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.$\Box$