<s>[INST] Solve ODE $y'' = y^3 -y y'$ I want to solve  $y'' = y^3 -y y'$ with boundary conditions $y(1) = 1/2$ and $y(2) = 1/3$ but not sure how to start. If someone could give me a complete step by step explanation, it would be greatly appreciated because I want to fully understand it.
Edit:
By the hint $(y'+\frac{y^2}{2})'=y^3$
$y' + \frac{y^2}{2} = \frac{y^4}{4} + c1$
$\frac{dy}{dx} = \frac{y^4}{4} - \frac{y^2}{2} + c1$
Integrate $\frac{dy}{\frac{y^4}{4} - \frac{y^2}{2} + c1} = dx$
Integration looks very messy.
I know that the answer is $y = \frac{1}{(x+1)}$
 [/INST] Hint:
$$(y'+\frac{y^2}{2})'=y^3$$
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