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
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.