Assuming $y=\sum_{n\geq 0}a_nx^{n+r}$ (Fröbenius method), $xy''+y'-xy=0$ reduces to $$\sum_{n\geq 0}a_n(n+r)(n+r-1)x^{n+r-1}+\sum_{n\geq 0}a_n(n+r)x^{n+r-1}-\sum_{n\geq 0}a_nx^{n+r+1}=0$$ Reindexing the last sum like $n\to n-2$ and combining the first two sums, we get $$\sum_{n\geq 0}(n+r)^2a_nx^{n+r-1}-\sum_{n\geq 2}a_{n-2}x^{n+r-1}=0$$ $$r^2a_0x^{r-1}+(1+r)^2a_1x^r+\sum_{n\geq 2}\left[(n+r)^2a_n-a_{n-2}\right]x^{n+r-1}=0$$ Thus, $$n=0: r^2a_0=0\overset{a_0\neq 0}{\iff}r^2=0$$ $$n=1: (1+r)^2a_1=0\overset{r= 0}{\iff} a_1=0$$ $$n\geq 2:(n+r)^2a_n-a_{n-2}=0\overset{r= 0}{\iff}a_n=\frac{a_{n-2}}{n^2}$$
For the recursive equation, if $n$ is odd, then $a_n=f(n)a_1=0$; if $n$ is even, $a_n=g(n)a_0\neq 0$, meaning that we'll work with even indexes: $$a_{2n}=\frac{a_{2(n-1)}}{(2n)^2}=\frac{a_{2(n-2)}}{2^4 n^2(n-1)^2}=...=\frac{a_0}{2^{2n}(n!)^2}$$ The first particular solution is then $$y_1=\sum_{n\geq 0}\frac{a_0}{2^{2n}(n!)^2}x^{2n}$$ Concerning the second particular solution, the solution sheet directly uses this expression which I don't know exactly where it comes from and that I want to derive (goal of this post): $$y_2=y_1\ln x+\sum_{n\geq 0}c_nx^n=\left(\ln x+\frac{\sum_{n\geq 0}c_nx^n}{y_1}\right)y_1=uy_1$$
$\mathbf{MY\ APPROACH:}$
An approach to get to get to the above relation would be to assume $y_2=uy_1$ and substitute in the ODE: $$\begin{aligned} \left. \begin{aligned} &y_2=uy_1\\ &y_2'=u'y_1+uy_1'\\ &y_2''=u''y_1++2u'y_1'+uy_1'' \end{aligned} \right\} \implies &x(u''y_1+2u'y_1'+uy_1'')+u'y_1+uy_1'-xuy_1=0\\ \implies &xu''y_1+2xu'y_1'+xuy_1''+u'y_1+uy_1'-xuy_1=0\\ \implies &xu''y_1+2xu'y_1'+u'y_1+u\underbrace{(xy_1''+y_1'-xy_1)}_0=0\\ \implies &xy_1u''+(2xy_1'+y_1)u'=0 \end{aligned} $$ Defining $v\equiv u'$, we can separate variables: $$\int\frac{\text dv}{v}=-\int\frac{2xy_1'(x)+y_1(x)}{xy_1(x)}\text dx$$ $$\ln v=-2\int\frac{y_1'(x)}{y_1(x)}\text dx-\int\frac{1}{x}\text dx$$ $$\ln v=-2\ln y_1-\ln x+C_1$$ $$v=\frac{\text du}{\text dx}=\frac{K_1}{x(y_1)^2}$$ $$u=K_1\int\frac{\text dx}{x(y_1)^2}+K_2\overset{?}{=}\ln x+\frac{\sum_{n\geq 0}c_nx^n}{y_1}$$
Am I going in the right path?