How to find the general solution of $x^2y'' - x(x+2)y' + (x+2)y = 0$

Question

$$x^2y'' - x(x+2)y' + (x+2)y = 0$$

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.

Answer 1

I'm not sure how you got that result, since substituting $y=x$ would only give $0=0$, due to it being a solution of the ODE.

What you need to do is to apply reduction of order. That is, to look for a solution of the form $$ y(x) = y_1(x)v(x) = xv $$

This gives the equation in $v(x)$ $$ x^2(2v' + xv'') - x(x+2)(v + xv') + (x+2)xv = 0 $$

Simplifying we get $$ v'' - v' = 0 $$ or $$ v(x) = c_1 + c_2e^x $$

Therefore $$ y(x) = c_1x + c_2xe^x $$

Answer 2

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.