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. 
 A: 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 $$x(y)=\frac{y^3}{2}-y,$$
A: Hint:
Moving the denominator, we have
$$xy'+y^3y'=y.$$
Then we notice the terms $y-xy'$ which can be the numerator of the derivative of $\dfrac xy$. So we rewrite
$$\frac{y-xy'}{y^2}=yy'$$ or
$$\left(\frac xy\right)'=\frac12(y^2)'$$
A: 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 $$x=\frac {1}{2} y^3-y$$
A: It's not a homogeneous differential equation to use that substitution $(y=tx)$. You can consider $x'=\dfrac {dx}{dy}$ instead of $y'$,  then the DE becomes linear. Or you can try this :
$$\frac{dy}{dx} = \frac{y}{x+y^3}\\$$
$$({x+y^3}){dy} = {y}{dx}$$
$$ {y}{dx}-x{dy}=y^3{dy}$$
$$\frac {{y}{dx}-x{dy}}{y^2}=y{dy}$$
$$d \left ( \frac  x y\right )=y{dy}$$
After integration it gives:
$$ \frac x y =\dfrac 12 y^2 +C$$
$x$ as a function of $y$:
$$ x(y) =\dfrac 12 y^3 +Cy$$
Apply initial condition:
$$y(2)=2 \implies C=-1$$
Finally:
$$ \boxed { x(y) =\dfrac { y^3}2 -y}$$
