutions: $y(x) = a_0\left(1 + \frac{x^3}{6} + \frac{x^6}{180} + \dots\right) + a_1\left(x + \frac{x^4}{12} + \frac{x^7}{504} + \dots\right).$ Since $P(x) = 0$ and $Q(x) = -x$ have no singularities anywhere in $\mathbb{C}$ , the radius of convergence is $R = \infty$ .

Worked example 8.B — Hermite's Equation & Polynomial Truncation

Find the recurrence relation for Hermite's equation $y'' - 2x y' + 2\lambda y = 0$ about $x_0 = 0$ , and show that if $\lambda = m$ is an integer, the series terminates into a polynomial.

Step 1: Series substitution: $\sum_{n=2}^\infty n(n-1)a_n x^{n-2} - 2 \sum_{n=1}^\infty n a_n x^n + 2\lambda \sum_{n=0}^\infty a_n x^n = 0.$ Shifting first sum $k = n - 2$ : $\sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k - \sum_{k=0}^\infty 2k a_k x^k + \sum_{k=0}^\infty 2\lambda a_k x^k = 0$ $\sum_{k=0}^\infty \Big[(k+2)(k+1)a_{k+2} - 2(k - \lambda)a_k\Big]x^k = 0.$

Step 2: Recurrence relation: $a_{k+2} = \frac{2(k - \lambda)}{(k+2)(k+1)} a_k.$

Step 3: Polynomial truncation: If $\lambda = m \in \{0, 1, 2, \dots\}$ , setting $k = m$ makes $a_{m+2} = 0$ . Consequently, all subsequent coefficients of the same parity vanish: $a_{m+4} = a_{m+6} = \dots = 0$ . If $m$ is even, setting $a_1 = 0$ produces an even polynomial of degree $m$ ; if $m$ is odd, setting $a_0 = 0$ produces an odd polynomial of degree $m$ . These are the orthogonal Hermite polynomials $H_m(x)$ fundamental to the quantum harmonic oscillator.

▲ Level 3 University Rigor — Theorem 8.2: Regular Singular Points, Method of Frobenius & Indicial Equations

1. Classification of Singular Points: A singular point $x_0$ of $y'' + P(x)y' + Q(x)y = 0$ is a regular singular point if both $p_0 = \lim_{x\to x_0} (x - x_0)P(x) \qquad \text{and} \qquad q_0 = \lim_{x\to x_0} (x - x_0)^2 Q(x)$ exist and are finite (i.e. $(x - x_0)P(x)$ and $(x - x_0)^2 Q(x)$ are analytic at $x_0$ ). Otherwise, $x_0$ is an irregular singular point.

2. The Method of Frobenius: If $x_0 = 0$ is a regular singular point, there exists at least one solution of the form: $y(x) = x^r \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty a_n x^{n+r}, \qquad a_0 \neq 0,$ where $r$ is a root of the indicial equation: $F(r) = r(r - 1) + p_0 r + q_0 = 0.$

3. Structure of the Second Solution: Let $r_1, r_2$ be the roots of the indicial equation with $\operatorname{Re}(r_1) \ge \operatorname{Re}(r_2)$ :

Case 1 ( $r_1 - r_2 \notin \mathbb{Z}$ ): Two linearly independent Frobenius series exist: $y_1(x) = x^{r_1}\sum_{n=0}^\infty a_n x^n, \qquad y_2(x) = x^{r_2}\sum_{n=0}^\infty b_n x^n.$
Case 2 ( $r_1 = r_2$ ): The second solution contains a logarithmic singularity: $y_2(x) = y_1(x)\ln x + x^{r_1}\sum_{n=1}^\infty b_n x^n.$
Case 3 ( $r_1 - r_2 = N \in \mathbb{Z}^+$ ): The second solution takes the form: $y_2(x) = C y_1(x)\ln x + x^{r_2}\sum_{n=0}^\infty b_n x^n,$ where $C$ may or may not be zero (e.g. Bessel's equation of integer order).

Polynomial truncation approximates transcendental functions with exceptional precision within the radius of convergence, as shown below for the Taylor series polynomials $P_2, P_4, P_6$ converging to $\cos(x)$ .

Progressive convergence of Taylor polynomials

P_2(x), P_4(x), P_6(x)

y(x) = \cos(x)

. Each higher degree extends the interval of high-fidelity agreement.

⚠️ Common Traps & University Pitfalls — Shifting Summation Indices & Initial Terms

Index shifting misalignments: When changing indices from $n$ to $k$ , verify both the exponent and the starting index. In $\sum_{n=2}^\infty n(n-1)a_n x^{n-2}$ , letting $k = n - 2$ means when $n = 2$ , $k = 0$ , so the sum starts at $k = 0$ , NOT $k = 2$ .

Radius of convergence and complex poles: For $(x^2 + 4)y'' + y = 0$ , the coefficients are continuous for all real $x$ . However, the roots of $A(x) = x^2 + 4 = 0$ are $x = \pm 2i$ . The distance from $x_0 = 0$ to $\pm 2i$ is $|2i| = 2$ . Hence, Theorem 8.1 guarantees convergence ONLY for $ |x| < 2 $, not on all of $ \mathbb{R} $! Complex singularities dictate real convergence boundaries.

Worked Example

C

— Power Series Solutions, Ordinary Points & the Method of Frobenius (Hard Application)

Problem (Hard): Consider the second-order linear homogeneous differential equation: $x^2 y'' + x y' + (x^2 - \nu^2) y = 0$ , where $\nu$ is a constant. This is Bessel's equation of order $\nu$ . Assume $\nu = 1/2$ and that $x > 0$ . $\textbf{Part 1: Ordinary Point Analysis}\\. x=0$ is a regular singular point. Determine the indicial equation for this differential equation and find its roots.

Step-by-Step Solution: The given differential equation is $x^2 y'' + x y' + (x^2 - \nu^2) y = 0$ . We can rewrite this in the standard form $y'' + P(x)y' + Q(x)y = 0$ by dividing by $x^2$ : $y'' + \frac{1}{x} y' + \frac{x^2 - \nu^2}{x^2} y = 0$ . Thus, $P(x) = \frac{1}{x}$ and $Q(x) = \frac{x^2 - \nu^2}{x^2} = 1 - \frac{\nu^2}{x^2}$ . $x=0$ is a singular point. To determine if it's a regular singular point, we examine $x P(x)$ and $x^2 Q(x)$ . $x P(x) = x \cdot \frac{1}{x} = 1$ , which is analytic at $x=0$ . $x^2 Q(x) = x^2 \cdot \frac{x^2 - \nu^2}{x^2} = x^2 - \nu^2$ , which is also analytic at $x=0$ . Therefore, $x=0$ is a regular singular point. The method of Frobenius assumes a solution of the form $y(x) = \sum_{n=0}^{\infty} c_n x^{n+r}$ , where $c_0 \neq 0$ . Substituting this into the differential equation, we get the indicial equation, which is a quadratic equation in $r$ . For Bessel's equation, the indicial equation is given by $r(r-1) + p_0 r + q_0 = 0$ , where $p_0 = \lim_{x \to 0} xP(x)$ and $q_0 = \lim_{x \to 0} x^2 Q(x)$ . In our case, $p_0 = 1$ and $q_0 = -\nu^2$ . So the indicial equation is $r(r-1) + 1 \cdot r - \nu^2 = 0$ , which simplifies to $r^2 - r + r - \nu^2 = 0$ , or $r^2 - \nu^2 = 0$ . Given $\nu = 1/2$ , the indicial equation becomes $r^2 - (1/2)^2 = 0$ , which is $r^2 - \frac{1}{4} = 0$ . The roots of this equation are $r = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$ .

Worked Example

D

— Power Series Solutions, Ordinary Points & the Method of Frobenius (Easy Application)

Problem (Easy): Consider the differential equation $y'' - xy = 0$ . The point $x=0$ is an ordinary point. If we assume a power series solution of the form $y(x) = \sum_{n=0}^{\infty} c_n x^n$ , find the recurrence relation for the coefficients $c_n$ .

Step-by-Step Solution: We are given the differential equation $y'' - xy = 0$ and assume a power series solution of the form $y(x) = \sum_{n=0}^{\infty} c_n x^n$ . \newline Differentiating twice with respect to $x$ , we get: $y'(x) = \sum_{n=1}^{\infty} n c_n x^{n-1}$ and $y''(x) = \sum_{n=2}^{\infty} n(n-1) c_n x^{n-2}$ . \newline Substituting these into the differential equation: $\sum_{n=2}^{\infty} n(n-1) c_n x^{n-2} - x \sum_{n=0}^{\infty} c_n x^n = 0$ . \newline We can rewrite the second term as $\sum_{n=0}^{\infty} c_n x^{n+1}$ . To combine the series, we need the powers of $x$ to be the same. Let $k = n-2$ in the first series, so $n = k+2$ . When $n=2, k=0$ . The first series becomes $\sum_{k=0}^{\infty} (k+2)(k+1) c_{k+2} x^k$ . \newline Let $k = n+1$ in the second series, so $n = k-1$ . When $n=0, k=1$ . The second series becomes $\sum_{k=1}^{\infty} c_{k-1} x^k$ . \newline So the equation is $\sum_{k=0}^{\infty} (k+2)(k+1) c_{k+2} x^k - \sum_{k=1}^{\infty} c_{k-1} x^k = 0$ . \newline We can separate the $k=0$ term from the first sum: $2(1)c_2 + \sum_{k=1}^{\infty} (k+2)(k+1) c_{k+2} x^k - \sum_{k=1}^{\infty} c_{k-1} x^k = 0$ . \newline For this equation to hold for all $x$ , the coefficients of each power of $x$ must be zero. \newline For $k=0$ : $2c_2 = 0$ implies $c_2 = 0$ . \newline For $k \geq 1$ : $(k+2)(k+1) c_{k+2} - c_{k-1} = 0$ . \newline Thus, the recurrence relation is $c_{k+2} = \frac{c_{k-1}}{(k+2)(k+1)}$ for $k \geq 1$ . Replacing $k$ with $n$ , we get $c_{n+2} = \frac{c_{n-1}}{(n+2)(n+1)}$ for $n \geq 1$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Power Series Solutions, Ordinary Points & the Method of Frobenius)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — Power Series Solutions, Ordinary Points & the Method of Frobenius

Find the recurrence relation and the first three non-zero terms of the power series solution about

x_0 = 0

y'' + x y' + 2y = 0, \qquad y(0) = 1, \quad y'(0) = 0.

💡 Hint 1 of 3

Series substitution:

\sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=1}^\infty n a_n x^n + 2\sum_{n=0}^\infty a_n x^n = 0

💡 Hint 2 of 3

Shift first sum

k = n - 2

\sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k + \sum_{k=1}^\infty k a_k x^k + \sum_{k=0}^\infty 2a_k x^k = 0

Recurrence relation: For

k = 0

2(1)a_2 + 2a_0 = 0 \implies a_2 = -a_0

💡 Hint 3 of 3

Particular solution:

y(x) = 1 - x^2 + \frac{1}{3}x^4 - \frac{1}{15}x^6 + \dots

Step 1: Series substitution: $\sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=1}^\infty n a_n x^n + 2\sum_{n=0}^\infty a_n x^n = 0.$ Shift first sum $k = n - 2$ : $\sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k + \sum_{k=1}^\infty k a_k x^k + \sum_{k=0}^\infty 2a_k x^k = 0.$

Step 2: Recurrence relation: For $k = 0$ : $2(1)a_2 + 2a_0 = 0 \implies a_2 = -a_0$ .
For $k \ge 1$ : $(k+2)(k+1)a_{k+2} + (k + 2)a_k = 0 \implies a_{k+2} = -\frac{a_k}{k + 1}.$

Step 3: Initial conditions: $a_0 = y(0) = 1$ , $a_1 = y'(0) = 0$ .
Since $a_1 = 0$ , all odd coefficients vanish: $a_3 = a_5 = a_7 = 0$ .
For even coefficients:
$k = 0 \implies a_2 = -a_0 = -1$ .
$k = 2 \implies a_4 = -\frac{a_2}{2 + 1} = -\frac{-1}{3} = \frac{1}{3}$ .
$k = 4 \implies a_6 = -\frac{a_4}{4 + 1} = -\frac{1/3}{5} = -\frac{1}{15}$ .

Step 4: Particular solution: $y(x) = 1 - x^2 + \frac{1}{3}x^4 - \frac{1}{15}x^6 + \dots$

Find the indicial equation and roots about

x = 0

for the differential equation

2x^2 y'' - x y' + (x - 5)y = 0,

and state the form of the two linearly independent Frobenius solutions.

💡 Hint 1 of 3

Check regular singularity at $x_0 = 0$ :

P(x) = -\frac{1}{2x}, \quad Q(x) = \frac{x - 5}{2x^2}

💡 Hint 2 of 3

Rewrite and simplify to obtain:

p_0 = \lim_{x\to 0} x P(x) = -\frac{1}{2}, \qquad q_0 = \lim_{x\to 0} x^2 Q(x) = -\frac{5}{2}

Both limits are finite, confirming

x = 0

is a regular singular point.

