Finding a Second Linear Solution using Reduction of Order Method for ODEs

[Mangoes]

Homework Statement

Use the method of reduction of order to find another independently linear solution y₂(x) when given one solution.

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

$$y_1(x) = x$$

The Attempt at a Solution

Hopefully y₂(x) will take the form of v(x)y₁(x) or I have no idea how to solve the ODE. I start by finding y'₂(x) and y''₂(x).

$$y_2 = vx$$
$$y'_2= v + xv'$$
$$y''_2 = v' + v' + xv'' = 2v' + xv''$$

I substitute the above into the equation:

$$x^2(2v' + xv'') - x(x+2)(v + xv') + (x+2)vx = 0$$

$$2x^2v' + x^3v'' - x^2v - x^3v' - 2xv - 2x^2v' + x^2v + 2xv = 0$$

And this is where I've gotten to. Everything cancels out and I can't see how I'll find my v...

 

[vela]

You have two terms left over: $x^3 (v'' - v')=0$.

 

[Mangoes]

Oh, wow. I overlooked the same thing three times...

Might be worth switching notation to avoid this from happening again.

Thanks.