Answer
$$ y=\frac{x}{2}\ln x-\frac{x}{4}+\frac{1}{4x}$$
Work Step by Step
Given $$x y^{\prime}+y=x \ln x, \quad y(1)=0$$
Rewriting the equation as
$$ y^{\prime}+\frac{1}{x} y= \ln x $$
Since the equation is linear and $p(x)= \frac{1}{x} $ and $q(x)=\ln x$, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{\int \frac{1}{x} dx}\\
&=e^{\ln x}\\
&=x
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
x y&= \int x \ln x dx\ \ \ , \ \text{Integrate by parts }\\
&=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C
\end{align*}
Then
$$ y=\frac{x}{2}\ln x-\frac{x}{4}+\frac{C}{x}$$
Since $y(1)=0 $, then $C= \frac{1}{4}$, hence
$$ y=\frac{x}{2}\ln x-\frac{x}{4}+\frac{1}{4x}$$
