Answer
$$y=\frac{1}{x} \ln x-\frac{1}{x}+\frac{3}{x^{2}}$$
Work Step by Step
Given $$x^{2} y^{\prime}+2 x y=\ln x, \quad y(1)=2$$
Rewriting the equation as
$$ y^{\prime}+\frac{2}{x} y=\frac{\ln x}{x^2}$$
Since the equation is linear and $p(x)= \frac{2}{x} $ and $q(x)= \frac{\ln x}{x^2}$, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{\int \frac{2}{x} dx}\\
&=e^{2\ln x}\\
&=x^2
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
x^2 y&= \int x^2 \frac{\ln x}{x^2}dx\\
&=\int \ln xdx\ \ \ , \ \text{Integrate by parts }\\
&= x\ln x-x+C
\end{align*}
Then
$$ y= \frac{\ln x}{x}-\frac{1}{x}+\frac{C}{x^2}$$
Since $y(1)=2 $, then $C= 3$, hence
$$y=\frac{1}{x} \ln x-\frac{1}{x}+\frac{3}{x^{2}}$$
