Ordinary Differential Equations/Separable equations: Separation of variables

Definition

A separable ODE is an equation of the form

$x^{\prime }(t)=g(t)f(x(t))$

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

We often write for this ODE

$x^{\prime }=g(t)f(x)$

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

$x^{\prime }=h(t,x)$

for some $h:ℝ\times {ℝ}^n\to ℝ$; 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

$x^{\prime }=g(t)f(x)$

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

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

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

Proof:

By the inverse and chain rules,

$\frac{d}{dt}{F}^{-1}(G(t))=\frac{1}{\frac{1}{f(F^{-1}(G(t)))}}{G}^{\prime }(t)=f(F^{-1}(G(t)))g(t)$;

since $f$ is never zero, the fraction occurring above involving $f$ is well-defined.$◻$