I just wanted to confirm I did this right.

"Verify by substitution that the given function is a solution of the given differential equation."

Given:

x^2 y'' - xy' + 2y = 0

y = x*cos( ln(x) )

My work:

y' = cos(ln(x)) - sin(ln(x))

y'' = -(1/x) * ( sin(ln(x)) + cos(ln(x)) )

Substituting back in to the differential equation:

x^2 (-(1/x)(sin(ln(x)) + cos(ln(x)))) - x(cos(ln(x)) - sin(ln(x))) - 2(xcos(ln(x)) = 0

-xsin(ln(x)) - xcos(ln(x)) - xcos(ln(x)) + xsin(ln(x)) + 2x*cos(ln(x)) = 0

0 = 0

Does this mean I verified the solution? See I'm confused because in first order diff. equations I would just separate out the x's and y's and integrate both sides and get my y equation. This is a little different apparently.