Solve $y + \frac{d}{{dx}}(xy) = x(\sin x + \log x)$.

Given differential equation is
$y + \frac{d}{{dx}}(xy) = x(\sin x + \log x)$

$\Rightarrow$ $y + x\frac{{dy}}{{dx}} + y = x(\sin x + \log x)$

$\Rightarrow$ $x\frac{{dy}}{{dx}} + 2y = x(\sin x + \log x)$

$\Rightarrow$ $\quad \frac{{dy}}{{dx}} + \frac{2}{x}y = \sin x + \log x$

which is a linear differential equation.
On comparing it with

$\frac{{dy}}{{dx}} + Py = Q,$

we get
$P = \frac{2}{x},Q = \sin x + \log x$
$IF = {e^{\int {\frac{2}{x}} dx}} = {e^{2\log x}} = {x^2}$

The general solution is
$y \cdot {x^2} = \int {(\sin x + \log x)} {x^2}dx + C$

$\Rightarrow$ $\quad y \cdot {x^2} = \int {\left( {{x^2}\sin x + {x^2}\log x} \right)} dx + C$

$\Rightarrow$ $y \cdot {x^2} = \int {{x^2}} \sin xdx + \int {{x^2}} \log xdx + C$

$\Rightarrow$ $y \cdot {x^2} = {I_1} + {I_2} + C$

…….(i)
Now, ${I_1} = \int {{x^2}} \sin xdx$

$= {x^2}( - \cos x) + \int 2 x\cos xdx$

$= - {x^2}\cos x + \left[ {2x(\sin x) - \int 2 \sin xdx} \right]$

${I_1} = - {x^2}\cos x + 2x\sin x + 2\cos x$
…..(ii)

and ${I_2} = \int {{x^2}} \log xdx$

$= \log x \cdot \frac{{{x^3}}}{3} - \int {\frac{1}{x}} \cdot \frac{{{x^3}}}{3}dx$

$= \log x \cdot \frac{{{x^3}}}{3} - \frac{1}{3}\int {{x^2}} dx$

$= \log x \cdot \frac{{{x^3}}}{3} - \frac{1}{3} \cdot \frac{{{x^3}}}{3}$

……..(iii)

On substituting the value of ${I_1}$ and ${I_2}$ in Eq. (i),

we get
$y \cdot {x^2} = - {x^2}\cos x + 2x\sin x + 2\cos x + \frac{{{x^3}}}{3}\log x - \frac{1}{9}{x^3} + C$

$\therefore y = - \cos x + \frac{{2\sin x}}{x} + \frac{{2\cos x}}{{{x^2}}} + \frac{x}{3}\log x - \frac{x}{9} + C{x^{ - 2}}$