If $f(x)=\int_{0}^{x}\sqrt {f(t)}dt$, then find $f(6)$.

Question

Let $f:[0,\infty)\to[0,\infty)$ be continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. If $f(x)=\int_{0}^{x}\sqrt {f(t)}dt$, then find $f(6)$. $$f(x)=\int_{0}^{x}\sqrt {f(t)}dt$$ $$g(x):=\sqrt {f(x)}\implies(g(x))^2=f(x)=\int_{0}^{x}g(t)dt\implies2g(x)g'(x)=g(x)$$ [by FTC-1] $$\implies g=0 \vee g'(x)=\frac{1}{2} $$ $$(g(0))^2=f(0)=0\implies g(x)=\frac{x}{2}\vee g=0 \implies f(6)=9 \vee f(6)=0$$ Is this correct? Also, how do I rule out $f(6)=0$ since my source only gives $9$ as the answer. I found this post after writing mine but I still don't think that $f(6)=0$ can be ruled out.

Answer 1

$f(x)=\frac {x^{2}} 4$ and $f(x)=0$ both satisfy the given equation so $f(6)$ is not uniquely determined.

Answer 2

If you differentiate the equality you get $f'(x)=\sqrt{f(x)}$. Now you can solve this separable ODE (note that $f(0)=0$) and get your answer.