How do I solve this initial value problem?
$y'=2x-2\sqrt{\max(y,0)}$
$y(0)=0$
I've never seen a problem of this type with the "max" and I don't know how to begin.
It must have a unique solution $y:[0,\infty)\rightarrow\mathbb{R}$
I know the solution must be of the type $y(x)=cx^2$.
Assume first that $y(x) \le 0 \ \forall x \in [0, \infty)$. The your equation will look like $y' = 2x$, so that $y = x^2 + c$; since $y(0) = 0$, then $c=0$, so $y=x^2$, but this one is not negative, so we have reached a contradiction. Therefore, there must exist $x_0 >0$ with $y(x_0) > 0$.
Let $(a,b) \subset [0, \infty)$ be the maximal (with respect to inclusion) interval containing $x_0$ on which $y > 0$, with $b$ possibily $\infty$. Maximality and the continuity of $y$ imply that $y(a) = y(b) = 0$.
Assume that $b$ is finite. If $y(b) = 0$, then $y'(b) = 2b > 0$. By the continuity of $y'$, there must exist some interval $(b - \varepsilon, b + \varepsilon)$ on which $y' > 0$, which means that on this interval $y$ must strictly increase. But $b - \varepsilon \in (a,b)$ on which $y$ was already strictly positive, so $0 < y(b - \varepsilon) < y(b) = 0$, which is a contradiction, therefore $b = \infty$.
So far, we have discovered that $y > 0$ on $(a, \infty)$, with $y(a) = 0$, and that there exist $\varepsilon > 0$ such that and $y(x) < 0$ on $(a - \varepsilon, a) \subset [0,a)$; consider the largest $\varepsilon \in (0, a)$ with this property. Think of what happens in $a - \varepsilon$: the maximality in the choice of $\varepsilon$ implies that $y(a - \varepsilon) = 0$, so $y' (a - \varepsilon) = 2 (a - \varepsilon) > 0$, so there must exist a small $r>0$ such that $y' > 0$ on the interval $(a - \varepsilon - r, a - \varepsilon + r)$, which means that $y$ is strictly increasing on it. But this means that $0 = y(a - \varepsilon) < y(a - \varepsilon + r) < 0$, which is a contradiction.
We have two possibilities: either $\varepsilon = a$, so that $y < 0$ on $(0, a)$; or $a = 0$ and there is no $\varepsilon$ as assumed. To rule out the first possibility, use the same argument as in the first paragraph: you will get $y(x) = x^2$, which is not negative as it should on $(0,a)$, so this case is not possible.
Therefore, we have proved that $y(x) > 0$ on $(0, \infty)$, with $y(0) = 0$. Thus, your equation becomes $y' = 2x - \sqrt y$. Solving this equation is far from easy, though. The change of unknown $y = u^2$ turns it into $uu' = x - u$, which is sorter but not simpler.
In any case, in order to show that a solution exists you don't need any of the above: just assume it to be of the form $y = c x^2$ and, after plugging this into the equation you will get that $c$ must verify $c + \sqrt c - 1 = 0$, which has a single positive solution: $c = \dfrac {3-\sqrt 5} 2$. Showing that this solution is unique, though, seems very difficult, and I do not even know whether this is true (everything boils down to the fact that one cannot use the Picard-Lindelöf theorem because the left-hand side is not uniformly continuous, its derivative with respect to $y$ being infinite in $y=0$).