Integration of $\int_{x(0)}^{x(t)}\frac{1}{\sqrt{|y|} }dy$ for $y<0$ I tried to solve $x'(t)=\sqrt{|x(t)|}$ by using separation of variables.
So I did $$\int_{0}^{t}\frac{x'(s)}{\sqrt{|x(s)|}}ds$$
and used the substituion $y=x(s)$, which gave me the integral
$$\int_{x(0)}^{x(t)}\frac{1}{\sqrt{|y|} }dy$$
If we have $y>0$, I dont't have problems solving this: $2\sqrt{x(t)}-2\sqrt{x(0)}$
But how can I solve the integral if we have $y<0$.
 A: Do you have any more information about the initial condition? For instance, $x'$ is always non-negative so if $x(0)>0$, then $x(t)>0$ for all $t>0$, and so $y$ is always positive as well.
It's also perhaps worth noting that this differential equations is one of the typical 'pathological' examples from existence and uniqueness theory for ODE's. Observe that the function $\sqrt{|x|}$ is not Lipschitz on any interval including $x=0$, and if you try to solve this problem with the initial condition $x(0) = 0$ (for instance), you can find infinitely many $C^1$ solutions. (the 'basic' solution is $x = \dfrac{1}{4}t^2 : t \geq 0; x = 0 : t< 0$, and you can get other solutions by translating this to the right any number of units).
Further, if you try and solve this with an initial condition $x(0) < 0$, there's going to be some $t_0 > 0$ so that $x(t_0) = 0$ (this doesn't follow directly from $x' > 0$, but that's at least reasonable intuition), and then you have the uniqueness problem all over again relative to that point.
So, in some sense, it only makes sense to talk about 'solving' this problem with initial data $x(0) > 0$.
