# Differential equation

I'm new to differential equations and I can't get the correct thinking. I successfully solved $y' = 2 \sqrt{y}$ as $x^2$ which wasn't that hard but I'm stuck at a more general form $y' = a \sqrt{y}$. The solution can't be that hard but I cannot find it.

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Please note that for $y'=2\sqrt{y}$ the general solution is $(x+C)^2$. – André Nicolas Jun 29 '11 at 19:33
@user6312: I know that but for the moment just a special solution will suffice. When I have the solution itself it's easier to find the general one. However I suck at finding a special solution. – TomK Jun 29 '11 at 19:34

A more general situation is a first order separable equation $f(y)\frac{dy}{dx}=g(x)$ which you can integrate and (potentially) solve for $y$ as a function of $x$. In the problem you give, we can integrate wrt $x$ to get $$\frac{y'}{\sqrt{y}}=a,\qquad 2\sqrt{y}=ax+C, \qquad y=(ax/2+C)^2$$ (abusing the constant $C$). As noted in the comments, don't forget the constant since it is important when you have some initial data (there are many solutions to a given differential equation and you might want to single one out using other data).
Thank you for your answer. However, I wonder what the solution for $y' = -a \sqrt{y}$ would be. – TomK Jun 29 '11 at 20:01
Also note that if your initial value is $y(0)=0$, then the solution is not unique; you can take $y(x)=0$ for all $x$, or take an arbitrary $x_0 \ge 0$ and let $y(x)=0$ for $x \le x_0$, $y(x)=(a/2)^2 (x-x_0)^2$ for $x>x_0$. (Assuming that $a>0$.) – Hans Lundmark Jun 29 '11 at 22:02