The Equation is:

$\displaystyle t^2y'+2ty-y^3=0$

I have the solution, which thanks to Mathematica, I know to be correct, it is:

$\displaystyle \pm \sqrt{\frac{5t}{2+5t^5c}}$

I have to do a solution check, and keep getting terribly lost in the algebra. I am wondering if anyone has any suggestions or tricks on how to plug this mess back in without it turning into the never-ending algebra problem. Thanks.