$(a)$ $\\$
First we divide both sides by $-x^2(ln(x)-1)$:
$$y’’-{x \over x^2(ln(x)-1)}y’+{1 \over x^2(ln(x)-1)}y=0$$
Let $p(t)=-{x \over x^2(ln(x)-1)}$ and $q(t)={1 \over x^2(ln(x)-1)}.$ $\\$
Noting $p(t)$ and $q(t)$ are continuous everywhere except when $x^2(ln(x)-1)\neq0$.
$\\$
By Abel’s Theorem:$$\begin{align}W(y_1,y_2)(x)&=cexp(-\int{p(x)dx})\\&=cexp(-\int-{x \over x^2(ln(x)-1)}dx)\\&=ce^{ln|ln|x|-1|}\\&=c(ln|x|-1)\end{align}$$
Let $c=1$, then $W(y_1,y_2)(x)=ln|x|-1$. $\\$
$(b)$ $\\$
Since $y_1(x)=x,$ then
$$\cases{y_1’(x)=1\\y_1’’(x)=0}$$
Substitute these values back to the ODE:
$$\begin{align}0-{x \over x^2(ln(x)-1)}\cdot 1 + {1 \over x^2(ln(x)-1)}\cdot x=0\end{align}$$
Thus $y_1=(x)$ is indeed a solution. $\\$
Now we need to find $y_2(x)$ $\\$
Since we already know: $$\begin{align}W(y_1,y_2)(x)&=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=ln|x|-1\\&=\begin{array}{|c c|}x& y_2(x) \\ 1& y_2’(x)\end{array}=xy_2’(x)-y_2(x)=ln|x|-1\end{align}$$
Hence, we have $$xy_2’-y_2=ln|x|-1$$
Divide both sides by $x$:
$$y_2’-{1\over x}y_2={ln|x|\over x}-{1 \over x}$$
$$\mu(x)=exp(\int-{1\over x}dx)=e^{-ln|x|}=x^{-1}$$
Multiply both sides by $x^{-1}:$ $\\$
$$\begin{align}x^{-1}y_2’-x^{-2}y_2&=x^{-2}ln|x|-x^{-2}\end{align}$$
Hence, $$(x^{-1}y_2)’=x^{-2}ln|x|-x^{-2}$$
Integrating both sides:
$$\int{(x^{-1}y_2)’}dx=\int{x^{-2}\ln|x|-x^{-2}}dx$$
$$\begin{align}x^{-1}y_2&=\int{x^{-2}\ln|x|}dx-\int{x^{-2}}dx\end{align}$$
For $\int{x^{-2}\ln|x|}dx$, we use Integral by parts: $\\$
Let $\cases{u=\ln|x|\\du={1\over x}dx}$ and $\cases{dv=x^{-2}dx\\v=-x^{-1}}$ $\\$
Thus:$$\begin{align}\int{x^{-2}\ln|x|}dx&=-x^{-1}\ln|x|-\int{-x^{-1}\cdot {1\over x}}dx\\&=-x^{-1}\ln|x|+\int{x^{-2}}dx \\&= -x^{-1}\ln|x|-x^{-1}+c\end{align}$$
Therefore,$$\begin{align}x^{-1}y_2&=\int{x^{-2}\ln|x|}dx-\int{x^{-2}}dx\\&=-x^{-1}\ln|x|-x^{-1}+c-\int{x^{-2}}dx\\&=-x^{-1}\ln|x|-x^{-1}+c+x^{-1}\\&=-x^{-1}\ln|x|+c\end{align}$$
Hence,$$y_2=-\ln|x|+cx$$
Let $c=1$, we have $$y_2=-\ln|x|+x$$
Since we already know that $W(y_1,y_2)(x)=\ln|x|-1\neq 0$,
Hence $y_2(x)$ is indeed another linearly independent solution. $\\$
$(c)$ $\\$
For $W(y_1,y_2)(x)=\ln|x|-1\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(-\ln|x|+x)$$ where $c_1$ and $c_2$ are some arbitrary constants.
Now, $$y’=c_1-{c_2\over x}+c_2$$
Let $x=1$ and $y=1$, $x=1$ and $y’=0$, we have
$$\cases{c_1+c_2=1\\c_1=0}\implies \cases{c_1=0\\c_2=1}$$
Therefore, the solution for the IVP is $$y(x)=-\ln|x|+x$$
