I'm stuck on solving $$y' y''=x^2+x$$ with $(x,y,y')=(0,1,2)$.
Here's what I've tried
$$\int y' y'' \ dx = \int (x^2+x)\ dx$$ by substitution $$\frac{1}{2}\left(\frac{dy}{dx}\right)^2=\frac{x^3}{3}+\frac{x^2}{2}+C$$
from initial conditions $C=2$
$$\frac{1}{2}\left(\frac{dy}{dx}\right)^2=\frac{x^3}{3}+\frac{x^2}{2}+2$$
now I am supposed to obtain $dy/dx$ but I am stuck. What to do? Thank you in advance!
$y'=\sqrt{\frac{2x^3}{3}+x^2+4}$. Off hand I don't think you can get a nice integral.