Hmm am I missing something obvious but how would I solve y"=ay^2?
For this type of second order ODE the 'standard approach' is based on the identity...
$\displaystyle \displaystyle y^{''} = \frac{d y^{'}}{dx}= \frac{d y^{'}}{dy}\ \frac{dy}{dx} = y^{'}\ \frac{d y^{'}}{dy}$ (1)
... so that the equation you have prosed $\displaystyle y^{''}= a\ y^{2}$ becomes...
$\displaystyle \displaystyle y^{'}\ dy^{'} = a\ y^{2}\ dy$ (2)
... that is 'separated variables' and the solution of which is...
$\displaystyle \displaystyle y^{'\ 2}= \frac{2\ a}{3} y^{3} + c_{1} \implies y^{'} = \pm \sqrt{\frac{2\ a}{3}\ y^{3} + c_{1}}$ (3)
Also (3) is 'sepatated variables' and its solution is...
$\displaystyle \displaystyle x = \int \frac{d y}{\pm \sqrt{\frac{2\ a}{3}\ y^{3} + c_{1}}} + c_{2}$ (4)
... even if the integral in (4) is not very 'confortable'...
Merry Christmas from Italy
$\displaystyle \chi$ $\displaystyle \sigma$