A special type of differential equation III

Question:
Solve $(x^2\cos x - y)dx + xdy = 0$.

Answer:
Rewrite the equation.
\[
\frac{dy}{dx}-\frac{y}{x}=-x\cos x
\]
When the right hand side is zero, the solution is $y=Cx$. Using variation of constants, we make $C$ as a function of $x$. Namely, $y=C(x)x$ and plug it in the above equation.
\begin{eqnarray}
& & \frac{d(C(x)x)}{dx}-\frac{C(x)x}{x}=-x\cos x  \\
& & C'(x)x+C(x)-C(x)=-x\cos x  \\
& & C'(x) = - \cos x  \\
& & C(x) = - \sin x + C
\end{eqnarray}
We know $C(x)=\frac{y}{x}$, so
\begin{eqnarray}
& & \frac{y}{x} = - \sin x + C \\
& & y = x(-\sin x + C)
\end{eqnarray}

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