How to solve $\int_0^x f(t) dt = g(f(x)) $?

Question

How to solve an equation of this type :

$$\int_0^x f(t) dt = g(f(x)) $$

for a given function $g$.

Im not so good with differential equations and integral equations.

I know how to solve $h'(x) = j(h(x)) $ in general and some approximations of specific cases of $k'(x) = k(l(x))$. But even my knowledge of those is limited.

However for this $$\int_0^x f(t) dt = g(f(x)) $$ I do not even know how to begin.

It seems very fundamental and simple though so my intuition assumes there must be a way.

I considered differentiating both sides but that got me to

$$ f(x) = g'(f(x)) f'(x) $$

Many thanks !

Answer 1

Let $F(x):=\int_0^xf(t)dt=g(dF/dx)$ so, if $g$ has a functional inverse $g^{-1}$,$$dF/dx=g^{-1}(F)\implies x+C=\int dx=\int\frac{dF}{g^{-1}(F)}$$with $C$ an integration constant. In theory, this obtains $x+C$ as a function of $F$, say $h(F)$. If $h$ is invertible, $F(x)=h^{-1}(x+C)$.

Answer 2

As you wrote, $$ f(x) = g'(f(x)) f'(x)$$ That is, $y = f(x)$ satisfies the differential equation $$ \dfrac{dy}{dx} = \frac{y}{g'(y)}$$ at least when $g'(y) \ne 0$. For $x=0$ we have $g(y(0)) = 0$. Depending on $g$, this may have no solution, or one, or several. If it happens that $g(0) = 0$, one solution of the integral equation is $f(x) = 0$.

The differential equation is separable, with solutions in implicit form

$$\int \frac{g'(y)}{y}\; dy = x + c$$