Find all continuous real valued function such that $(f(x))^2+C=\int\limits_0^xf(t)dt$

Question

Find all continuous real valued function such that $$(f(x))^2+C=\int\limits_0^xf(t)dt$$ for some $C\in\mathbb{R}$

If I set $F(x)=\int\limits_0^xf(t)dt$ then $F$ is differentiable and $F'(x)=f(x)$,

so $(f(x))^2+C$ is diferentible as it does $\sqrt{F(x)-C}=|f(x)|$,

and then $F'(x)=(|f(x)|^2+C)'=2|f(x)||f(x)|'=f(x)$.

and then I don't know how to continue. I just see the solution $f(x)=0$ whith $C=0$.

Answer 1

suppose $f$ is differentiable. differentiating $$ (f(x))^2+C=\int\limits_0^xf(t)dt \tag 1$$ gives us $$2ff' = f \to f' = \frac12, f = \frac12x + k$$ puuting it back in $(1)$ we get $$\left(\frac12x + k\right)^2 + C= \frac14 x^2 + kx $$ putting $x = 0$ connect $$ C = -k^2, \quad f = \frac12x+k.$$

Answer 2

You have on differentiating both sides, $2f(x)f'(x)=f(x)$ for all $x$. Choose some $x$ for which $f(x)\neq0$ then $2f'(x)=1\implies f(x)=\dfrac{x}{2}+K$ where K is another constant.

Note that $f(0)=K$ in our solution. In your given equation, setting $x=0$, we have $f(0)^2+C=0\implies K^2+C=0\implies K=\sqrt{-C}$

Note that $C\leq0$ so that it makes sense. Hence $f(x)=\dfrac{x}{2}+\sqrt{-C}$