Solve $y^2y'= y^3-3x$ for $y(0)=2$

Question

$$y^2y'= y^3-3x$$ dividing by $y^2$: $$y' = y-\frac {3x}{y^2}$$ From here, I am not sure what sort of substitution to use. Might one possibility be $v = \frac {1}{y^2}$?

Answer 1

$u = y^3$

$\frac{u^{'}}{3} = u-3x$

$\implies \frac{u^{'}}{3} - u=-3x$

$u(x) = ce^{3x}+(3x+1)$

$y(0) = 2\implies 8=c+1 \implies u(x)=7e^{3x}+(3x+1)$

Answer 2

Hint: $y^2y'={1\over 3}(y^3)'=y^3-x$. Write $h(x)=y^3$, ${1\over 3}h'(x)-h(x)=-x$.