2
$\begingroup$

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.

  • 0
    Please note that for $y'=2\sqrt{y}$ the general solution is $(x+C)^2$.2011-06-29
  • 0
    @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

1 Answers 1