How to find the general solution of $x^2y'' - x(x+2)y' + (x+2)y = 0$ $$x^2y'' - x(x+2)y' + (x+2)y = 0$$ 
where a particular solution is $y_1(x) = x$
So, can I start off by subbing in $x$ for $y$ since I have a particular solution?
Which would give me:
$\frac{x}{x+2}y'' - y' + y = 0$
If this is a valid approach, then solving this DE would not be too difficult.

Here $y_1(x)=x$ is your known integral. For finding complete solution in terms of known integral you put: $y=vx$, where $v$ is also function of $x$; then the above ODE becomes $\dfrac{d^2v}{dx^2}-\dfrac{dv}{dx}=0$, which can be easily solve for $v$. Then substitute  $v$ in $y=vx$ you got general solution.
