First-Order Differential Equations & Separable Equations

A differential equation relates a function to its own rate of change — and rates of change are how most of the physical world is actually described: a population's growth rate depends on its current size, a cooling object's temperature falls at a rate proportional to how far above room temperature it still is, a radioactive sample decays at a rate proportional to how much is left. Calculus gave you derivatives; this chapter is about running that machinery in reverse — starting from a stated relationship between a function and its derivative, and recovering the function itself.

● Definition 1.1 — Differential equation, order, solution

A differential equation is an equation relating an unknown function $y=y(x)$ to one or more of its derivatives. The order of the equation is the order of the highest derivative that appears; this chapter deals only with first-order equations, involving $y'$ but no higher derivative.

A solution on an interval $I$ is a differentiable function $y(x)$ that satisfies the equation for every $x\in I$ . The general solution is the whole family of solutions, typically written with one arbitrary constant $C$ ; if a condition $y(x_0)=y_0$ is also specified, the problem is called an initial value problem (IVP) — it asks for a solution passing through that point. An IVP doesn't automatically pick out a unique member of the general solution's family: existence and uniqueness require additional hypotheses, which this chapter's examples all happen to satisfy but which aren't checked here in general. When a solution to an IVP is found, it's called a particular solution.

Worked example 1.A — Verifying a solution

Verify that $y=Ce^{2x}$ is a solution of $\dfrac{dy}{dx}=2y$ for any constant $C$ , show that these are the only solutions, then find the particular solution satisfying $y(0)=5$ .

These are solutions: differentiating, $\dfrac{dy}{dx}=\dfrac{d}{dx}\big[Ce^{2x}\big]=2Ce^{2x}=2\big(Ce^{2x}\big)=2y$ — the equation holds for every $C$ .

These are the only solutions: if $y$ is any solution of $y'=2y$ , then $\frac{d}{dx}\big[e^{-2x}y\big]=e^{-2x}\big(y'-2y\big)=e^{-2x}(0)=0,$ so $e^{-2x}y$ is constant on the interval — call it $C$ — meaning $y=Ce^{2x}$ after all. So $y=Ce^{2x}$ is genuinely the general solution, not just a family of solutions.

Applying the initial condition: $y(0)=Ce^{0}=C=5$ , so the particular solution is $y=5e^{2x}$ .

● Definition 1.2 — Separable equation

A first-order equation is separable if it can be written in the form $\frac{dy}{dx}=f(x)\,g(y),$ where the right-hand side factors into a function of $x$ alone times a function of $y$ alone. Every example in this chapter is separable — it's the first and most widely applicable technique for solving a differential equation in closed form.

Solving a separable equation

1. Check for equilibrium solutions

Before dividing by

g(y)

, find its zeros: every constant

a

with

g(a)=0

gives a constant solution

y(x)=a

, which the division in the next step would otherwise silently lose

2. Separate

Rewrite

\dfrac{dy}{dx}=f(x)g(y)

as

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

, valid wherever

g(y)\neq0

3. Integrate both sides

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

— one arbitrary constant is enough, since combining both sides' constants just renames it

4. Solve for y, if possible

An implicit relation between

x

and

y

already counts as a solution just as validly as an explicit

y(x)

— the remaining work is identifying a differentiable branch of it, and the interval on which the original equation actually makes sense

▲ Theorem 1.1 — Separation on a nonzero branch

Let $I$ and $J$ be open intervals. Suppose $f$ is continuous on $I$ , and $g$ is continuous and nonzero on $J$ . Let $F$ be an antiderivative of $f$ on $I$ , and $G$ an antiderivative of $\dfrac{1}{g}$ on $J$ (so $G'(y)=\dfrac{1}{g(y)}$ for $y\in J$ ). Then a differentiable function $y:I\to J$ solves $\frac{dy}{dx}=f(x)\,g(y)$ if and only if $G(y(x))=F(x)+C$ for some constant $C$ . (Since $g$ is nonzero throughout $J$ by assumption, this covers every solution staying on one nonzero branch of $g$ — a constant solution sitting at a zero of $g$ is a separate case, found directly rather than through this formula.)

Proof

(⇒) Suppose $y(x)$ solves the equation, with $g(y(x))\neq0$ . Dividing both sides by $g(y(x))$ : $\frac{1}{g(y(x))}\,y'(x)=f(x).$ By the chain rule, $\dfrac{d}{dx}\big[G(y(x))\big]=G'(y(x))\,y'(x)=\dfrac{1}{g(y(x))}\,y'(x)$ , so the left side above is exactly $\dfrac{d}{dx}\big[G(y(x))\big]$ , and the equation becomes $\frac{d}{dx}\big[G(y(x))\big]=f(x)=\frac{d}{dx}\big[F(x)\big].$ So $G(y(x))-F(x)$ has zero derivative on the interval, hence is a constant $C$ : $G(y(x))=F(x)+C$ .

