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 $y = \frac{1}{(x+1)}$

The integration is not messy. After the hint, you have
$$dx=\frac {dy}{\frac {y^4}4 -\frac {y^2} 2+c_1}=4\frac {dy}{y^4 -2y^2+c_1}=4 \frac {dy}{(y^2-a)(y^2-b)}$$ Now, use partial fraction decomposition
$$\frac {1}{(y^2-a)(y^2-b)}=\frac 1{a-b}\left(\frac 1{y^2-a}-\frac 1{y^2-b} \right)$$ Just continue and, when done, apply the conditions to get $a$ and $b$ which define $c_1$ and the next constant of integration.
