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, so leave it as an 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.

When separation fails, an Exact equation (hidden potential function) is the next thing to try.