(⇐) Conversely, suppose $G(y(x))=F(x)+C$ . Differentiating both sides with respect to $x$ (chain rule on the left): $G'(y(x))\,y'(x)=F'(x) \;\Longrightarrow\; \frac{1}{g(y(x))}\,y'(x)=f(x).$ Since $g(y(x))\neq0$ by hypothesis, multiplying both sides by $g(y(x))$ gives $y'(x)=f(x)\,g(y(x))$ — exactly the original equation.

Worked example 1.B — Solving an initial value problem

Solve $\dfrac{dy}{dx}=xy$ , $y(0)=3$ .

First, the equilibrium: $g(y)=y$ has the single zero $y=0$ , giving the constant solution $y\equiv0$ (indeed $\frac{d}{dx}[0]=0=x\cdot0$ ).

For $y\neq0$ , separate: $\frac{dy}{y}=x\,dx.$ Integrate both sides: $\ln|y|=\frac{x^2}{2}+C.$ Exponentiate: $|y|=e^{x^2/2+C}=e^{C}e^{x^2/2}$ ; writing $A=\pm e^{C}$ (necessarily nonzero, since $e^{C}>0$ ) absorbs the sign, giving the nonzero-branch family $y=Ae^{x^2/2}$ , $A\neq0$ . Combined with the equilibrium above — exactly this formula's $A=0$ case — the complete general solution is $y=Ae^{x^2/2},\qquad A\in\mathbb{R}.$ Applying $y(0)=3$ : $A\cdot e^{0}=A=3$ . So $y=3e^{x^2/2}.$

Worked example 1.C — Application: radioactive decay

A radioactive sample decays at a rate proportional to the amount present: $\dfrac{dN}{dt}=-kN$ for a constant $k>0$ , with $N(0)=N_0$ and $N_0>0$ . Find $N(t)$ , and express the half-life $T$ (the time at which $N=N_0/2$ ) in terms of $k$ .

Separate: $\dfrac{dN}{N}=-k\,dt$ (valid since $N>0$ ), so $\ln N=-kt+C$ , giving $N=Ae^{-kt}$ ; $N(0)=A=N_0$ , so $N(t)=N_0e^{-kt}.$ Setting $N(T)=N_0/2$ : $e^{-kT}=\tfrac12$ , so $-kT=-\ln2$ , and $T=\frac{\ln2}{k}.$ The half-life doesn't depend on $N_0$ — a hallmark of exponential decay (the assumption $N_0>0$ matters: if $N_0=0$ the sample is identically zero, and "half of nothing" isn't a meaningful time to solve for) — and the same reasoning (with $k<0$ reinterpreted as growth) drives unconstrained population-growth models.

Every example above reduced to the same moves: check for equilibrium solutions, separate, integrate, solve for $y$ (or accept an implicit relation when that last step isn't possible in closed form). Not every first-order equation is separable — the next stage of this track extends the method to equations that aren't, starting with linear first-order equations and the integrating-factor technique.

📝 Chapter Quiz

Questions Level

§

Practice set

Four free-response problems on separable equations, Basic to Advanced. Try each one, then reveal the worked solution.

D1

Solve $\dfrac{dy}{dx}=-2y$ , $y(0)=4$ . Find $y(1)$ .

Basic

$\dfrac{dy}{y}=-2\,dx \Rightarrow \ln|y|=-2x+C \Rightarrow y=Ae^{-2x}$ . $y(0)=A=4$ , so $y=4e^{-2x},\qquad y(1)=4e^{-2}\approx0.541.$

D2

Solve $\dfrac{dy}{dx}=x^2$ , $y(1)=2$ . Find $y(2)$ .

Basic

$dy=x^2\,dx \Rightarrow y=\dfrac{x^3}{3}+C$ . $y(1)=\dfrac13+C=2\Rightarrow C=\dfrac53$ . So $y=\frac{x^3}{3}+\frac53,\qquad y(2)=\frac83+\frac53=\frac{13}{3}.$

D3

Solve $\dfrac{dy}{dx}=\dfrac{y}{x}$ for $x,y>0$ , with $y(1)=2$ . Find $y(4)$ .

Intermediate

$\dfrac{dy}{y}=\dfrac{dx}{x} \Rightarrow \ln y=\ln x+C \Rightarrow y=Ax$ . $y(1)=A=2$ , so $y=2x,\qquad y(4)=8.$

D4

Solve $\dfrac{dy}{dx}=xy^2$ , $y(0)=\dfrac12$ . Find $y(1)$ .

Advanced

$\dfrac{dy}{y^2}=x\,dx \Rightarrow -\dfrac1y=\dfrac{x^2}{2}+C$ . $y(0)=\tfrac12 \Rightarrow -2=C$ . So $-\dfrac1y=\dfrac{x^2}{2}-2 \Rightarrow y=\dfrac{-1}{\tfrac{x^2}{2}-2}=\dfrac{2}{4-x^2}$ , and $y(1)=\frac{2}{3}.$ This solution's denominator vanishes at $x=\pm2$ , so it's only valid on the interval $-2<x<2$ containing the initial point $x=0$ — it can't be extended past either pole.

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