💡 Hint 3 of 3

Because the roots do not differ by an integer, Case 1 of Frobenius' Theorem applies. Two linearly independent Frobenius series solutions exist without any logarithmic terms:

y_1(x) = x^{5/2}\sum_{n=0}^\infty a_n x^n, \qquad y_2(x)

Step 1: Check regular singularity at $x_0 = 0$ : $P(x) = -\frac{1}{2x}, \quad Q(x) = \frac{x - 5}{2x^2}.$ $p_0 = \lim_{x\to 0} x P(x) = -\frac{1}{2}, \qquad q_0 = \lim_{x\to 0} x^2 Q(x) = -\frac{5}{2}.$ Both limits are finite, confirming $x = 0$ is a regular singular point.

Step 2: Indicial equation: $r(r - 1) + p_0 r + q_0 = r^2 - r - \frac{1}{2}r - \frac{5}{2} = 0 \implies 2r^2 - 3r - 5 = 0.$ Factoring: $(2r - 5)(r + 1) = 0 \implies r_1 = \frac{5}{2}, \quad r_2 = -1.$

Step 3: Root difference: $r_1 - r_2 = \frac{5}{2} - (-1) = \frac{7}{2} otin \mathbb{Z}.$ Because the roots do not differ by an integer, Case 1 of Frobenius' Theorem applies. Two linearly independent Frobenius series solutions exist without any logarithmic terms: $y_1(x) = x^{5/2}\sum_{n=0}^\infty a_n x^n, \qquad y_2(x) = x^{-1}\sum_{n=0}^\infty b_n x^n \quad (a_0, b_0 eq 0).$

[Practice 3 · Medium] Find

\mathcal L\{u(t-2)\sin(3(t-2))\}

💡 Hint 1 of 3

Second shifting theorem with

f(t)=\sin(3t)

💡 Hint 2 of 3

Rewrite and simplify to obtain:

\frac{3 e^{- 2 s}}{s^{2} + 9}

💡 Hint 3 of 3

Evaluate and simplify

\frac{3 e^{- 2 s}}{s^{2} + 9}

carefully (checking signs and units) to obtain the final answer.

Second shifting theorem with

f(t)=\sin(3t)

\frac{3 e^{- 2 s}}{s^{2} + 9}

[Practice 4 · Hard] If

\mathcal L\{y\}=Y(s)

y(0)=-3

and

y'(0)=-1

, find

\mathcal L\{y''\}

💡 Hint 1 of 3

Start with the expression:

\mathcal L\{y''\}

💡 Hint 2 of 3

Rewrite and substitute:

\mathcal L\{y''\} = s^2Y(s)-sy(0)-y'(0)

💡 Hint 3 of 3

Evaluate

s^2Y(s)-sy(0)-y'(0)

to obtain the final answer.

\mathcal L\{y''\}=s^2Y(s)-sy(0)-y'(0)

=

Y s^{2} + 3 s + 1

📝 Chapter Quiz

Questions Level

Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability

In scientific computing, weather forecasting, aerospace navigation, and chemical reactor design, nonlinear differential equations almost never possess closed-form analytical solutions. Instead, mathematicians and engineers rely on numerical integration algorithms to advance the solution forward step by step along discrete time grids. Understanding numerical ODE algorithms requires analyzing two competing forces: truncation error (the accuracy of the local Taylor polynomial approximation) and numerical stability (whether propagation errors attenuate or blow up uncontrollably). This chapter investigates Forward Euler, Heun's Predictor-Corrector (RK2), Fourth-Order Runge–Kutta (RK4), and the phenomenon of stiffness.

● Level 1 Foundation — Definition 9.1: Forward Euler's Method & Discrete Vector Fields

Consider the initial value problem: $\frac{dy}{dt} = f(t, y), \qquad y(t_0) = y_0.$ Discretize the time domain into uniform intervals of step size $h > 0$ , generating discrete grid points $t_n = t_0 + n h$ . Forward Euler's Method approximates the continuous solution $y(t_{n+1})$ by extrapolating along the tangent line evaluated at the beginning of the step: $y_{n+1} = y_n + h f(t_n, y_n), \qquad n = 0, 1, 2, \dots$

Slope field for

\frac{dy}{dx} = x -y

with the integral curve passing through initial condition

(0, 1)

▲ Level 1 Foundation — Theorem 9.1: Local Truncation Error

O(h^{2})

vs Global Error

O(h)

1. Local Truncation Error (LTE): By Taylor's Theorem with Lagrange remainder, assuming $y \in C^2$ : $y(t_n + h) = y(t_n) + h y'(t_n) + \frac{h^2}{2}y''(\xi_n) = y(t_n) + h f(t_n, y(t_n)) + \frac{h^2}{2}y''(\xi_n)$ for some $\xi_n \in (t_n, t_{n+1})$ . The local error committed in a single step is: $e_{n+1} = y(t_{n+1}) - \big(y(t_n) + h f(t_n, y(t_n))\big) = \frac{h^2}{2}y''(\xi_n) = O(h^2).$

2. Global Truncation Error (GTE): To integrate across a fixed interval $[t_0, T]$ , the total number of steps is $N = \frac{T - t_0}{h} = O(h^{-1})$ . The accumulated global error is: $E(T) \le N \cdot O(h^2) = \left(\frac{T - t_0}{h}\right) \cdot C h^2 = O(h).$ Forward Euler is therefore a first-order method: halving the step size $h$ only halves the global error.

● Level 2 Analytical Methods — Definition 9.2: Improved Euler (Heun / Predictor-Corrector) & RK2

Forward Euler's error arises because it evaluates slope exclusively at the beginning of the interval $t_n$ . Improved Euler (Heun's Method) eliminates first-order bias by averaging the slope at the start with a predicted slope at the end (analogous to the Trapezoidal Rule in numerical quadrature):

Predictor step (Euler trial): $\tilde{y}_{n+1} = y_n + h f(t_n, y_n).$
Corrector step (Trapezoidal average): $y_{n+1} = y_n + \frac{h}{2}\Big[f(t_n, y_n) + f(t_{n+1}, \tilde{y}_{n+1})\Big].$

Heun's method has local truncation error $O(h^3)$ and global error $O(h^2)$ , making it a second-order Runge–Kutta method (RK2).

Hierarchy of Numerical Integration Methods

Forward Euler (RK1)

Local error

O(h^2)

, Global error

O(h)

. Single function evaluation per step:

y_{n+1} = y_n + h f_n

Heun's Method (RK2)

Local error

O(h^3)

, Global error

O(h^2)

. Two evaluations per step: predictor

\tilde{y}

and average slope.

Classical RK4

Local error

O(h^5)

, Global error

O(h^4)

. Four evaluations per step with Simpson-rule weighting.

Backward (Implicit) Euler

y_{n+1} = y_n + h f(t_{n+1}, y_{n+1})

. Solves implicit equation at each step; unconditionally stable for stiff systems.

◆ Why It Works — First-Principles Derivation (Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability)

\int u\,\frac{dv}{dx}\,dx = u v - \int v\,\frac{du}{dx}\,dx

Worked example 9.A — Manual Step Comparison: Euler vs Heun

Apply one step of step size $h = 0.2$ to approximate $y(0.2)$ for $\frac{dy}{dt} = y - t^2 + 1, \qquad y(0) = 0.5.$ Compare Forward Euler and Heun's method with the exact analytical solution $y(t) = t^2 + 2t + 1 - 0.5e^t$ (at $t = 0.2$ , $y(0.2) = 0.2^2 + 2(0.2) + 1 - 0.5e^{0.2} \approx 1.44 - 0.6107 = 0.8293$ ).

Forward Euler:
$f(t_0, y_0) = f(0, 0.5) = 0.5 - 0^2 + 1 = 1.5$ .
$y_1 = y_0 + h f(0, 0.5) = 0.5 + 0.2(1.5) = 0.8000$ .
Absolute error: $|0.8293 - 0.8000| = 0.0293$ .

Heun's Method (Improved Euler):
Predictor: $\tilde{y}_1 = 0.8000$ .
Slope at endpoint: $f(t_1, \tilde{y}_1) = f(0.2, 0.8) = 0.8 - (0.2)^2 + 1 = 0.8 - 0.04 + 1 = 1.76$ .
Average slope: $\frac{f_0 + f_1}{2} = \frac{1.5 + 1.76}{2} = 1.63$ .
Corrector: $y_1 = 0.5 + 0.2(1.63) = 0.5 + 0.326 = 0.8260$ .
Absolute error: $|0.8293 - 0.8260| = 0.0033$ .
Notice that Heun's method reduces the error by nearly a factor of 10 in a single step!

⊙ Explorer · Euler's method vs the exact solution

interactive

step size $h$ : 20

use improved Euler (Heun's method)

approx

y(2)

4.866

exact

y(2)

5.305

error

0.440

The IVP is Worked examples 9.A/9.B's $y' = y -x^{2} + 1$ , $y(0) = 0.5$ , run out to $x = 2$ with $h = 0.2 by$ default — matching plain Euler's known error of about 0.44 at that step size. Drag $h$ down toward its smallest value and watch the amber polyline hug the indigo exact curve; tick the improved-Euler checkbox at the same $h$ and watch the error drop by roughly an order of magnitude, exactly as Definition 9.2 predicts.

▲ Level 3 University Rigor — Theorem 9.2: Classical Fourth-Order Runge–Kutta (RK4) & Numerical Stiffness /

A

-Stability

1. Classical Fourth-Order Runge–Kutta (RK4): The gold standard explicit solver evaluates four slope samples per time step, weighted according to Simpson's Rule: $k_1 = f(t_n, y_n),$ $k_2 = f\left(t_n + \frac{h}{2}, \, y_n + \frac{h}{2}k_1\right),$ $k_3 = f\left(t_n + \frac{h}{2}, \, y_n + \frac{h}{2}k_2\right),$ $k_4 = f(t_n + h, \, y_n + h k_3).$ $y_{n+1} = y_n + \frac{h}{6}\big(k_1 + 2k_2 + 2k_3 + k_4\big).$ RK4 achieves local truncation error $O(h^5)$ and global error $O(h^4)$ . Halving $h$ reduces global error by a factor of $2^4 = 16$ .

2. Numerical Stiffness & Dahlquist Absolute Stability: Consider Dahlquist's linear test equation: \[ \frac{dy}{dt} = \lambda y, \qquad \operatorname{Re}(\lambda) < 0. \] The true solution $ y(t) = y_0 e^{\lambda t} $ decays exponentially to zero as $ t \to \infty $. Applying Forward Euler: \[ y_{n+1} = y_n + h(\lambda y_n) = (1 + h\lambda)y_n \implies y_n = (1 + h\lambda)^n y_0. \] For numerical stability (ensuring $ y_n \to 0 $), the amplification factor must satisfy: \[ |1 + h\lambda| < 1 \implies -1 < 1 + h\lambda < 1 \implies h < \frac{2}{|\lambda|}. \] If $ h > \frac{2}{|\lambda|} $, the numerical solution diverges to infinity with alternating signs, even though the true solution is rapidly decaying! Systems possessing rapid decaying transients (large $|\lambda|$ ) are termed stiff. Stiff systems require Implicit (Backward) Euler: $y_{n+1} = y_n + h\lambda y_{n+1} \implies y_{n+1} = \frac{1}{1 - h\lambda}y_n.$ Since $|1 - h\lambda| > 1$ for all $h > 0$ whenever $ \operatorname{Re}(\lambda) < 0 $, Backward Euler is unconditionally A-stable.

Forward Euler creates a piecewise linear approximation that diverges from the true curve due to accumulating curvature truncation error, as compared below.

Tangent extrapolation in Forward Euler for

y' = y, y(0) = 1

. A single large step

h = 1

yields

y(1) \approx 2

(error

\approx 0.718

); refining to

h = 0.5

brings the polyline closer to the true exponential curve.

⚠️ Common Traps & University Pitfalls — Global vs. Local Error and Step Size Stiff Instability

Confusing Local and Global error orders: Forward Euler has local truncation error $O(h^2)$ , but its global error is $O(h)$ . Students often confuse the two and mistakenly expect Euler's method to have second-order global accuracy.

Rounding error vs. truncation error tradeoff: Reducing $h$ reduces truncation error, but dramatically increases the total number of floating-point arithmetic operations $N$ . Eventually, machine roundoff error accumulates and overwhelms the solution if $h$ is made excessively tiny.

Stiff instability blowup: In systems with fast decay (e.g. $y' = -100y$ ), choosing $h = 0.05$ violates $ h < 2/100 = 0.02 $. Forward Euler produces wildly oscillating explosive numbers $ y_n = (-4)^n \to \pm \infty $, mistaking mathematical decay for physical explosion!

Worked Example

B

— Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability (Medium Application)

Problem (Medium): Consider the differential equation $\dfrac{dy}{dx} = x + y$ with the initial condition $y(0) = 1$ . You are asked to approximate $y(0.2)$ using Euler's method. If the step size $h = 0.1$ is used, and it is known that the exact solution at $x=0.2$ is $y(0.2) = 1.24$ , determine the value of $k$ such that $y(0.2)_{\text{Euler}} = 1.24 - k$ .

