= Solution
After division by $x$, the coefficient of $y'$ is $P(x)=1/x-1$. The <Abel identity> for the <Wronskian> gives
$$
W'=-P(x)W=\left(1-\frac1x\right)W,
$$
so
$$
\boxed{W(x)=C\frac{e^x}{x}}.
$$
For $\lambda=1$, direct substitution confirms that $y_1=1-x$. <Reduction of order> gives a second solution proportional to
$$
y_2=(1-x)\int\frac{e^x}{x(1-x)^2}\,dx.
$$
To extract the requested coefficients, put
$$
y_2=(1-x)\log x+b_1x+b_2x^2+\cdots.
$$
For the differential operator
$$
L[y]=xy''+(1-x)y'+y,
$$
one finds
$$
L[(1-x)\log x]=x-3.
$$
The analytic correction $h=\sum_{n\geq1}b_nx^n$ must therefore satisfy $L[h]=3-x$. Its constant and linear coefficients give
$$
b_1=3,\qquad 4b_2=-1.
$$
Hence
$$
\boxed{
y_2=(1-x)\log x+3x-\frac14x^2+\cdots}.
$$
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