# Ordinary Differential Equations/Separable equations: Separation of variables

## Definition

A separable ODE is an equation of the form

${\displaystyle x'(t)=g(t)f(x(t))}$

for some functions ${\displaystyle g:\mathbb {R} \to \mathbb {R} }$, ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}$. In this chapter, we shall only be concerned with the case ${\displaystyle n=1}$.

We often write for this ODE

${\displaystyle x'=g(t)f(x)}$

for short, omitting the argument of ${\displaystyle x}$.

[Note that the term "separable" comes from the fact that an important class of differential equations has the form

${\displaystyle x'=h(t,x)}$

for some ${\displaystyle h:\mathbb {R} \times \mathbb {R} ^{n}\to \mathbb {R} }$; hence, a separable ODE is one of these equations, where we can "split" the ${\displaystyle h}$ as ${\displaystyle h(t,x)=g(t)f(x)}$.]

## Informal derivation of the solution

Using Leibniz' notation for the derivative, we obtain an informal derivation of the solution of separable ODEs, which serves as a good mnemonic.

Let a separable ODE

${\displaystyle x'=g(t)f(x)}$

be given. Using Leibniz notation, it becomes

${\displaystyle {\frac {dx}{dt}}=g(t)f(x)}$.

We now formally multiply both sides by ${\displaystyle dt}$ and divide both sides by ${\displaystyle f(x)}$ to obtain

${\displaystyle {\frac {dx}{f(x)}}=g(t)dt}$.

Integrating this equation yields

${\displaystyle \int {\frac {dx}{f(x)}}=\int g(t)dt}$.

Define

${\displaystyle F(x):=\int {\frac {dx}{f(x)}}}$;

this shall mean that ${\displaystyle F}$ is a primitive of ${\displaystyle {\frac {1}{f(x)}}}$. If then ${\displaystyle F}$ is invertible, we get

${\displaystyle x=F^{-1}\left(\int g(t)dt\right)=F^{-1}\circ G}$,

where ${\displaystyle G}$ is a primitive of ${\displaystyle g}$; that is, ${\displaystyle x(s)=F^{-1}(G(s))}$, now inserting the variable of ${\displaystyle x}$ back into the notation.

Now the formulae in this derivation don't actually mean anything; it's only a formal derivation. But below, we will prove that it actually yields the right result.

## General solution

Theorem 2.1:

Let a separable, one-dimensional ODE

${\displaystyle x'(t)=g(t)f(x(t))}$

be given, where ${\displaystyle f}$ is never zero. Let ${\displaystyle F}$ be an antiderivative of ${\displaystyle f}$ and ${\displaystyle G}$ an antiderivative of ${\displaystyle g}$. If ${\displaystyle F}$ is invertible, the function

${\displaystyle x(t):=F^{-1}(G(t))}$

solves the ODE under consideration.

Proof:

By the inverse and chain rules,

${\displaystyle {\frac {d}{dt}}F^{-1}(G(t))={\frac {1}{\frac {1}{f(F^{-1}(G(t)))}}}G'(t)=f(F^{-1}(G(t)))g(t)}$;

since ${\displaystyle f}$ is never zero, the fraction occuring above involving ${\displaystyle f}$ is well-defined.${\displaystyle \Box }$