Solve the equation: $y''-y'x^{-1}+2yx^{-2}=(\ln x+1)\cos(\ln x)+\ln x\sin(\ln x)$. Firstly, I noticed that it is an Euler-Cauchy equation, so I multiplied it by $x^2$. Then I tackled the homogeneous part and obtained $\lambda(\lambda-1)-\lambda+2=0$, or $\lambda^2-2\lambda+2=0$, from which it follows $\lambda_{1,2}=1\pm i$. Therefore, the solution of the homogeneous part is $y=x(A\cos(\ln x)+B\sin(\ln x))$. However, I'm struggling with the particular solution part. I tried using the method of variation of parameters, so $y(x)=x(A(x)\cos(\ln x)+B(x)\sin(\ln x))$, then I took the derivatives, equated them to zero, and differentiated once more, but I'm unable to progress through the problem.
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