Step-by-Step Solution: Euler's method for approximating the solution to $\dfrac{dy}{dx} = f(x, y)$ with initial condition $y(x_0) = y_0$ and step size $h$ is given by $y_{n+1} = y_n + h f(x_n, y_n)$ . In this problem, $f(x, y) = x + y$ , $x_0 = 0$ , $y_0 = 1$ , and $h = 0.1$ . We want to approximate $y(0.2)$ with $h=0.1$ . This requires two steps: Step 1: From $x_0 = 0$ to $x_1 = 0.1$ . $y_1 = y_0 + h f(x_0, y_0)$ $y_1 = 1 + 0.1 (0 + 1)$ $y_1 = 1 + 0.1 (1)$ $y_1 = 1 + 0.1$ $y_1 = 1.1$ Step 2: From $x_1 = 0.1$ to $x_2 = 0.2$ . $y_2 = y_1 + h f(x_1, y_1)$ $y_2 = 1.1 + 0.1 (0.1 + 1.1)$ $y_2 = 1.1 + 0.1 (1.2)$ $y_2 = 1.1 + 0.12$ $y_2 = 1.22$ So, the approximation using Euler's method is $y(0.2)_{\text{Euler}} = 1.22$ . We are given that the exact solution is $y(0.2) = 1.24$ and $y(0.2)_{\text{Euler}} = 1.24 - k$ . Substituting the value of $y(0.2)_{\text{Euler}}$ , we get: $1.22 = 1.24 - k$ $k = 1.24 - 1.22$ $k = 0.02$ Therefore, the value of $k$ is 0.02.

Worked Example

C

— Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability (Hard Application)

Problem (Hard): Consider the differential equation $\frac{dy}{dx} = x + y$ with the initial condition $y(0) = 1$ . Use the Euler method with a step size $h = 0.1$ to approximate the value of $y(0.2)$ .

Step-by-Step Solution: The Euler method approximates the solution to an initial value problem $\frac{dy}{dx} = f(x, y)$ , $y(x_0) = y_0$ using the iterative formula $y_{n+1} = y_n + h f(x_n, y_n)$ , where $h$ is the step size and $x_{n+1} = x_n + h$ . Given $f(x, y) = x + y$ , $x_0 = 0$ , $y_0 = 1$ , and $h = 0.1$ . For the first step (approximating y(0.1)): $x_1 = x_0 + h = 0 + 0.1 = 0.1$ $y_1 = y_0 + h f(x_0, y_0) = 1 + 0.1 (0 + 1) = 1 + 0.1(1) = 1.1$ . For the second step (approximating y(0.2)): $x_2 = x_1 + h = 0.1 + 0.1 = 0.2$ $y_2 = y_1 + h f(x_1, y_1) = 1.1 + 0.1 (0.1 + 1.1) = 1.1 + 0.1 (1.2) = 1.1 + 0.12 = 1.22$ . Therefore, the approximate value of $y(0.2)$ using the Euler method is 1.22.

Worked Example

D

— Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability (Easy Application)

Problem (Easy): Consider the differential equation $\dfrac{dy}{dx} = x + y$ with the initial condition $y(0) = 1$ . Use the Euler method with a step size of $h = 0.1$ to approximate $y(0.2)$ .

Step-by-Step Solution: Using Euler's method $y_{n+1} = y_n + h(x_n + y_n)$ with $x_0 = 0, y_0 = 1, h = 0.1$ : 1. $y_1 = 1 + 0.1(0 + 1) = 1.100$ at $x_1 = 0.1$ . 2. $y_2 = 1.1 + 0.1(0.1 + 1.1) = 1.1 + 0.12 = 1.220$ at $x_2 = 0.2$ .

Error Diagnosis — Common Student Traps vs. Exam-Board Method (Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability)

Common Student Trap	Why It Fails Mathematically	Correct Exam-Board Method
Writing $\ln\|y\| = x^2 + C \implies y = e^{x^2} + C$	Exponentiation distributes over addition as multiplication ( $e^{a+b} = e^a e^b$ ), not addition.	Write $y = e^{x^2 + C} = Ae^{x^2}$ and solve for $A$ using the initial condition.
Leaving original $x$ -limits on a $u$ -substituted definite integral	Evaluates the antiderivative $F(u)$ at $x$ -values instead of $u(x)$ -values.	Convert bounds immediately: $u_{\text{low}} = u(a)$ and $u_{\text{high}} = u(b)$ .
Computing total distance as $\left\|\int_{t_1}^{t_2} v(t)\,dt\right\|$	Cancels forward and backward motion (net displacement) instead of accumulating total path length.	Split the integral at every root where $v(t) = 0$ and integrate $\|v(t)\|$ .

▲ Exam Pacing & Verification Strategy — Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability

Apply two steps of Heun's method (Improved Euler) with step size

h = 0.1

to approximate

y(0.2)

for the IVP

\frac{dy}{dt} = 2t y, \qquad y(0) = 1.

Compare your approximation with the exact analytical value

y(0.2) = e^{(0.2)^2} = e^{0.04} \approx 1.04081077

💡 Hint 1 of 3

First interval $[0, 0.1]$ : Initial point:

t_0 = 0, y_0

💡 Hint 2 of 3

Slope at start:

k_1 = f(0, 1) = 2(0)(1) = 0

. Predictor:

\tilde{y}_1 = y_0 + h k_1 = 1 + 0.1(0)

💡 Hint 3 of 3

Absolute error:

|1.04081077 - 1.040704|

, demonstrating second-order accuracy.

Step 1: First interval $[0, 0.1]$ :
Initial point: $t_0 = 0, y_0 = 1$ .
Slope at start: $k_1 = f(0, 1) = 2(0)(1) = 0$ .
Predictor: $\tilde{y}_1 = y_0 + h k_1 = 1 + 0.1(0) = 1$ .
Slope at endpoint: $k_2 = f(0.1, 1) = 2(0.1)(1) = 0.2$ .
Corrector: $y_1 = y_0 + \frac{h}{2}(k_1 + k_2) = 1 + \frac{0.1}{2}(0 + 0.2) = 1 + 0.01 = 1.0100.$

Step 2: Second interval $[0.1, 0.2]$ :
Point: $t_1 = 0.1, y_1 = 1.01$ .
Slope at start: $k_1 = f(0.1, 1.01) = 2(0.1)(1.01) = 0.2020$ .
Predictor: $\tilde{y}_2 = y_1 + h k_1 = 1.01 + 0.1(0.202) = 1.01 + 0.0202 = 1.0302$ .
Slope at endpoint: $k_2 = f(0.2, 1.0302) = 2(0.2)(1.0302) = 0.41208$ .
Corrector: $y_2 = 1.01 + \frac{0.1}{2}(0.2020 + 0.41208) = 1.01 + 0.05(0.61408) = 1.01 + 0.030704 = 1.040704.$

Step 3: Error evaluation:
Exact value: $y(0.2) = e^{0.04} \approx 1.04081077$ .
Absolute error: $|1.04081077 - 1.040704| = 0.000107$ , demonstrating second-order accuracy.

Consider the stiff chemical decay differential equation

\frac{dy}{dt} = -80y, \qquad y(0) = 1.

(a) Determine the critical maximum step size

h_{\text{crit}}

for Forward Euler to remain stable. (b) Calculate the numerical solution after 2 steps with

h = 0.03

versus

h = 0.02

💡 Hint 1 of 3

(a) Critical step size: For

y' = \lambda y

with

\lambda = -80

, Forward Euler amplification factor is

1 + h\lambda = 1 - 80h

💡 Hint 2 of 3

Absolute stability requires $ |1 - 80h| < 1 \implies -1 < 1 - 80h < 1 \implies 0 < 80h < 2 \implies h < \frac{2}{80} = 0.025 $. Thus $ h_{\text{crit}} = 0.025 $.

💡 Hint 3 of 3

Use the resulting form to finish:

y_1 = -0.6

y_2 = (-0.6)^2

y_{10} = (-0.6)^{10}

(stable decay to equilibrium zero).

(a) Critical step size: For $y' = \lambda y$ with $\lambda = -80$ , Forward Euler amplification factor is $1 + h\lambda = 1 - 80h$ .
Absolute stability requires $ |1 - 80h| < 1 \implies -1 < 1 - 80h < 1 \implies 0 < 80h < 2 \implies h < \frac{2}{80} = 0.025 $.
Thus $h_{\text{crit}} = 0.025$ .

(b) Numerical comparison:
With $h = 0.03 > h_{\text{crit}}$ : $1 - 80(0.03) = 1 - 2.4 = -1.4$ .
$y_1 = -1.4(1) = -1.4$ , $y_2 = (-1.4)^2 = 1.96$ , $y_{10} = (-1.4)^{10} \approx 28.9$ (explosive oscillation to $\pm \infty$ ).
With $ h = 0.02 < h_{\text{crit}} $: $ 1 - 80(0.02) = 1 - 1.6 = -0.6 $.
$y_1 = -0.6$ , $y_2 = (-0.6)^2 = 0.36$ , $y_{10} = (-0.6)^{10} \approx 0.0060 \to 0$ (stable decay to equilibrium zero).

[Practice 3 · Medium] The first-order differential equation

\dfrac{df}{dx} = 12x - 2f

has its slope field plotted below. Determine the slope

\dfrac{df}{dx}

at the point

A(6, 23)

💡 Hint 1 of 3

Evaluating at

(6, 23)

gives.

💡 Hint 2 of 3

Set up the numerical expression:

12(6) - 2(23)

💡 Hint 3 of 3

Evaluate

12(6) - 2(23)

step by step to obtain the final answer.

Evaluating at

(6, 23)

gives

12(6) - 2(23) = 26

[Practice 4 · Hard] Find a particular solution of

y''-y'-6y=e^{2x}+18

💡 Hint 1 of 3

Split the forcing (the superposition note): for

e^{2x}

r=2

isn't a root of

r^2-r-6=0

(roots

3,-2

), so trial

Ae^{2x}

gives

(4-2-6)A=1\Rightarrow-4A=1\Rightarrow A=-\tfrac14

💡 Hint 2 of 3

For the constant

18

, trial

B

gives

-6B = 18\Rightarrow B

💡 Hint 3 of 3

Adding the two pieces:

y_p=-\tfrac14e^{2x}-3

Split the forcing (the superposition note): for

e^{2x}

r=2

isn't a root of

r^2-r-6=0

(roots

3,-2

), so trial

Ae^{2x}

gives

(4-2-6)A=1\Rightarrow-4A=1\Rightarrow A=-\tfrac14

. For the constant

18

, trial

B

gives

-6B=18\Rightarrow B=-3

. Adding the two pieces:

y_p=-\tfrac14e^{2x}-3

📝 Chapter Quiz

Questions Level

Practice set

Fifty university-rigor problems spanning all nine chapters of Ordinary Differential Equations (separable equations and existence theory, linear first-order integrating factors and Bernoulli equations, second-order homogeneous auxiliary roots and Wronskian theory, undetermined coefficients and resonance, variation of parameters and Green's kernels, Laplace transforms and discontinuous forcing, systems of linear ODEs and phase portraits, power series and Frobenius method, and numerical methods and stiff stability). Try each problem, then reveal the complete analytical derivation.

Solve the initial value problem $\dfrac{dy}{dx} = 2x y^2$ with $y(0) = 1$ . Find $y(1/2)$ and state its maximal interval of valid existence.

Intermediate

Step 1: Separate variables: $\dfrac{dy}{y^2} = 2x\,dx$ .
Step 2: Integrate: $-\dfrac{1}{y} = x^2 + C$ .
Step 3: Initial condition: $-\dfrac{1}{1} = 0 + C \implies C = -1$ .
Thus: $-\dfrac{1}{y} = x^2 - 1 \implies y(x) = \dfrac{1}{1 - x^2}$ .
At $x = 1/2$ : $y(1/2) = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}.$
Step 4: Maximal interval: Vertical asymptotes occur at $x = \pm 1$ . The maximal open interval containing the initial point $x_0 = 0$ is $(-1, 1)$ . As $x \to 1^-$ , $y(x) \to +\infty$ (finite-time blowup).

Find all equilibrium solutions of $\dfrac{dy}{dt} = y^2 - 4$ , classify each as an attractor (sink) or repeller (source), and determine $\lim_{t\to\infty} y(t)$ for the solution with $y(0) = 0$ .

