(a) reformat equation into

$$y'' + p(x)y' + q(x)$$

where

$$p(x) = -\frac{x}{x-1}$$

by equation of wronskian

\begin{gather*}

W(x) = \exp(-\int p(x) dx)\\

W(x) = \exp(\int \frac{x}{x-1} dx) = \exp(\int \frac{u+1}{u}du) = c_0ue^u = c_0(xe^{x-1} - e^{x-1})

\end{gather*}

choose $c_0 = e$

$$W(x) = xe^x - e^x$$

(b) Plug in $y = x, y' = 1, y'' = 0$, equation becomes

$$-x + x = 0$$

therefore $y=x$ is a solution, let second solution be $y_2$ and let $y_1 = x$ since

$$W(x) = y_1y_2' - y_1'y_2$$

Plug in $y_1 = x, y_1' = 1$

$$xe^x - e^x = xy_2' - y_2$$

By inspection

$$y_2 = e^x$$

Hence, general solution is

$$y(x) = c_1x + c_2e^x$$

c) Set $y(0) = 1$

$$c_2 = 1$$

Set $y'(0) = 0, y'(0) = c_1 + c_2e^x$

$$c_1 + c_2 = 0, c_1 = -c_2 = -1$$

Hence the solution is

$$y(x) = -x + e^x$$