# Ordinary Differential Equations/Separable equations: Separation of variables

## Definition

A separable ODE is an equation of the form

$x'(t)=g(t)f(x(t))$

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

We often write for this ODE

$x'=g(t)f(x)$

for short, omitting the argument of $x$ .

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

$x'=h(t,x)$

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

## 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

$x'=g(t)f(x)$

be given. Using Leibniz notation, it becomes

${\frac {dx}{dt}}=g(t)f(x)$ .

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

${\frac {dx}{f(x)}}=g(t)dt$ .

Integrating this equation yields

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

Define

$F(x):=\int {\frac {dx}{f(x)}}$ ;

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

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

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

$x'(t)=g(t)f(x(t))$

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

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

solves the ODE under consideration.

Proof:

By the inverse and chain rules,

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

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