Verify that that the initial value problem $y'=2\sqrt{y} \ \ , y(0)=0$ has two two different solutions $y_1(x)=x^2$ and $y_2(x)=0$.

Question

Solving for the general solution, I got

$y(x)= x^2 +c_1x + \frac{{c_1}^2}{4}$

How is $y(x)=0$ a solution for $y'=2\sqrt{y}$ ?

Answer 1

Sometimes there exist some solutions that are not particular cases of the general solution. These singular solutions can take place when uniqueness of a solution is not guaranteed from the usual conditions (here $f(x,y)=2\sqrt y$ fails to satisfy some condition; check the theorems you have to see which one).

In any case, for what I see you just are asked to check that these are indeed solutions for an IVP (I assume that $y=0$ is actually the condition $y(0)=0$. It should also be the case that the domain of $y$ is taken to be $[0,\infty)$ or something smaller, because otherwise $y_1$ won't be a solution.

If that is so, then you just have to see that $$y_1(0)=0$$ (which is obvious) and that $$y_1'(x)=2\sqrt {y_1(x)}, \quad \forall x\in [0,\infty),$$ which is also true; then do the same for $y_2(x)=0$, which is straightforward.