I am trying to solve the following:
$$y''(x)=\frac{4}{3} y(x)^3 y'(x)$$
given that $y(0)=1$ and $y'(0)=1/3$. This is a link to Wolfram Alpha.
My idea was that because when $y=1$, $y^3 = 1$ it can be solved for the private case only by putting $1$ instead of $y^3$
the question is to find y I need the way as I have no clue what to do.