A special type of differential equation II

Question:
(1) Solve
$x\frac{dy}{dx}+y=0$
(2) Solve
$x\frac{dy}{dx}+y=x\ln x$ by using the result from (1).

Answer:
(1) The differential equation can be expressed as
\[
x\frac{dy}{dx}+\frac{dx}{dx}y =0, \qquad \rightarrow  x'y + xy' =0
\]
Namely,
\[
\frac{d(xy)}{dx}=0 \qquad \rightarrow xy=C \quad \rightarrow \rightarrow y=\frac{C}{x}
\]

(2) Use variation of constants. When $x \ln x = 0$, $y=\frac{C(x)}{x}$. Then, plug it in the original equation.
\begin{eqnarray*}
& & x\frac{d}{dx}\frac{C(x)}{x}+\frac{C(x)}{x}=x \ln x  \\
& & \frac{xC'(x)-C(x)}{x^2}x+\frac{C(x)}{x}=x\ln x  \\
& & C'(x) = x \ln x
\end{eqnarray*}
Integrate both sides.
\begin{eqnarray*}
C(x) &=& \int x\ln x dx  \\
C(x) &=& \frac{x^2}{2}\ln x - \int \left(\frac{1}{x} \frac{x^2}{2}\right)dx  \\
C(x) &=&  \frac{x^2}{2}\ln x - \frac{x^2}{4}+ C
\end{eqnarray*}
Since $C=xy$, we have
\begin{eqnarray*}
xy &=&  \frac{x^2}{2}\ln x - \frac{x^2}{4} + C \\
y &=& \frac{x}{2}\ln x - \frac{x}{4} + \frac{C}{x}
\end{eqnarray*}

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