Q:

Second order differential equation with variation of parameters

I'd like to solve $y''+\frac{2}{x}y'=\frac{1}{x^2}$
I identify this as Euler-Cauchy and rewrite to $x^2y''+2xy'=1$
So I have a homogenous solution of $y_h=C_1+\frac{C_2}{x}$
Second order Variation of parameters:  
$C_1(x)'=\frac{1\frac{1}{x}}{\frac{1}{x^2}}=x \rightarrow C_1 = \frac{x^2}{2}$  
$C_1(x)'=-\frac{1\cdot1}{\frac{1}{x^2}}=-x^2 \rightarrow C_2 = -\frac{x^3}{3}$
Is this procedure correct?

A:

No. You can see that $C_1+\frac {C_2} x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+\frac {C_2} x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions. 

