Solving a differential-initial value equation: $y'(x) = \frac{y}{x+y^3}$, $ y(2)=2$ 
The equation and given values are
  $$
\begin{split}
\frac{dy}{dx} &= \frac{y}{x+y^3}\\
y(2) &= 2
\end{split}
$$

At first I thought it was a separation of variables problem, but I then was told later on by a tutor that it could be solved with substitution. 
I then subbed-in with the following variables:
$$u+x\frac{du}{dx}=\frac{ux}{x+(ux)^3}$$
Then subtracted the $u$ from both sides and multiplied by the denominator:
$$x\frac{du}{dx}=\frac{ux-u(x+(ux)^3)}{x+(ux)^3}$$
$$x\frac{du}{dx}=\frac{ux-ux-u^4x^3}{x+u^3x^3}$$
$$x\frac{du}{dx}=\frac{-u^4x^3}{x+u^3x^3}$$
Now If I factor out an x from the quotient 
$$x\frac{du}{dx}=-\frac{x(u^4x^2)}{x(1+u^3x^3)}$$
$$x\frac{du}{dx}=-\frac{u^4x^2}{1+u^3x^3}$$
Now can I further simplify the quotient till I can integrate or should I go back and do something different to make it easier. 

HINT Invert both sides of the equation to make it a linear first order equation and solve. 
ANSWER After inverting we get, $$\frac{dx}{dy}=\frac{x}{y}+y^2,$$ which equates to :$$\frac{dx}{dy}-\frac{x}{y}=y^2.$$
Multiply both of the equation with $e^{\int \frac{-1}{y}dy}=\frac{1}{y}$.
Integrating the total derivative we then get $$\frac{x}{y}=\int ydy,$$ which equates to $$x(y)=\frac{y^3}{2}+cy,$$ for some integration constant $c$. 
Solve with the initial conditions to get the answer from there, which is that $c=-1$ and so $$x(y)=\frac{y^3}{2}-y,$$
