Separable equation

A separable equation is a first-order ODE that can be written so that all $x$-stuff is on one side and all $y$-stuff is on the other. Once separated, you integrate both sides independently.

The form: $\frac{dy}{dx}=g(x)p(y)$.

If $g$ depends only on $x$ and $p$ depends only on $y$, the equation is separable. You can rearrange:

$\frac{1}{p(y)}dy=g(x)dx$

(assuming $p(y)\ne 0$), then integrate both sides:

$\int \frac{1}{p(y)}dy=\int g(x)dx+c$

This gives an implicit relation between $x$ and $y$. Solve for $y$ explicitly if possible.

Why this works

Let $H(y)$ be an antiderivative of $\frac{1}{p(y)}$ and $G(x)$ be an antiderivative of $g(x)$. By the chain rule:

$\frac{d}{dx}H(y(x))=H^{\prime }(y)\cdot y^{\prime }(x)=\frac{1}{p(y)}\cdot \frac{dy}{dx}$

The original ODE $\frac{dy}{dx}=g(x)p(y)$ rearranges to $\frac{1}{p(y)}\frac{dy}{dx}=g(x)$, so

$\frac{d}{dx}H(y(x))=g(x)=\frac{d}{dx}G(x)$

Two functions with equal derivatives differ by a constant: $H(y(x))=G(x)+c$.

This is the implicit solution.

Special case: $p(y)\equiv 0$

If $p(y)=0$, then $\frac{dy}{dx}=0$, so $y(x)=$ constant on the interval. This is the trivial constant solution.

Worked example 1

$y^{\prime }=\frac{x}{y}{e}^{x+2y}$.

Rearrange:

$y{e}^{2y}dy=x{e}^{-x}dx$

(That’s $g(x)=x{e}^{-x}$ and $\frac{1}{p(y)}=y{e}^{2y}$.)

Integrate both sides:

$\int y{e}^{2y}dy=\int x{e}^{-x}dx$

By parts on the left ($u=y$, $dv=e^{2y}dy$): $\frac{1}{2}y{e}^{2y}-\frac{1}{4}{e}^{2y}$.

By parts on the right ($u=x$, $dv=e^{-x}dx$): $-x{e}^{-x}-e^{-x}=-(x+1)e^{-x}$.

So:

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

Multiplying through by 4 and rearranging:

$e^{2y}(2y-1)=-4e^{-x}(x+1)+c$

This is the implicit solution. You can’t isolate $y$ in elementary form — leave it implicit. See Implicit solution to ODE.

Worked example 2

$y^{\prime }=(1+y^2)\tan x$.

Rearrange:

$\frac{1}{1+y^2}dy=\tan xdx$

Integrate:

$\arctan y=\int \tan xdx$

For $\int \tan xdx$: substitute $u=\cos x$, $du=-\sin xdx$, so $\int \tan xdx=-\int \frac{du}{u}=-\ln |u|+C=-\ln |\cos x|+C$.

Result:

$\arctan (y(x))=-\ln |\cos x|+c$

Solve for $y$:

$y(x)=\tan (c-\ln |\cos x|)$

Explicit, this time.

Recognizing separability

Some ODEs aren’t obviously separable until you rearrange. Tactics:

• Move all $y$-terms to one side, all $x$-terms to the other.
• Watch for $f(x,y)$ that factors as $g(x)p(y)$ — often hidden behind algebraic manipulation.

Examples that aren’t separable:

• $y^{\prime }=x+y$ — sum, not product. Use Integrating factor instead.
• $y^{\prime }=(x+y{)}^2$ — can’t separate. Try a substitution like $u=x+y$.

When integration is hard

Sometimes the integrals you produce after separating are themselves hard. Don’t stop at the separation step — make sure the integrals can actually be computed (or at least left as $\int g(x)dx$ for explicit numerical evaluation later).

For other first-order solution methods, see Integrating factor (when separation fails because of additive structure) and Exact equation (when there’s a hidden potential function).