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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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    @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.2011-06-29

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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).

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    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.)2011-06-29