<s>[INST] 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. 
 [/INST] The differential equation, $$\frac{dy}{dx}=\frac{y}{x+y^3}$$ is not homogenous so the substitution $y=ux$ is not helpful.
If you change your eqaution into $$\frac{dx}{dy}=\frac{x+y^3}{y}$$
then it is linear and you may solve it using integrating factor method.
$$\frac{dx}{dy}=\frac{x+y^3}{y}=(1/y)x+y^2$$
The integration factor is $1/y$ and after multipluing by $1/y$ and integrating you get the solution $$x=\frac {1}{2} y^3 +cy$$
The initial condition of $y(2)=2$ gives you $$x=\frac {1}{2} y^3-y$$
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