Basic

Step 1: Locate equilibria: Set $y^2 - 4 = 0 \implies y_1 = -2, \, y_2 = 2$ .
Step 2: Classify stability: Let $f(y) = y^2 - 4$ . Then $f'(y) = 2y$ .
At $y = -2$ : $f'(-2) = -4 \lt 0 \implies y = -2$ is a stable attractor (sink).
At $y = 2$ : $f'(2) = +4 \gt 0 \implies y = 2$ is an unstable repeller (source).
Step 3: Trajectory with $y(0) = 0$ : Since $y(0) = 0$ lies in the interval $(-2, 2)$ , $f(y) = y^2 - 4 \lt 0$ , so $y(t)$ is monotonically decreasing towards the lower stable equilibrium. Therefore: $\lim_{t\to\infty} y(t) = -2.$

A harvested population evolves according to $\dfrac{dP}{dt} = P\left(1 - \dfrac{P}{10}\right) - 2.1$ . Determine the two equilibrium populations, identify which is the extinction threshold, and determine whether a starting population $P(0) = 2$ survives or collapses.

Advanced

Step 1: Equilibrium quadratic: Set $\dfrac{dP}{dt} = 0$ : $-\frac{P^2}{10} + P - 2.1 = 0 \implies P^2 - 10P + 21 = 0.$ Factoring: $(P - 3)(P - 7) = 0 \implies P_1 = 3, \, P_2 = 7$ .
Step 2: Stability: $f'(P) = 1 - \frac{P}{5}$ .
$f'(3) = 1 - 0.6 = +0.4 \gt 0 \implies P = 3$ is an unstable extinction threshold.
$f'(7) = 1 - 1.4 = -0.4 \lt 0 \implies P = 7$ is the stable carrying capacity.
Step 3: Dynamics for $P(0) = 2$ : Since $P(0) = 2 \lt 3$ , $\dfrac{dP}{dt} \lt 0$ . The population strictly declines and collapses to $0$ in finite time, proving that $P = 3$ is the minimum viable population needed to avoid extinction.

Solve the first-order linear initial value problem $x \dfrac{dy}{dx} + 3y = \dfrac{\sin x}{x^2}$ for $x \gt 0$ with $y(\pi) = 0$ . Find $\lim_{x\to 0^+} y(x)$ .

Intermediate

Step 1: Standard form: Divide by $x$ : $\frac{dy}{dx} + \frac{3}{x}y = \frac{\sin x}{x^3}.$ Step 2: Integrating factor: $\mu(x) = \exp\left(\int \frac{3}{x}\,dx\right) = e^{3\ln x} = x^3$ .
Step 3: Multiply and integrate: $\frac{d}{dx}\big[x^3 y\big] = x^3 \left(\frac{\sin x}{x^3}\right) = \sin x \implies x^3 y = -\cos x + C.$ Step 4: Initial condition $y(\pi) = 0$ : $\pi^3 (0) = -\cos\pi + C \implies 0 = -(-1) + C \implies C = -1$ .
Thus: $y(x) = \frac{-\cos x - 1}{x^3} = -\frac{1 + \cos x}{x^3}.$
Step 5: Limit as $x \to 0^+$ : By Taylor expansion, $1 + \cos x = 1 + (1 - x^2/2 + \dots) = 2 - x^2/2 + \dots$ .
Hence: $\lim_{x\to 0^+} y(x) = \lim_{x\to 0^+} -\frac{2}{x^3} = -\infty.$

Solve the nonlinear Bernoulli differential equation $\dfrac{dy}{dx} - \dfrac{1}{x}y = 2x^2 y^3$ for $x \gt 0$ with $y(1) = 1$ . Find $y(2)$ .

Advanced

Step 1: Bernoulli transformation: Here $n = 3$ . Divide by $y^3$ : $y^{-3}y' - \frac{1}{x}y^{-2} = 2x^2$ .
Substitute $v = y^{1-n} = y^{-2} \implies v' = -2y^{-3}y' \implies y^{-3}y' = -\frac{1}{2}v'$ .
Substituting yields: $-\frac{1}{2}v' - \frac{1}{x}v = 2x^2 \implies v' + \frac{2}{x}v = -4x^2.$ Step 2: Integrating factor for $v$ : $\mu(x) = e^{\int 2/x\,dx} = x^2$ .
Multiply and integrate: $\frac{d}{dx}\big[x^2 v\big] = x^2(-4x^2) = -4x^4 \implies x^2 v = -\frac{4}{5}x^5 + C \implies v(x) = -\frac{4}{5}x^3 + \frac{C}{x^2}.$ Step 3: Initial condition $y(1) = 1 \implies v(1) = 1$ : $1 = -\frac{4}{5}(1) + C \implies C = \frac{9}{5}.$ Thus: $v(x) = \frac{9 - 4x^5}{5x^2} \implies y(x) = \dfrac{1}{\sqrt{v(x)}} = \sqrt{\frac{5x^2}{9 - 4x^5}}$ .
At $x = 1$ , $y(1) = 1$ . Note that at $x = 2$ , $9 - 4(32) = 9 - 128 = -119 \lt 0$ , so the solution ceases to exist before $x = 2$ ; the singularity occurs at $x = (9/4)^{1/5} \approx 1.176$ .

An RC circuit with resistance $R = 2 \,\Omega$ and capacitance $C = 0.5 \text{ F}$ is driven by a delayed step voltage $E(t) = 10 u(t - 2)$ . With $q(0) = 0$ , find the exact charge $q(t)$ for $t \ge 0$ .

Intermediate

Step 1: Governing ODE: $R \frac{dq}{dt} + \frac{1}{C}q = E(t) \implies 2q' + 2q = 10u(t - 2) \implies q' + q = 5u(t - 2)$ .
Step 2: Integrating factor: $\mu(t) = e^t$ .
For $0 \le t \lt 2$ , $E(t) = 0 \implies q(t) = 0$ .
For $t \ge 2$ : $\frac{d}{dt}[e^t q] = 5e^t \implies e^t q = 5e^t + K \implies q(t) = 5 + K e^{-t}.$ By continuity at $t = 2$ : $q(2) = 5 + K e^{-2} = 0 \implies K = -5e^2$ .
Thus for $t \ge 2$ : $q(t) = 5\big(1 - e^{-(t - 2)}\big)$ .
Expressed with unit step notation: $q(t) = 5\big(1 - e^{-(t - 2)}\big)u(t - 2).$

Solve the second-order homogeneous initial value problem $y'' - 7y' + 10y = 0$ with $y(0) = 1$ and $y'(0) = -1$ . Find $y(1)$ .

Basic

Step 1: Auxiliary roots: $r^2 - 7r + 10 = (r - 2)(r - 5) = 0 \implies r_1 = 2, \quad r_2 = 5.$ General solution: $y(x) = C_1 e^{2x} + C_2 e^{5x}$ .
Step 2: Derivative: $y'(x) = 2C_1 e^{2x} + 5C_2 e^{5x}$ .
Step 3: Apply initial conditions: $\begin{cases} C_1 + C_2 = 1 \\ 2C_1 + 5C_2 = -1 \end{cases}$ From first: $C_1 = 1 - C_2$ . Substitute: $2(1 - C_2) + 5C_2 = 2 + 3C_2 = -1 \implies 3C_2 = -3 \implies C_2 = -1, \, C_1 = 2$ .
Step 4: Particular solution: $y(x) = 2e^{2x} - e^{5x}, \qquad y(1) = 2e^2 - e^5.$

Solve $y'' + 4y' + 4y = 0$ with $y(0) = 2$ and $y'(0) = 1$ . Determine the exact time $t \gt 0$ at which $y(t)$ attains its absolute maximum value.

Intermediate

Step 1: Characteristic roots: $r^2 + 4r + 4 = (r + 2)^2 = 0 \implies r = -2$ (repeated).
General solution: $y(t) = (C_1 + C_2 t)e^{-2t}$ .
Step 2: Initial conditions: $y(0) = C_1 = 2$ .
$y'(t) = C_2 e^{-2t} - 2(C_1 + C_2 t)e^{-2t} = \big((C_2 - 2C_1) - 2C_2 t\big)e^{-2t}$ .
$y'(0) = C_2 - 2(2) = 1 \implies C_2 = 5$ .
Particular solution: $y(t) = (2 + 5t)e^{-2t}.$
Step 3: Find maximum: Set $y'(t) = 0$ : $\big((5 - 4) - 10t\big)e^{-2t} = (1 - 10t)e^{-2t} = 0 \implies 1 - 10t = 0 \implies t_{\max} = \frac{1}{10} = 0.1.$
Maximum value: $y(0.1) = (2 + 0.5)e^{-0.2} = 2.5e^{-0.2} \approx 2.0468$ .

Solve the Cauchy-Euler differential equation $x^2 y'' - x y' + 5y = 0$ for $x \gt 0$ with $y(1) = 2$ and $y'(1) = 4$ .

Advanced

Step 1: Auxiliary equation: Substitute $y = x^m$ : $m(m - 1) - m + 5 = m^2 - 2m + 5 = 0 \implies m = \frac{2 \pm \sqrt{4 - 20}}{2} = 1 \pm 2i.$ For complex roots $\alpha \pm i\beta$ , the real fundamental basis is $\{x^\alpha \cos(\beta\ln x), x^\alpha \sin(\beta\ln x)\}$ .
General solution: $y(x) = x\big(C_1 \cos(2\ln x) + C_2 \sin(2\ln x)\big).$ Step 2: Apply $y(1) = 2$ : Since $\ln 1 = 0$ : $y(1) = 1(C_1 \cos 0 + 0) = C_1 = 2$ .
Step 3: Derivative: $y'(x) = \big(C_1 \cos(2\ln x) + C_2 \sin(2\ln x)\big) + x\left(-\frac{2C_1}{x}\sin(2\ln x) + \frac{2C_2}{x}\cos(2\ln x)\right).$ $y'(1) = (C_1 + 0) + 2C_2 = 2 + 2C_2 = 4 \implies 2C_2 = 2 \implies C_2 = 1.$ Step 4: Particular solution: $y(x) = x\big(2\cos(2\ln x) + \sin(2\ln x)\big).$

D10

Find the general solution of $y'' - 4y' + 3y = 2x + e^{2x}$ .

Intermediate

Step 1: Homogeneous: $r^2 - 4r + 3 = (r - 1)(r - 3) = 0 \implies y_c = C_1 e^x + C_2 e^{3x}$ .
Step 2: Particular trial form: Neither $2x$ nor $e^{2x}$ is in $y_c$ , so trial is $y_p = (Ax + B) + C e^{2x}$ .
Step 3: Match polynomial coefficients: $y_{p1} = Ax + B \implies y_{p1}' = A, y_{p1}'' = 0$ . $0 - 4A + 3(Ax + B) = 3Ax + (3B - 4A) = 2x \implies 3A = 2 \implies A = 2/3,$ $3B - 4(2/3) = 0 \implies 3B = 8/3 \implies B = 8/9.$ Step 4: Match exponential: $y_{p2} = C e^{2x} \implies y_{p2}'' - 4y_{p2}' + 3y_{p2} = (4 - 8 + 3)C e^{2x} = -C e^{2x} = e^{2x} \implies C = -1$ .
Step 5: General solution: $y(x) = C_1 e^x + C_2 e^{3x} + \frac{2}{3}x + \frac{8}{9} - e^{2x}.$

D11

Solve the undamped resonant initial value problem $y'' + 9y = 6\cos(3t)$ with $y(0) = 0$ and $y'(0) = 1$ . Identify the secular growth term.

Advanced

Step 1: Homogeneous: $r^2 + 9 = 0 \implies y_c = C_1 \cos 3t + C_2 \sin 3t$ .
Step 2: Trial with resonance: Forcing $6\cos 3t$ matches $y_c$ (multiplicity $s=1$ ). Trial: $y_p(t) = t(A\cos 3t + B\sin 3t).$ Differentiating and substituting: $y_p'' + 9y_p = -6A\sin 3t + 6B\cos 3t = 6\cos 3t \implies B = 1, \quad A = 0.$ Thus $y_p(t) = t\sin(3t)$ .
Step 3: Apply initial conditions: $y(t) = C_1 \cos 3t + C_2 \sin 3t + t\sin 3t$ .
$y(0) = C_1 = 0$ .
$y'(t) = 3C_2 \cos 3t + \sin 3t + 3t\cos 3t \implies y'(0) = 3C_2 = 1 \implies C_2 = 1/3$ .
Step 4: Particular solution: $y(t) = \frac{1}{3}\sin(3t) + t\sin(3t).$ The term $t\sin(3t)$ is the secular growth term whose amplitude increases linearly with $t$ .

D12

Solve the double-resonance equation $y'' - 6y' + 9y = 4e^{3t}$ with $y(0) = 1$ and $y'(0) = 3$ .

Advanced

Step 1: Homogeneous: $r^2 - 6r + 9 = (r - 3)^2 = 0 \implies y_c(t) = (C_1 + C_2 t)e^{3t}$ .
Step 2: Resonance multiplication: Both $e^{3t}$ and $t e^{3t}$ are in $y_c$ (multiplicity $s = 2$ ). Multiply trial by $t^2$ : $y_p(t) = A t^2 e^{3t}.$ Using operator $(D - 3)^2[A t^2 e^{3t}] = e^{3t} D^2[A t^2] = e^{3t}(2A) = 4e^{3t} \implies 2A = 4 \implies A = 2$ .
Thus $y_p(t) = 2t^2 e^{3t}$ .
Step 3: Initial conditions: $y(t) = (C_1 + C_2 t + 2t^2)e^{3t}$ .
$y(0) = C_1 = 1$ .
$y'(t) = (C_2 + 4t)e^{3t} + 3(1 + C_2 t + 2t^2)e^{3t} \implies y'(0) = C_2 + 3 = 3 \implies C_2 = 0$ .
Particular solution: $y(t) = (1 + 2t^2)e^{3t}.$

D13

Find a particular solution of $y'' + 4y = \sec(2x)$ for $x \in (-\pi/4, \pi/4)$ using Variation of Parameters.

