Population dynamics, square root and zero derivative Hy everybody ! 

I'm studying population dynamics for my calculus exam, and I don't understand something that seems really easy, so I thought you might be able to help me out ;)

Here's the thing. I have this differential equation $\frac{dN}{dt} = \sqrt{N}$.

Our book makes us realize that both $N(t) = 0$ and $N(t) = \frac{t^2}{4}$ are solution, which makes sense so far, simply by replacing in the original equation.

Now suppose we start at $N(0) = 0$. How can $N$ start growing like $\frac{t^2}{4}$ if it's derivative is $0$ at $t = 0$ ? Because zero derivative should mean no growth, so $\sqrt{N}$ should remain zero, which means still no growth, and so on. My brain is melting right now.
Zero derivative at one point does not mean no growth at other points. 

For example the  derivative of function $$N(t)=\frac {t^2}{4}$$ is $$N'(t)=t/2$$ That is the derivative is zero at $t=0$ but it is not zero at other points. The point of the problem is that in this case the solution is not unique.
You are correct. There are indeed two solutions. If the population is zero it will stay zero. The only way it will grow is if the initial conditions are positive. However, the population zero solution is "unstable" in the sense that even if the initial condition is slightly positive, the $t^2$ solution will kick in and push the population away from zero.