$(a)$ $\\$
First we divide both sides by $x^2-1$:
$$y’’-{2x\over x^2-1}y’+{2 \over x^2-1}y=0$$
Let $p(t)=-{2x\over x^2-1}$ and $q(t)={2 \over x^2-1}.$ $\\$
Noting $p(t)$ and $q(t)$ are continuous everywhere except when $x=1$ or $x=-1.$
By Abel’s Theorem:$$\begin{align}W(y_1,y_2)(x)&=cexp(-\int{p(x)dx})\\&=cexp(-\int{-{2x\over x^2-1}}dx)\\&=ce^{ln|x^2-1|}\\&=c(x^2-1)\end{align}$$
Let $c=1$, then $W(y_1,y_2)(x)=x^2-1$. $\\$
$(b)$ $\\$
Since $y_1(x)=x,$ then
$$\cases{y_1’(x)=1\\y_1’’(x)=0}$$
Substitute these values back to the ODE:
$$\begin{align}0-{2x\over x^2-1}\cdot 1+{2 \over x^2-1}\cdot x=0\end{align}$$
$$-{2x\over x^2-1}+{2x\over x^2-1}=0$$
Thus $y_1=(x)$ is indeed a solution. $\\$
Now we need to find $y_2(x)$ $\\$
Since we already know: $$\begin{align}W(y_1,y_2)(x)&=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=x^2-1\\&=\begin{array}{|c c|}x& y_2(x) \\ 1& y_2’(x)\end{array}=xy_2’(x)-y_2(x)=x^2-1\end{align}$$
Hence, we have $$xy_2’-y_2=x^2-1$$
Divide both sides by $x$:
$$y_2’-{1\over x}y_2=x-{1 \over x}$$
$$\mu(x)=exp(\int-{1\over x}dx)=e^{-ln|x|}=x^{-1}$$
Multiply both sides by $x^{-1}:$ $\\$
$$\begin{align}x^{-1}y_2’-x^{-2}y_2&=1-x^{-2}\end{align}$$
Hence, $$(x^{-1}y_2)’=1-x^{-2}$$
Integrating both sides:
$$\int{(x^{-1}y_2)’}dx=\int{1-x^{-2}}dx$$
$$\begin{align}x^{-1}y_2&=\int{1dx}-\int{x^{-2}}dx\\&=x+x^{-1}+c\end{align}$$
Hence,$$y_2=x^2+1+cx$$
Let $c=1$, we have $$y_2=x^2+x+1$$
Since we already know that $W(y_1,y_2)(x)=x^2-1\neq 0$,
Hence $y_2(x)$ is indeed another linearly independent solution. $\\$
$(c)$ $\\$
For $W(y_1,y_2)(x)=x^2-1\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(x^2+x+1)$$ where $c_1$ and $c_2$ are some arbitrary constants.
Now, $$y’=c_1+2c_2x+c_2$$
Let $x=0$ and $y=1$, $x=0$ and $y’=2$, we have
$$\cases{c_2=1\\c_1+c_2=2}\implies \cases{c_1=1\\c_2=1}$$
Therefore, the solution for the IVP is $$y(x)=x^2+2x+1$$