Advanced

Step 1: Basis & Wronskian: $y_1 = \cos 2x, \, y_2 = \sin 2x$ .
$W = \cos 2x(2\cos 2x) - (-2\sin 2x)\sin 2x = 2$ .
Step 2: Parameter integrals: $u_1'(x) = -\frac{y_2(x)g(x)}{W} = -\frac{\sin 2x \sec 2x}{2} = -\frac{1}{2}\tan 2x \implies u_1(x) = \frac{1}{4}\ln|\cos 2x|.$ $u_2'(x) = \frac{y_1(x)g(x)}{W} = \frac{\cos 2x \sec 2x}{2} = \frac{1}{2} \implies u_2(x) = \frac{x}{2}.$ Step 3: Particular solution: $y_p(x) = u_1 y_1 + u_2 y_2 = \frac{1}{4}\cos(2x)\ln|\cos 2x| + \frac{x}{2}\sin(2x).$

D14

Find a particular solution of $y'' + 2y' + y = \dfrac{e^{-x}}{x}$ for $x \gt 0$ using Variation of Parameters.

Advanced

Step 1: Basis & Wronskian: $y_1 = e^{-x}, \, y_2 = x e^{-x}$ .
$W = e^{-x}(e^{-x} - x e^{-x}) - (-e^{-x})x e^{-x} = e^{-2x}$ .
Step 2: Parameter integrals: $u_1'(x) = -\frac{x e^{-x}(e^{-x}/x)}{e^{-2x}} = -1 \implies u_1(x) = -x.$ $u_2'(x) = \frac{e^{-x}(e^{-x}/x)}{e^{-2x}} = \frac{1}{x} \implies u_2(x) = \ln x.$ Step 3: Particular solution: $y_p(x) = (-x)e^{-x} + (\ln x)x e^{-x} = x e^{-x}(\ln x - 1).$

D15

Construct the Green's kernel $G(x, t)$ for $L[y] = y'' - y$ , and express the particular solution of $y'' - y = \dfrac{1}{e^x + 1}$ satisfying $y(0) = 0, y'(0) = 0$ as an explicit definite integral.

Advanced

Step 1: Basis & Wronskian: $y_1(x) = e^x, \, y_2(x) = e^{-x}$ . $W(t) = e^t(-e^{-t}) - e^t(e^{-t}) = -2$ .
Step 2: Green's kernel: $G(x, t) = \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(t)} = \frac{e^t e^{-x} - e^x e^{-t}}{-2} = \frac{e^{x - t} - e^{-(x - t)}}{2} = \sinh(x - t).$ Step 3: Integral representation: $y(x) = \int_0^x \sinh(x - t) \cdot \frac{1}{e^t + 1}\,dt.$ Verifying: $G(t, t) = \sinh(0) = 0$ , and $\left.\frac{\partial G}{\partial x}\right|_{x=t} = \cosh(0) = 1$ , satisfying the universal Cauchy impulse properties.

D16

Solve the initial value problem with step forcing using the Laplace transform: $y'' + 4y = 8u(t - 2), \qquad y(0) = 0, \quad y'(0) = 0.$

Intermediate

Step 1: Transform equation: $(s^2 + 4)Y(s) = \frac{8e^{-2s}}{s} \implies Y(s) = 8e^{-2s} \cdot \frac{1}{s(s^2 + 4)}.$ Step 2: Partial fractions on transfer factor: $\frac{1}{s(s^2 + 4)} = \frac{1/4}{s} - \frac{s/4}{s^2 + 4} \implies \mathcal{L}^{-1}\left\{\frac{8}{s(s^2 + 4)}\right\} = 2 - 2\cos(2t).$ Step 3: Apply Second Shift Theorem: $y(t) = u(t - 2)\big(2 - 2\cos(2(t - 2))\big).$

D17

Solve $y'' + 2y' + 2y = 3\delta(t - \pi)$ with $y(0) = 1$ and $y'(0) = 0$ using the Laplace transform.

Advanced

Step 1: Transform: $(s^2 + 2s + 2)Y(s) - s(1) - 2(1) = 3e^{-\pi s} \implies (s^2 + 2s + 2)Y(s) = s + 2 + 3e^{-\pi s}.$ Complete square in denominator: $s^2 + 2s + 2 = (s + 1)^2 + 1$ .
Write numerator as $(s + 1) + 1$ : $Y(s) = \frac{s + 1}{(s + 1)^2 + 1} + \frac{1}{(s + 1)^2 + 1} + 3e^{-\pi s} \cdot \frac{1}{(s + 1)^2 + 1}.$ Step 2: Inverse transform:
For unshifted terms: $e^{-t}\cos t + e^{-t}\sin t$ .
For shifted term: $3u(t - \pi)e^{-(t - \pi)}\sin(t - \pi) = -3u(t - \pi)e^{-(t - \pi)}\sin t$ .
Thus: $y(t) = e^{-t}(\cos t + \sin t) - 3u(t - \pi)e^{-(t - \pi)}\sin t.$

D18

Use the Laplace transform and the Convolution Theorem to solve the integral equation $y(t) = t^2 + \int_0^t e^{t - \tau} y(\tau)\,d\tau.$

Advanced

Step 1: Recognize convolution: The equation is $y(t) = t^2 + e^t * y(t)$ .
Step 2: Transform both sides: $Y(s) = \frac{2}{s^3} + \left(\frac{1}{s - 1}\right)Y(s) \implies Y(s)\left(1 - \frac{1}{s - 1}\right) = \frac{2}{s^3}.$ $Y(s)\left(\frac{s - 2}{s - 1}\right) = \frac{2}{s^3} \implies Y(s) = \frac{2(s - 1)}{s^3(s - 2)}.$ Step 3: Partial fractions: $\frac{2(s - 1)}{s^3(s - 2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s - 2}.$ By cover-up, $D = \frac{2(2 - 1)}{2^3} = \frac{2}{8} = \frac{1}{4}$ .
$C = \frac{2(0 - 1)}{0 - 2} = 1$ .
Matching terms gives $A = -1/4$ , $B = -1/2$ .
Thus: $Y(s) = -\frac{1/4}{s} - \frac{1/2}{s^2} + \frac{1}{s^3} + \frac{1/4}{s - 2}.$ Step 4: Inverse transform: $y(t) = -\frac{1}{4} - \frac{1}{2}t + \frac{1}{2}t^2 + \frac{1}{4}e^{2t}.$

D19

Find the general solution of the linear system $\mathbf{x}' = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\mathbf{x}$ . Classify the equilibrium at the origin and determine the cartesian equation of the stable separatrix line.

Intermediate

Step 1: Characteristic polynomial: $\det\begin{pmatrix} 1 - \lambda & 3 \\ 3 & 1 - \lambda \end{pmatrix} = (1 - \lambda)^2 - 9 = \lambda^2 - 2\lambda - 8 = (\lambda - 4)(\lambda + 2) = 0.$ $\lambda_1 = 4, \qquad \lambda_2 = -2.$ Since $\Delta = \det(A) = -8 \lt 0$ , the origin is a saddle point.
Step 2: Eigenvectors:
For $\lambda_1 = 4$ : $\begin{pmatrix} -3 & 3 \\ 3 & -3 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .
For $\lambda_2 = -2$ : $\begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .
Step 3: General solution: $\mathbf{x}(t) = C_1 e^{4t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}.$ The stable separatrix corresponds to $C_1 = 0$ , lying along the line spanned by $(1, -1)^T$ : $y = -x.$

D20

Find the general real solution of $\mathbf{x}' = \begin{pmatrix} -2 & -5 \\ 1 & 0 \end{pmatrix}\mathbf{x}$ , classify the origin, and determine whether trajectories rotate clockwise or counterclockwise.

Intermediate

Step 1: Characteristic polynomial: $\det\begin{pmatrix} -2 - \lambda & -5 \\ 1 & -\lambda \end{pmatrix} = \lambda(\lambda + 2) + 5 = \lambda^2 + 2\lambda + 5 = 0 \implies \lambda = -1 \pm 2i.$ Since $\alpha = -1 \lt 0$ , the origin is an asymptotically stable spiral sink.
Step 2: Eigenvector for $\lambda = -1 + 2i$ : $\begin{pmatrix} -1 - 2i & -5 \\ 1 & 1 - 2i \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0} \implies v_1 + (1 - 2i)v_2 = 0 \implies \mathbf{v} = \begin{pmatrix} -1 + 2i \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} + i \begin{pmatrix} 2 \\ 0 \end{pmatrix}.$ Step 3: Real and imaginary parts: $e^{(-1 + 2i)t}\mathbf{v} = e^{-t}(\cos 2t + i\sin 2t)\left[\begin{pmatrix} -1 \\ 1 \end{pmatrix} + i \begin{pmatrix} 2 \\ 0 \end{pmatrix}\right]$ $= e^{-t}\left[\cos 2t \begin{pmatrix} -1 \\ 1 \end{pmatrix} - \sin 2t \begin{pmatrix} 2 \\ 0 \end{pmatrix}\right] + i e^{-t}\left[\sin 2t \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \cos 2t \begin{pmatrix} 2 \\ 0 \end{pmatrix}\right].$ Fundamental real solutions: $\mathbf{x}_1(t) = e^{-t}\begin{pmatrix} -\cos 2t - 2\sin 2t \\ \cos 2t \end{pmatrix}, \qquad \mathbf{x}_2(t) = e^{-t}\begin{pmatrix} -\sin 2t + 2\cos 2t \\ \sin 2t \end{pmatrix}.$ $\mathbf{x}(t) = C_1 \mathbf{x}_1(t) + C_2 \mathbf{x}_2(t).$ Step 4: Rotational sense: At $(1, 0)$ : $\mathbf{x}' = A(1, 0)^T = (-2, 1)^T$ . Since $x_2' = +1 \gt 0$ , rotation is counterclockwise.

D21

Solve the defective linear system $\mathbf{x}' = \begin{pmatrix} 1 & -1 \\ 1 & 3 \end{pmatrix}\mathbf{x}$ with $\mathbf{x}(0) = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ .

Advanced

Step 1: Eigenvalues: $\det\begin{pmatrix} 1 - \lambda & -1 \\ 1 & 3 - \lambda \end{pmatrix} = (1 - \lambda)(3 - \lambda) + 1 = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 = 0 \implies \lambda = 2.$ Step 2: Eigenvector: $(A - 2I)\mathbf{v}_1 = \begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix}\mathbf{v}_1 = \mathbf{0} \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.$ Step 3: Generalized eigenvector: Solve $(A - 2I)\mathbf{v}_2 = \mathbf{v}_1$ : $\begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \implies -v_{21} - v_{22} = 1 \implies \mathbf{v}_2 = \begin{pmatrix} 0 \\ -1 \end{pmatrix}.$ Step 4: General solution: $\mathbf{x}(t) = C_1 e^{2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 e^{2t}\left[\begin{pmatrix} 0 \\ -1 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \end{pmatrix}\right].$ Apply $\mathbf{x}(0) = (2, 1)^T$ : $C_1 \begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 \begin{pmatrix} 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \implies C_1 = 2, \quad -C_1 - C_2 = 1 \implies C_2 = -3.$ Particular solution: $\mathbf{x}(t) = 2e^{2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} - 3e^{2t}\begin{pmatrix} t \\ -1 - t \end{pmatrix} = e^{2t}\begin{pmatrix} 2 - 3t \\ 1 + 3t \end{pmatrix}.$

D22

Find the power series recurrence relation about $x_0 = 0$ for Hermite's differential equation with parameter 2: $y'' - 2x y' + 4y = 0, \qquad y(0) = 1, \quad y'(0) = 0.$ Prove that the solution terminates to a quadratic polynomial and state the polynomial.

Advanced

Step 1: Series substitution: $\sum_{n=2}^\infty n(n-1)a_n x^{n-2} - 2\sum_{n=1}^\infty n a_n x^n + 4\sum_{n=0}^\infty a_n x^n = 0.$ Shift first index: $\sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k - \sum_{k=0}^\infty 2k a_k x^k + \sum_{k=0}^\infty 4a_k x^k = 0.$ Step 2: Recurrence relation: $(k+2)(k+1)a_{k+2} - 2(k - 2)a_k = 0 \implies a_{k+2} = \frac{2(k - 2)}{(k+2)(k+1)} a_k.$ Step 3: Initial conditions: $a_0 = 1, \, a_1 = 0$ .
Since $a_1 = 0$ , all odd coefficients vanish: $a_3 = a_5 = a_7 = \dots = 0$ .
For $k = 0$ : $a_2 = \frac{2(0 - 2)}{(2)(1)} a_0 = -2(1) = -2$ .
For $k = 2$ : $a_4 = \frac{2(2 - 2)}{(4)(3)} a_2 = 0$ .
Since $a_4 = 0$ , the recurrence terminates identically: $a_6 = a_8 = \dots = 0$ .
Step 4: Exact solution: $y(x) = 1 - 2x^2.$ (Up to scaling, this is the 2nd Hermite polynomial $H_2(x) = 4x^2 - 2 = -2(1 - 2x^2)$ ).

