I am working on the following exercise (Source is unknown, unfortunately):
Find all continuous, non-negative functions $f:\Bbb{R}\to\Bbb{R}$ satisfying $$\int_0^xf(t)dt=(f(x))^\alpha +C$$ for some constant $C\neq 0$ and $\alpha >1$.
Here is my attempt: Because of FTC, we know that $(f(x))^\alpha$ is a differentiable function (thus continuous), and because the composition of two continuous is also continuous, $f$ is continuous. Applying FTC we have that \begin{align*} \frac{d}{dx}\int_0^xf(t)dt & = \frac{d}{dx}\Big[(f(x))^\alpha+C\Big] \\ f(x) & =\alpha f(x)^{\alpha-1}f'(x) \\ f(x)(\alpha f(x)^{\alpha-2}f'(x)-1) & = 0 \end{align*} so $f(x)=0$ or $f'(x)=\alpha f(x)^{2-\alpha}$. Now, because this equation is in the form $y'+p(t)y=0$, we can probably do something with integrating factors and solve this D.E. However, I don't want to use any differential equation knowledge to solve this. How can I proceed?
$\frac{df}{dx}=\alpha f^{2-\alpha}$. By separating variables, we get $f^{\alpha-2}df=\alpha dx$. By integrating both sides, we have $\frac{1}{\alpha-1}f^{\alpha-1}=\alpha x+C$. Therefore, $f(x)=(\alpha(\alpha-1)x+C)^{\frac{1}{\alpha-1}}$ if $\alpha\neq1$. When $\alpha=1$ then $\frac{df}{f}=dx$. Therefore we have $ln f=x+C$. Hence $f(x)=Ce^{x}$.