Separable ODEs Put Each Variable on Its Own Side

An equation is separable when it can be written

$\frac{dy}{dt}=g(t)h(y).$

Where $h(y)\ne0$, rearrange it as

$\frac{1}{h(y)}\,dy=g(t)\,dt$

and integrate both sides.

Example: proportional decay

For $y'=-ky$,

$\frac{dy}{y}=-k\,dt,$

so $\ln|y|=-kt+C$ and $y=Ce^{-kt}$. An initial value determines $C$.

Do not lose equilibria

Dividing by $h(y)$ excludes values where $h(y)=0$. Check these first: each root $y^*$ of $h$ gives the constant solution $y(t)=y^*$.

The chain-rule justification

Suppose $H'(y)=1/h(y)$ and $G'(t)=g(t)$. Along a solution,

$\frac{d}{dt}H(y(t))=H'(y)y'=\frac{1}{h(y)}g(t)h(y)=g(t).$

Therefore $H(y(t))=G(t)+C$. The familiar movement of $dy$ and $dt$ is shorthand for this chain-rule argument.

Use qualitative reasoning as an error check

Before integrating, determine where the derivative is positive or negative. A derived formula that moves opposite to the phase-line arrows contains an algebra or sign error.

Check your understanding

For $y'=y^2$, which initial data produce the equilibrium solution, and why does dividing by $y^2$ lose it?

Show the reasoning

$y(0)=0$ produces $y(t)=0$. Dividing by $y^2$ assumes $y\ne0$, so that branch is excluded before integration.