D23

For the differential equation with a regular singular point at $x_0 = 0$ : $2x^2 y'' + x y' - (x + 1)y = 0,$ find the indicial equation, determine the two roots $r_1 \ge r_2$ , and find the first three non-zero terms of the Frobenius series solution corresponding to $r_1$ .

Advanced

Step 1: Indicial equation: $p_0 = \lim_{x\to 0} x P(x) = \lim_{x\to 0} x \left(\frac{x}{2x^2}\right) = \frac{1}{2}, \qquad q_0 = \lim_{x\to 0} x^2 Q(x) = \lim_{x\to 0} x^2 \left(-\frac{x + 1}{2x^2}\right) = -\frac{1}{2}.$ $F(r) = r(r - 1) + \frac{1}{2}r - \frac{1}{2} = r^2 - \frac{1}{2}r - \frac{1}{2} = 0 \implies 2r^2 - r - 1 = (2r + 1)(r - 1) = 0.$ Roots: $r_1 = 1, \, r_2 = -1/2$ . Difference $r_1 - r_2 = 3/2 \notin \mathbb{Z}$ .
Step 2: Frobenius series for $r_1 = 1$ : $y(x) = \sum_{n=0}^\infty a_n x^{n+1}$ .
Substituting into $2x^2 y'' + x y' - y = x y$ : $\sum_{n=0}^\infty \big[2(n+1)n + (n+1) - 1\big]a_n x^{n+1} = \sum_{n=0}^\infty a_n x^{n+2}$ $\sum_{n=0}^\infty n(2n + 3)a_n x^{n+1} = \sum_{k=1}^\infty a_{k-1} x^{k+1}.$ For $n = 0$ : $0 \cdot 3 a_0 = 0$ (satisfied for arbitrary $a_0 \neq 0$ ).
For $n \ge 1$ : $a_n = \frac{a_{n-1}}{n(2n + 3)}.$ Step 3: Coefficients:
$n = 1$ : $a_1 = \frac{a_0}{1(5)} = \frac{a_0}{5}$ .
$n = 2$ : $a_2 = \frac{a_1}{2(7)} = \frac{a_0}{5 \cdot 14} = \frac{a_0}{70}$ .
Step 4: Series solution: $y_1(x) = x\left(1 + \frac{x}{5} + \frac{x^2}{70} + \dots\right).$

D24

Consider the stiff decay equation $\dfrac{dy}{dt} = -100y$ with $y(0) = 1$ . Determine the stability threshold $h_{\text{crit}}$ for Forward Euler, and compute the value of $y(0.1)$ using Backward (Implicit) Euler with step size $h = 0.05$ .

Advanced

Step 1: Forward Euler stability threshold: For $y' = \lambda y$ with $\lambda = -100$ , the amplification factor is $1 + h\lambda = 1 - 100h$ . Stability requires: $|1 - 100h| \lt 1 \implies -1 \lt 1 - 100h \lt 1 \implies h \lt \frac{2}{100} = 0.02.$ Thus $h_{\text{crit}} = 0.02$ . If $h \gt 0.02$ , Forward Euler explodes uncontrollably.
Step 2: Backward (Implicit) Euler formula: $y_{n+1} = y_n + h(-100y_{n+1}) \implies (1 + 100h)y_{n+1} = y_n \implies y_{n+1} = \frac{1}{1 + 100h}y_n.$ With $h = 0.05$ : $1 + 100(0.05) = 1 + 5 = 6$ .
Amplification factor per step is $\frac{1}{6} \lt 1$ .
Step 3: Two steps to $t = 0.1$ :
Step 1 ( $t = 0.05$ ): $y_1 = \frac{1}{6}(1) = \frac{1}{6} \approx 0.1667$ .
Step 2 ( $t = 0.10$ ): $y_2 = \frac{1}{6}\left(\frac{1}{6}\right) = \frac{1}{36} \approx 0.0278$ .
Notice that while Forward Euler with $h = 0.05$ produces $y_2 = (1 - 5)^2 = (-4)^2 = 16$ (false physical explosion), Backward Euler stably and monotonically decays to $1/36$ , remaining unconditionally stable.

D25

Consider the separable differential equation $\dfrac{dy}{dx} = ky$ where $k$ is a positive constant. A solution curve passes through the points $(0, 5)$ and $(2, 20)$ . Determine the value of $k$ and then find the value of $y$ when $x = 3$ .

Intermediate

The differential equation $\dfrac{dy}{dx} = ky$ is separable. Separating variables gives $\frac{dy}{y} = k \, dx$ . Integrating both sides, we get $\int \frac{dy}{y} = \int k \, dx$ , which yields $\ln|y| = kx + C_1$ . Exponentiating both sides gives $|y| = e^{kx+C_1} = e^{C_1}e^{kx}$ . Let $C = \pm e^{C_1}$ , so $y = Ce^{kx}$ . \nWe are given that the solution curve passes through $(0, 5)$ . Substituting these values into $y = Ce^{kx}$ : $5 = Ce^{k(0)} \Rightarrow 5 = C$ . Thus, the particular solution is $y = 5e^{kx}$ . \nWe are also given that the solution curve passes through $(2, 20)$ . Substituting these values into $y = 5e^{kx}$ : $20 = 5e^{k(2)} \Rightarrow 4 = e^{2k}$ . Taking the natural logarithm of both sides: $\ln 4 = \ln(e^{2k}) \Rightarrow \ln 4 = 2k \Rightarrow k = \dfrac{\ln 4}{2} = \dfrac{2 \ln 2}{2} = \ln 2$ . \nSo, the value of $k$ is $\ln 2$ . The particular solution is $y = 5e^{(\ln 2)x} = 5(e^{\ln 2})^x = 5(2^x)$ . \nFinally, we need to find the value of $y$ when $x = 3$ . Substituting $x = 3$ into $y = 5(2^x)$ : $y(3) = 5(2^3) = 5(8) = 40$ . \nTherefore, $k = \ln 2$ and $y(3) = 40$ .

Final Answer: $k = \ln 4, y(3) = 80$

D26

Solve $y'+4y=4 e^{- t}$ , $y(0)=3$ with the Laplace transform. Find $y(1)$ .

Advanced

$Y(s)=\frac{3 s + 7}{\left(s + 1\right) \left(s + 4\right)}=\frac{5}{3 \left(s + 4\right)} + \frac{4}{3 \left(s + 1\right)}$ . Inverting, $y(t)=\left(\frac{4}{3} + \frac{5 e^{- 3 t}}{3}\right) e^{- t}$ , so $y(1)=\frac{5 + 4 e^{3}}{3 e^{4}}$ .

Final Answer: $\frac{5 + 4 e^{3}}{3 e^{4}}$

D27

Solve $x^2 y' + x y = \frac{1}{x^2 + 1}$ for $x \gt 0$ , with $y(1) = \frac{\pi}{4}$ . What is $y(\sqrt{3})$ ?

Advanced

Standard form: $y' + \frac{1}{x}y = \frac{1}{x(x^2 + 1)}$ . $\mu = x$ . Multiplying gives $(x y)' = \frac{1}{x(x^2 + 1)} = \frac{1}{x} - \frac{x}{x^2 + 1}$ . Integrating gives $x y = \ln x - \frac{1}{2}\ln(x^2 + 1) + C$ . Using $y(1) = \pi/4$ yields $\pi/4 = -\frac{1}{2}\ln 2 + C \implies C = \frac{\pi}{4} + \frac{1}{2}\ln 2$ . At $x = \sqrt{3}$ , $\sqrt{3}y = \frac{1}{2}\ln 3 - \frac{1}{2}\ln 4 + \frac{\pi}{4} + \frac{1}{2}\ln 2 = \frac{\pi}{4} + \frac{1}{2}\ln(3/2)$ . Thus $y(\sqrt{3}) = \frac{1}{\sqrt{3}}\left(\frac{\pi}{4} + \frac{1}{2}\ln(3/2)\right)$ .

Final Answer: $\frac{1}{\sqrt{3}}\left(\frac{\pi}{4} + \frac{1}{2}\ln(3/2)\right)$

D28

Consider the linear first-order differential equation $\frac{dy}{dx} + 2y = 6$ . If the initial condition is $y(0) = 1$ , find the value of $y(1)$ .

Basic

The given differential equation is a linear first-order ODE of the form $\frac{dy}{dx} + P(x)y = Q(x)$ , where $P(x) = 2$ and $Q(x) = 6$ . The integrating factor $\mu(x)$ is given by $\mu(x) = e^{\int P(x) dx}$ .

Calculating the integrating factor: $\mu(x) = e^{\int 2 dx} = e^{2x}$ .

Multiplying the differential equation by the integrating factor: $e^{2x}\frac{dy}{dx} + 2e^{2x}y = 6e^{2x}$ .
The left side is the derivative of the product $y\mu(x)$ : $\frac{d}{dx}(ye^{2x}) = 6e^{2x}$ .

Integrating both sides with respect to $x$ : $ye^{2x} = \int 6e^{2x} dx$ .
$ye^{2x} = 6 \cdot \frac{1}{2}e^{2x} + C$ .
$ye^{2x} = 3e^{2x} + C$ .

Solving for $y$ : $y = 3 + Ce^{-2x}$ .

Using the initial condition $y(0) = 1$ to find $C$ : $1 = 3 + Ce^{-2(0)}$ .
$1 = 3 + C$ .
$C = 1 - 3 = -2$ .

So, the particular solution is $y(x) = 3 - 2e^{-2x}$ .

Now, we need to find $y(1)$ : $y(1) = 3 - 2e^{-2(1)} = 3 - 2e^{-2}$ .

Final Answer: $3 - 2e^{-1}$

D29

Solve $y''+3y'+2y=20e^{3x}-4x$ , $y(0)=-1$ , $y'(0)=8$ . Find $y(1)$ .

Advanced

Homogeneous: $r^2+3r+2=0$ has roots $r=-1,-2$ , so $y_c=C_1e^{-x}+C_2e^{-2x}$ . Split the forcing (the superposition note): for $20e^{3x}$ , $3$ isn't a root, trial $Ae^{3x}$ ; for $-4x$ , trial $Bx+C$ (or just $C$ ). Adding the two pieces: $y_p=-2x+e^{3x}+3$ . With $C_1=-3$ , $C_2=-2$ (matching the stated initial conditions), $y=y_c+y_p$ , and $y(1)$ follows by substitution.

Final Answer: $-\frac{3}{e}-\frac{2}{e^{2}}+1+e^{3}$

D30

Consider the second-order nonhomogeneous linear differential equation $y'' - 2y' + y = e^x \sin(x)$ . Let $y_p(x)$ be a particular solution obtained using the method of undetermined coefficients. Determine the value of $y_p''(\pi) - 2y_p'(\pi) + y_p(\pi)$ .

Advanced

The problem asks for the value of the left-hand side of the differential equation $y'' - 2y' + y = e^x \sin(x)$ when evaluated at a particular solution $y_p(x)$ . By definition, a particular solution $y_p(x)$ must satisfy the nonhomogeneous differential equation for all $x$ . Therefore, when we substitute $y_p(x)$ into the differential equation, the left-hand side must equal the right-hand side.

Specifically, for any particular solution $y_p(x)$ , it must be true that $y_p''(x) - 2y_p'(x) + y_p(x) = e^x \sin(x)$ for all $x$ .

The question asks for the value of $y_p''(\pi) - 2y_p'(\pi) + y_p(\pi)$ . This is simply the evaluation of the left-hand side of the differential equation at $x = \pi$ . Since $y_p(x)$ is a particular solution, this expression must be equal to the right-hand side of the differential equation evaluated at $x = \pi$ .

So, $y_p''(\pi) - 2y_p'(\pi) + y_p(\pi) = e^\pi \sin(\pi)$ .

We know that $\sin(\pi) = 0$ .

Therefore, $y_p''(\pi) - 2y_p'(\pi) + y_p(\pi) = e^\pi \times 0 = 0$ .

Final Answer: $0$

D31

On $[0,1]$ , Euler's method with $h=0.1$ gives an error of about $\frac{1}{50}$ . Roughly how many steps are needed for an error of $\frac{1}{5000}$ ?

Intermediate

The error scales like $h^{1}$ ; reducing it by $100$ needs $h$ smaller by $100$ , i.e. $h=\frac{1}{1000}$ , which is 1000 steps.

Final Answer: $1000$

Key Mathematical Principle & Verification: In exam revision, always verify that the final value satisfies the domain constraints, units, and original governing equation by direct back-substitution into the prompt.

