Let $f:\Bbb{R}\to\Bbb{R}$ differentiable such that $f(0)=0$ and for all $x\in\Bbb{R}$ we have $f'(x)=(f(x))^2$. Show $f(x)=0$ for all $x\in\Bbb{R}$ Let $f:\Bbb{R}\to\Bbb{R}$ differentiable such that $f(0)=0$ and for all $x\in\Bbb{R}$ we have $f'(x)=(f(x))^2$. Show that $f(x)=0$ for all $x\in\Bbb{R}$
Just a Hint!
What can i use to solve this problem?
 A: Let $U = \{ x \in \Bbb{R} : f(x) = 0 \}$ be the set of zeros of $f$. Clearly $U$ is closed and non-empty. We claim that $U$ is also open. Since $\Bbb{R}$ is connected, this is possible only when $U = \Bbb{R}$.
To this end, let $a \in U$ and $\delta > 0$ be arbitrary and let $M = \sup_{|x-a| < \delta} |f(x)| $. Then by the mean value theorem, for $x \in (a-\delta,a+\delta)$ we have
$$ |f(x)| = |f(x) - f(a)| \leq M^2|x - a| \leq M^2 \delta. $$
From this we have $M \leq M^2\delta$. If $\delta$ is sufficiently small, then we have $M\delta < 1$ and the previous inequality yields $M = 0$. This shows that $(a-\delta, a+\delta) \subseteq U$ and hence the claim follows.
A: It's immediate from the existence and uniqueness theorem for ordinary differential equations.
If you don't have that, you might note that (if $f(x) \ne 0$) $\dfrac{d}{dx} \dfrac{1}{f(x)} = -1$. 
A: Hint: For $x \ge 0$
$$f(x) = \int_0^x f(t)^2 dt$$
from the fundamental theorem of calculus and the fact that $f(0)=0$.
Note also that $f$ is increasing and positive for $x \ge 0$ (why?).
Hence for all $x \ge 0$,
$$f(x) \le xf(x)^2.$$
If $f(x)$ is not $0$ for some $0 \le y \le 1$, we can divide the above inequality by $f(y)$ to see that $f(y) \ge 1$. This is a contradiction (why? Hint: IVT). We can deduce that $f(x) = 0$ for all $x \in [0,1].$ 
From this we can deduce the following inequality:
$$f(x) \le (x-1)f(x)^2.$$
Applying the above logic inductively we can prove that $f(x)=0$ for all $x \ge 0$. (i.e suppose there is $1 \le y \le  2$ such that $f(y)$ is not $0$.....etc).
A similar argument works for $x<0$. This shows $f$ must be everywhere $0$.
