Find a 2nd independent solution of the given equation (given one solution)? Find a second independent solution of the given equation and one solution. 

I'm confused by the hint. How does this help identify a second solution given a first solution? 
 A: Write the differential equation in the form $ y''+P(x)y'+Q(x)y=0$
Since $x>1 \implies x \neq 1$ so
$$y''-\frac x {x-1}y'+\frac y {x-1}=0$$
$$ \text {1)} \left (\frac {y_2} {y_1}\right )'=\frac {W(y_1,y_2)} {y_1^2}$$ 
$$\frac {y_2'y_1-y_1'y_2}{y_1^2}=\frac {W(y_1,y_2)} {y_1^2}$$
$$\frac {y_2'e^x-e^xy_2}{e^{2x}}=\frac {W(e^x,y_2)} {e^{2x}}$$
$${y_2'e^x-e^xy_2}= {W(e^x,y_2)}$$
$${y_2'e^x-e^xy_2}=\left|
\pmatrix{e^x & y_2\\
e^x & y'_2}
\right|$$
$${y_2'e^x-e^xy_2}= e^xy'_2-e^xy_2$$
2)Then use Abel's identity ... with $P(x)=-\dfrac x{x-1}$
$$W=W_0e^{-\int P(x)dx}=W_0e^{\int \frac x {x-1}}=Ke^{\int 1+\frac 1 {x-1}}=Ke^xe^{ln|x-1|}=Ke^x(x-1) \text { since x>1}$$
$$W= e^xy'_2-e^xy_2 \implies y'_2-y_2=K(x-1)$$
I let you finish...solve for $y_2$ the equation.It's a first order linear equation. 

Another way to integrate this equation...
$$(x-1)y''-xy'+y=0$$
$$(x-1)y''-xy'\color{red}{+y'-y'}+y=0$$
$$(x-1)y''\color{blue}{-xy'+y'}+\color{green}{y-y'}=0$$
$$(x-1)y''\color{blue}{-(x-1)y'}+y-y'=0$$
$$(x-1)(y''-y')-(y'-y)=0$$
Substitute $z=y'-y$
$$(x-1)z'-z=0$$
Integrate..
$$\int \frac {dz}{z}=\int \frac {dx}{x-1}$$
$$z=K_1(x-1)$$
$$y'-y=K_1(x-1)$$
$e^{-x}$ as inetgrating factor
$$(ye^{-x})'=e^{-x}K_1(x-1)$$
$$ye^{-x}=\int e^{-x}K_1(x-1)dx+K_2$$
$$y(x)=K_1e^{x}\int e^{-x}(x-1)dx+K_2e^x$$
$$\boxed{y(x)=K_1x+K_2e^x}$$