D32

Consider the differential equation $\dfrac{dy}{dx} = x + y$ with the initial condition $y(0) = 1$ . You are asked to approximate $y(0.2)$ using Euler's method. If the step size $h = 0.1$ is used, and it is known that the exact solution at $x=0.2$ is $y(0.2) = 1.24$ , determine the value of $k$ such that $y(0.2)_{\text{Euler}} = 1.24 - k$ .

Intermediate

Euler's method for approximating the solution to $\dfrac{dy}{dx} = f(x, y)$ with initial condition $y(x_0) = y_0$ and step size $h$ is given by $y_{n+1} = y_n + h f(x_n, y_n)$ .

In this problem, $f(x, y) = x + y$ , $x_0 = 0$ , $y_0 = 1$ , and $h = 0.1$ .

We want to approximate $y(0.2)$ with $h=0.1$ . This requires two steps:

Step 1: From $x_0 = 0$ to $x_1 = 0.1$ .
$y_1 = y_0 + h f(x_0, y_0)$
$y_1 = 1 + 0.1 (0 + 1)$
$y_1 = 1 + 0.1 (1)$
$y_1 = 1 + 0.1$
$y_1 = 1.1$

Step 2: From $x_1 = 0.1$ to $x_2 = 0.2$ .
$y_2 = y_1 + h f(x_1, y_1)$
$y_2 = 1.1 + 0.1 (0.1 + 1.1)$
$y_2 = 1.1 + 0.1 (1.2)$
$y_2 = 1.1 + 0.12$
$y_2 = 1.22$

So, the approximation using Euler's method is $y(0.2)_{\text{Euler}} = 1.22$ .

We are given that the exact solution is $y(0.2) = 1.24$ and $y(0.2)_{\text{Euler}} = 1.24 - k$ .
Substituting the value of $y(0.2)_{\text{Euler}}$ , we get:
$1.22 = 1.24 - k$
$k = 1.24 - 1.22$
$k = 0.02$

Therefore, the value of $k$ is 0.02.

D33

Find the coefficient $a_5$ of the power series solution $y=\sum a_nx^n$ of $y''-xy'-y=0$ with $y(0)=1$ , $y'(0)=2$ .

Advanced

Substitute the series and match powers of $x$ to get a recurrence; with $a_0=1,\ a_1=2$ the coefficients are $a_0=1, a_1=2, a_2=\frac{1}{2}, a_3=\frac{2}{3}, a_4=\frac{1}{8}, a_5=\frac{2}{15}$ .

Final Answer: $\frac{2}{15}$

D34

Solve $y''+4y'+13y=0$ , $y(0)=-3$ , $y'(0)=-6$ . Find $y(\frac{\pi}{9})$ .

Advanced

Auxiliary equation $r^2+4r+13=0$ has roots $r=-2\pm3i$ , so $y=e^{-2x}(C_1\cos3x+C_2\sin3x)$ . $y(0)=C_1=-3$ ; $y'(0)=-2C_1+3C_2=-6\Rightarrow C_2=-4$ . Evaluating at $x=\frac{\pi}{9}$ (where $3x=\frac{\pi}{3}$ ) gives the result.

Final Answer: $-\frac{2\sqrt{3}}{e^{\frac{2\pi}{9}}}-\frac{3}{2e^{\frac{2\pi}{9}}}$

D35

Consider the second-order linear homogeneous differential equation $y'' - 2y' + y = 0$ . \\ If one of the fundamental solutions is $y_1(x) = e^x$ , and the Wronskian of two linearly independent solutions $y_1(x)$ and $y_2(x)$ is given by $W(y_1, y_2)(x) = 5e^{2x}$ , find the unknown parameter $k$ in $y_2(x) = kxe^x$ .

Intermediate

The given differential equation is $y'' - 2y' + y = 0$ . The characteristic equation is $r^2 - 2r + 1 = 0$ , which factors as $(r-1)^2 = 0$ . Thus, the roots are repeated: $r_1 = r_2 = 1$ . \\
One fundamental solution is given as $y_1(x) = e^x$ . For a second-order linear homogeneous equation $y'' + p(x)y' + q(x)y = 0$ with a known solution $y_1(x)$ , the Wronskian is given by $W(y_1, y_2)(x) = y_1(x)y_2'(x) - y_1'(x)y_2(x) = C e^{-\int p(x) dx}$ . \\
In our equation, $p(x) = -2$ . So, $-\int p(x) dx = -\int (-2) dx = 2x$ . Therefore, $W(y_1, y_2)(x) = C e^{2x}$ . \\
We are given that $W(y_1, y_2)(x) = 5e^{2x}$ , which means $C = 5$ . \\
We are also given $y_1(x) = e^x$ and $y_2(x) = kxe^x$ . \\
Let's compute the Wronskian using these forms: \\
$y_1'(x) = e^x$ \\
$y_2'(x) = \frac{d}{dx}(kxe^x) = k(1 \cdot e^x + x \cdot e^x) = ke^x(1+x)$
\\
Now, substitute into the Wronskian formula: \\
$W(y_1, y_2)(x) = y_1(x)y_2'(x) - y_1'(x)y_2(x)$
$W(y_1, y_2)(x) = e^x \cdot [ke^x(1+x)] - e^x \cdot [kxe^x]$
$W(y_1, y_2)(x) = ke^{2x}(1+x) - kxe^{2x}$
$W(y_1, y_2)(x) = ke^{2x} + kxe^{2x} - kxe^{2x}$
$W(y_1, y_2)(x) = ke^{2x}$
\\
We equate this to the given Wronskian: \\
$ke^{2x} = 5e^{2x}$
\\
Therefore, $k = 5$ .

D36

Solve $\frac{dy}{dx} = \frac{1 + y^2}{x^2 - 4}$ on $(-2, 2)$ with $y(0) = 1$ . At what value of $x$ does $y(x) \to +\infty$ ?

Advanced

Separable: $\frac{dy}{1 + y^2} = \frac{dx}{x^2 - 4} \implies \arctan y = \frac{1}{4}\ln\left(\frac{2 - x}{2 + x}\right) + C$ . Using $y(0) = 1$ gives $\arctan(1) = \frac{\pi}{4} = 0 + C \implies C = \frac{\pi}{4}$ . Blow-up $y \to \infty$ occurs when $\arctan y \to \frac{\pi}{2}$ , so $\frac{1}{4}\ln\left(\frac{2 - x}{2 + x}\right) + \frac{\pi}{4} = \frac{\pi}{2} \implies \ln\left(\frac{2 - x}{2 + x}\right) = \pi \implies \frac{2 - x}{2 + x} = e^\pi$ . Solving for $x$ : $2 - x = 2e^\pi + xe^\pi \implies x(1 + e^\pi) = 2(1 - e^\pi) \implies x = -2\frac{e^\pi - 1}{e^\pi + 1} = -2\tanh(\pi/2)$ .

Final Answer: $-2\tanh(\pi/2)$

D37

Consider the system of linear differential equations: $\begin{aligned} \frac{dx}{dt} &= ax + by \\ \frac{dy}{dt} &= cx + dy \end{aligned}$ where $a, b, c, d$ are constants. The characteristic equation for the system is $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$ . We are given that the eigenvalues of the system are $\lambda_1 = 2$ and $\lambda_2 = -1$ . The trace of the coefficient matrix, $T = a+d$ , is $1$ . Determine the value of the determinant of the coefficient matrix, $D = ad-bc$ .

Intermediate

The characteristic equation of a 2x2 system of linear differential equations $\begin{aligned} \frac{dx}{dt} &= ax + by \\ \frac{dy}{dt} &= cx + dy \end{aligned}$ is given by $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$ . This can be written in terms of the trace (T = a+d) and determinant (D = ad-bc) of the coefficient matrix as $\lambda^2 - T\lambda + D = 0$ . We are given that the eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = -1$ . The sum of the roots of a quadratic equation $p\lambda^2 + q\lambda + r = 0$ is $-q/p$ , and the product of the roots is $r/p$ . For the characteristic equation $\lambda^2 - T\lambda + D = 0$ , the sum of the eigenvalues is $\lambda_1 + \lambda_2 = -(-T)/1 = T$ , and the product of the eigenvalues is $\lambda_1 \lambda_2 = D/1 = D$ . We are given that the trace $T = a+d = 1$ . Let's verify this with the given eigenvalues: $\lambda_1 + \lambda_2 = 2 + (-1) = 1$ , which matches the given trace. We need to find the determinant $D = ad-bc$ . Using the product of the eigenvalues: $D = \lambda_1 \lambda_2 = (2)(-1) = -2$ .

Final Answer: $-2$

D38

Solve $\mathbf x'=\left[\begin{matrix}-6 & -4\\8 & 6\end{matrix}\right]\mathbf x$ , $\mathbf x(0)=\langle 1, 0 \rangle$ . Find $x_1(1)$ .

Advanced

Eigenpairs: $\lambda=2$ with $\langle 1, -2 \rangle$ , $\lambda=-2$ with $\langle -1, 1 \rangle$ . Matching $\mathbf x(0)$ gives $c_1=-1,\ c_2=-2$ , so $\mathbf x(t)=\langle - e^{2 t} + 2 e^{- 2 t}, 2 e^{2 t} - 2 e^{- 2 t} \rangle$ and $x_1(1)=\frac{2 - e^{4}}{e^{2}}$ .

Final Answer: $\frac{2 - e^{4}}{e^{2}}$

D39

Consider the differential equation $y'' - 3y' + 2y = f(t)$ with initial conditions $y(0) = 0$ and $y'(0) = 0$ . If the forcing function is $f(t) = u(t-2)$ , where $u(t)$ is the Heaviside step function, what is the Laplace transform of the solution $Y(s) = \mathcal{L}\{y(t)\}$ ?

Basic

The Laplace transform of the differential equation is $s^2Y(s) - sy(0) - y'(0) - 3(sY(s) - y(0)) + 2Y(s) = \mathcal{L}\{f(t)\}$ . Given $y(0) = 0$ and $y'(0) = 0$ , this simplifies to $s^2Y(s) - 3sY(s) + 2Y(s) = \mathcal{L}\{u(t-2)\}$ , so $Y(s)(s^2 - 3s + 2) = \mathcal{L}\{u(t-2)\}$ . The Laplace transform of $u(t-a)$ is $\frac{e^{-as}}{s}$ . Therefore, $\mathcal{L}\{u(t-2)\} = \frac{e^{-2s}}{s}$ . Thus, $Y(s) = \frac{e^{-2s}}{s(s^2 - 3s + 2)}$ . Factoring the denominator, $s^2 - 3s + 2 = (s-1)(s-2)$ . So, $Y(s) = \frac{e^{-2s}}{s(s-1)(s-2)}$ .

Final Answer: $\frac{1}{s(s-1)(s-2)}e^{-2s}$

D40

Find a particular solution of $y''-10y'+25y=e^{5x}x\ln x$ (for $x \gt 0$ ) using variation of parameters.

Advanced

Homogeneous: $r^2-10r+25=0$ has roots $r=5 \text{ (repeated)}$ , so $y_1=e^{5x}$ , $y_2=xe^{5x}$ . $u_1=-\frac{x^{3}\log{\left(x\right)}}{3}+\frac{x^{3}}{9}$ , $u_2=\frac{x^{2}\log{\left(x\right)}}{2}-\frac{x^{2}}{4}$ . So $y_p=\frac{x^{3}e^{5x}\log{\left(x\right)}}{6}-\frac{5x^{3}e^{5x}}{36}$ .

Final Answer: $\frac{x^{3}e^{5x}\log{\left(x\right)}}{6}-\frac{5x^{3}e^{5x}}{36}$

D41

Consider the second-order linear non-homogeneous differential equation: $\frac{d^2y}{dx^2} + 4y = \sec(2x)$ . Suppose a particular solution is found using the method of variation of parameters. Let the fundamental set of solutions for the associated homogeneous equation be $y_1(x) = \cos(2x)$ and $y_2(x) = \sin(2x)$ . If the particular solution is expressed in the form $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$ , determine the value of $u_2'(x)$ .

Advanced

The given differential equation is $\frac{d^2y}{dx^2} + 4y = \sec(2x)$ . The associated homogeneous equation is $\frac{d^2y}{dx^2} + 4y = 0$ . The characteristic equation is $r^2 + 4 = 0$ , which has roots $r = \pm 2i$ . Thus, the fundamental set of solutions for the homogeneous equation is $y_1(x) = \cos(2x)$ and $y_2(x) = \sin(2x)$ .

We use the method of variation of parameters to find a particular solution of the form $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$ .

The Wronskian of $y_1$ and $y_2$ is:
$W(y_1, y_2)(x) = y_1y_2' - y_1'y_2 = \cos(2x)(2\cos(2x)) - (-\sin(2x))(2\sin(2x))$
$W(y_1, y_2)(x) = 2\cos^2(2x) + 2\sin^2(2x) = 2(\cos^2(2x) + \sin^2(2x)) = 2$ .

