This is the same problem as in my most recent question. I'll be picking up where that left off.
The given is:
$$x^2y^{"}+xy^{'}+y=\ln x$$
The Aux. equation is:
$$y=C_1\cos(\ln x)+C_2\sin(\ln x)$$
Now we can see that $y_1=\cos(\ln x)$ and $y_2=\sin(\ln x)$ and taking the wronskian of these two values and the given formula in standard form gives us:
$$W=\begin{bmatrix}\cos(\ln x)\ \sin(\ln x)\\ -\frac{\sin(\ln x)}{x}\ \frac{\cos(\ln x)}{x} \end{bmatrix}=\frac{\cos^2(\ln x)}{x}+\frac{\sin^2(\ln x)}{x}=\frac{1}{x}$$
$$W_1=\frac{\ln(x)\cos(\ln x)}{x^2}$$
$$W_2=\frac{\ln(x)\sin(\ln x)}{x^2}$$
Now finding $U_{1,2}^{'}$ is just $\frac{W_{n}}{W}$:
$$U_1^{'}=\frac{\ln(x)\cos(\ln x)}{x}$$
$$U_2^{'}=\frac{\ln(x)\sin(\ln x)}{x}$$
Then after integrating I have:
$$U_1=\ln(x)\sin(\ln x)+\cos(\ln x)$$
$$U_2=-\ln(x)\cos(\ln x)+\sin(\ln x)$$
The rest of the problem is easy enough, I just wanna make sure my work is correct before solving.
Best Answer
In the original question, there was an answer with the hint to substitute $x=e^t$. The equation is then written as $$y''(t)+y(t)=t$$ Hopefully you found $y_h(t)=a\cos(t)+b\sin(t)$ for the homogeneous equation. Here you find a particular solution $y_p(t)=At+B$ with $A=1$ and $B=0$.
Substitute back, then $y_h(x)=a\cos(\ln(x))+b\sin(\ln(x))$ and a particular solution is then given by $y_p(x)=\ln(x)$.
Finally $y(x)=y_h(x)+y_p(x)$.