Use the method of reduction of order to find another independently linear solution y2(x) when given one solution.
[tex] x^2y'' - x(x+2)y' + (x+2)y = 0 [/tex]
[tex] y_1(x) = x [/tex]
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).
[tex] y_2 = vx [/tex]
[tex] y'_2= v + xv' [/tex]
[tex] y''_2 = v' + v' + xv'' = 2v' + xv'' [/tex]
I substitute the above into the equation:
[tex] x^2(2v' + xv'') - x(x+2)(v + xv') + (x+2)vx = 0 [/tex]
[tex] 2x^2v' + x^3v'' - x^2v - x^3v' - 2xv - 2x^2v' + x^2v + 2xv = 0 [/tex]
And this is where I've gotten to. Everything cancels out and I can't see how I'll find my v...