The non-homogeneous term is $g(x) = \sec(2x)$ .

The formulas for $u_1'(x)$ and $u_2'(x)$ are:
$u_1'(x) = -\frac{y_2(x)g(x)}{W(x)}$
$u_2'(x) = \frac{y_1(x)g(x)}{W(x)}$

Substituting the given functions:
$u_2'(x) = \frac{\cos(2x) \cdot \sec(2x)}{2}$
$u_2'(x) = \frac{\cos(2x)}{2\cos(2x)}$
$u_2'(x) = \frac{1}{2}$ .

The value of $u_2'(x)$ is $\frac{1}{2}$ .

D42

Consider the differential equation $\frac{dy}{dx} = \frac{x^2 \cos(y)}{1 - x^3}$ with the initial condition $y(0) = \frac{\pi}{2}$ . If there exists a unique solution in a neighborhood of $x=0$ and $y=\frac{\pi}{2}$ , find the value of $y'(0)$ .

Advanced

The given differential equation is $\frac{dy}{dx} = \frac{x^2 \cos(y)}{1 - x^3}$ with the initial condition $y(0) = \frac{\pi}{2}$ .

We are asked to find $y'(0)$ . The differential equation directly gives us the expression for $\frac{dy}{dx}$ , which is $y'(x)$ .

To find $y'(0)$ , we substitute $x=0$ and $y=y(0)=\frac{\pi}{2}$ into the differential equation:

$y'(0) = \frac{(0)^2 \cos(y(0))}{1 - (0)^3}$

Substitute $y(0) = \frac{\pi}{2}$ :

$y'(0) = \frac{0 \cdot \cos(\frac{\pi}{2})}{1 - 0}$

We know that $\cos(\frac{\pi}{2}) = 0$ .

So, $y'(0) = \frac{0 \cdot 0}{1}$

$y'(0) = \frac{0}{1}$

$y'(0) = 0$

The existence and uniqueness theorem for first-order ODEs guarantees that if $f(x, y) = \frac{x^2 \cos(y)}{1 - x^3}$ and $\frac{\partial f}{\partial y} = \frac{x^2 (-\sin(y))}{1 - x^3}$ are continuous in a rectangular region containing $(0, \frac{\pi}{2})$ , then a unique solution exists. At $x=0$ , both $f(x, y)$ and $\frac{\partial f}{\partial y}$ are continuous since the denominator $1 - x^3$ is non-zero and the numerators are well-defined. Therefore, the condition for a unique solution is met.

D43

Solve $y'+2y=4 e^{3 t}$ , $y(0)=-3$ with the Laplace transform. Find $y(1)$ .

Advanced

$Y(s)=\frac{13 - 3 s}{\left(s - 3\right) \left(s + 2\right)}=- \frac{19}{5 \left(s + 2\right)} + \frac{4}{5 \left(s - 3\right)}$ . Inverting, $y(t)=\frac{4 e^{3 t}}{5} - \frac{19 e^{- 2 t}}{5}$ , so $y(1)=\frac{-19 + 4 e^{5}}{5 e^{2}}$ .

Final Answer: $\frac{-19 + 4 e^{5}}{5 e^{2}}$

D44

Solve $(x^2 - 1)y' + 2y = (x - 1)^2$ for $x \gt 1$ , with $y(2) = 0$ . What is $y(3)$ ?

Advanced

Standard form: $y' + \frac{2}{x^2 - 1}y = \frac{x - 1}{x + 1}$ . Integrating factor: $\mu = \frac{x - 1}{x + 1}$ . Multiplying gives $\left(\frac{x - 1}{x + 1}y\right)' = \left(\frac{x - 1}{x + 1}\right)^2 = 1 - \frac{4}{x + 1} + \frac{4}{(x + 1)^2}$ . Integrating gives $\frac{x - 1}{x + 1}y = x - 4\ln(x + 1) - \frac{4}{x + 1} + C$ . Using $y(2) = 0$ yields $0 = 2 - 4\ln 3 - 4/3 + C \implies C = 4\ln 3 - 2/3$ . At $x = 3$ , $\frac{1}{2}y(3) = 3 - 4\ln 4 - 1 + 4\ln 3 - 2/3 = 4/3 + 4\ln(3/4)$ . Multiplying by 2 gives $y(3) = \frac{8}{3} + 8\ln\left(\frac{3}{4}\right)$ .

Final Answer: $\frac{8}{3} + 8\ln\left(\frac{3}{4}\right)$

D45

Consider the first-order linear differential equation $\frac{dy}{dx} + 2y = 4$ . If the initial condition is $y(0) = 1$ , find the value of $y(1)$ .

Basic

The given differential equation is a first-order linear ODE of the form $\frac{dy}{dx} + P(x)y = Q(x)$ . Here, $P(x) = 2$ and $Q(x) = 4$ . The integrating factor is $I(x) = e^{\int P(x) dx} = e^{\int 2 dx} = e^{2x}$ . Multiplying the differential equation by the integrating factor, we get: $e^{2x} \frac{dy}{dx} + 2e^{2x}y = 4e^{2x}$ . The left side is the derivative of $y \cdot I(x)$ : $\frac{d}{dx}(ye^{2x}) = 4e^{2x}$ . Integrating both sides with respect to $x$ : $ye^{2x} = \int 4e^{2x} dx = 4 \cdot \frac{1}{2}e^{2x} + C = 2e^{2x} + C$ . Solving for $y$ : $y(x) = 2 + Ce^{-2x}$ . Using the initial condition $y(0) = 1$ : $1 = 2 + Ce^{0} \implies 1 = 2 + C \implies C = -1$ . Thus, the particular solution is $y(x) = 2 - e^{-2x}$ . Finally, we need to find $y(1)$ : $y(1) = 2 - e^{-2(1)} = 2 - e^{-2}$ .

Final Answer: $2 - e^{-2}$

D46

Solve $y''+4y'+3y=-35e^{4x}-9x$ , $y(0)=4$ , $y'(0)=-14$ . Find $y(1)$ .

Advanced

Homogeneous: $r^2+4r+3=0$ has roots $r=-1,-3$ , so $y_c=C_1e^{-x}+C_2e^{-3x}$ . Split the forcing (the superposition note): for $-35e^{4x}$ , $4$ isn't a root, trial $Ae^{4x}$ ; for $-9x$ , trial $Bx+C$ (or just $C$ ). Adding the two pieces: $y_p=-3x-e^{4x}+4$ . With $C_1=-2$ , $C_2=3$ (matching the stated initial conditions), $y=y_c+y_p$ , and $y(1)$ follows by substitution.

Final Answer: $-e^{4}-\frac{2}{e}+\frac{3}{e^{3}}+1$

D47

Consider the initial value problem $\frac{dy}{dx} = x + y$ with $y(0) = 1$ . Use the Euler method with step size $h = 0.5$ to approximate $y(1)$ .

Basic

The Euler method approximates the solution of an ODE $\frac{dy}{dx} = f(x, y)$ with initial condition $y(x_0) = y_0$ using the iterative formula $y_{i+1} = y_i + h f(x_i, y_i)$ , where $h$ is the step size and $x_i = x_0 + ih$ .

Given: $f(x, y) = x + y$ , $x_0 = 0$ , $y_0 = 1$ , and $h = 0.5$ .

We want to approximate $y(1)$ . Since $1 = 0 + 2 \times 0.5$ , we need two steps.

Step 1: Approximate $y(0.5)$ .
$x_1 = x_0 + h = 0 + 0.5 = 0.5$
$y_1 = y_0 + h f(x_0, y_0) = 1 + 0.5 (0 + 1) = 1 + 0.5(1) = 1.5$ .

Step 2: Approximate $y(1)$ .
$x_2 = x_1 + h = 0.5 + 0.5 = 1.0$
$y_2 = y_1 + h f(x_1, y_1) = 1.5 + 0.5 (0.5 + 1.5) = 1.5 + 0.5(2.0) = 1.5 + 1.0 = 2.5$ .

Therefore, the Euler approximation for $y(1)$ is 2.50.

D48

Consider the differential equation $\frac{dy}{dx} = \frac{\cos(x)}{y^2}$ for $y \gt 0$ and $x \in [0, \frac{\pi}{2}]$ . If $y(0) = 2$ , find the value of $y(\frac{\pi}{2})$ .

Advanced

The given differential equation is $\frac{dy}{dx} = \frac{\cos(x)}{y^2}$ . This is a separable differential equation. We can rewrite it as $y^2 dy = \cos(x) dx$ .
Integrating both sides, we get:
$\int y^2 dy = \int \cos(x) dx$
$\frac{y^3}{3} = \sin(x) + C$
We are given the initial condition $y(0) = 2$ . Substituting these values into the integrated equation:
$\frac{2^3}{3} = \sin(0) + C$
$\frac{8}{3} = 0 + C$
$C = \frac{8}{3}$
So, the particular solution is:
$\frac{y^3}{3} = \sin(x) + \frac{8}{3}$
We need to find the value of $y(\frac{\pi}{2})$ . Substitute $x = \frac{\pi}{2}$ into the particular solution:
$\frac{y(\frac{\pi}{2})^3}{3} = \sin(\frac{\pi}{2}) + \frac{8}{3}$
$\frac{y(\frac{\pi}{2})^3}{3} = 1 + \frac{8}{3}$
$\frac{y(\frac{\pi}{2})^3}{3} = \frac{3}{3} + \frac{8}{3}$
$\frac{y(\frac{\pi}{2})^3}{3} = \frac{11}{3}$
$y(\frac{\pi}{2})^3 = 11$
$y(\frac{\pi}{2}) = \sqrt[3]{11}$

D49

Consider the system of linear first-order differential equations given by: $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$ .\nFind the eigenvalues of the coefficient matrix.

Basic

The given system is of the form $\mathbf{x}' = A \mathbf{x}$ , where $A = \begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}$ is the coefficient matrix. To find the eigenvalues, we solve the characteristic equation det $(A - \lambda I) = 0$ , where $I$ is the identity matrix and $\lambda$ represents the eigenvalues.

$A - \lambda I = \begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1-\lambda & 2 \\ 3 & -\lambda \end{pmatrix}$
\ndet $(A - \lambda I) = (1-\lambda)(-\lambda) - (2)(3) = -\lambda + \lambda^2 - 6 = \lambda^2 - \lambda - 6$ \nSetting the determinant to zero:
$\lambda^2 - \lambda - 6 = 0$ \nFactoring the quadratic equation:
$(\lambda - 3)(\lambda + 2) = 0$ \nThus, the eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = -2$ .

D50

Consider the nonhomogeneous second-order linear differential equation $y'' - 5y' + 6y = 12x + 8$ . $y_h(x) = c_1e^{2x} + c_2e^{3x}$ is the general solution to the corresponding homogeneous equation. Find the particular solution $y_p(x)$ using the method of undetermined coefficients, and then determine the value of the constant $A$ such that $y_p(x) = Ax + B$ for some constant $B$ .

Intermediate

The given nonhomogeneous differential equation is $y'' - 5y' + 6y = 12x + 8$ . \nThe form of the nonhomogeneous term is $g(x) = 12x + 8$ , which is a polynomial of degree 1. Thus, we assume a particular solution of the form $y_p(x) = Ax + B$ .

We need to find the first and second derivatives of $y_p(x)$ :
$y_p'(x) = A$
$y_p''(x) = 0$

Substitute these into the differential equation:
$0 - 5(A) + 6(Ax + B) = 12x + 8$
$-5A + 6Ax + 6B = 12x + 8$
$6Ax + (6B - 5A) = 12x + 8$

Equating the coefficients of like powers of $x$ :
For the $x$ term: $6A = 12$ $\Rightarrow$ $A = 2$ .
For the constant term: $6B - 5A = 8$ .
Substitute $A = 2$ into the second equation:
$6B - 5(2) = 8$
$6B - 10 = 8$
$6B = 18$
$B = 3$

Therefore, the particular solution is $y_p(x) = 2x + 3$ . The value of the constant $A$ is 2.

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Differential Equations

First-Order Differential Equations, Separable Equations & Existence Theory

Analytical Procedure for Separable Equations

Linear First-Order Equations, Integrating Factors & Nonlinear Reductions

Analytical Procedure for First-Order Linear Equations

Second-Order Linear Homogeneous Equations, Auxiliary Roots & Wronskian Theory

Analytical Procedure for Homogeneous Second-Order Equations

Nonhomogeneous Equations, Undetermined Coefficients & Resonance

Analytical Procedure for Undetermined Coefficients

Variation of Parameters, Wronskian Systems & Green's Kernel

Analytical Procedure for Variation of Parameters

The Laplace Transform, Discontinuous Forcing & Convolution Theory

Operational Properties of the Laplace Transform

Systems of Linear Differential Equations, Phase Plane Topology & Matrix Exponentials

Power Series Solutions, Ordinary Points & the Method of Frobenius

Analytical Mechanics of Power Series Solutions

Numerical Methods: Euler, Improved Euler (RK2), RK4 & Numerical Stability

Hierarchy of Numerical Integration Methods

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