We have
$\text{y}+\frac{\text{d}}{\text{dx}}(\text{xy})=\text{x}(\sin\text{x}+\log\text{x})$
$\Rightarrow\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}\text{y}=\text{x}(\sin\text{x}+\log\text{x}).$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}(\sin\text{x}+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2}{\text{x}}\text{y}=\sin\text{x}+\log\text{x}$
This is a linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\frac{2}{\text{x}},\text{Q}=\sin\text{x}+\log\text{x}$
$\text{I.F.}=\text{e}^{\int\frac{2}{\text{x}}\text{dx}}=\text{e}^{2\log\text{x}}$
$\text{e}^{\log\text{x}^2}=\text{x}^2$
So, the reneral solition is,
$\text{y}.\text{x}^2=\int(\sin\text{x}+\log\text{x})\text{x}^2\text{dx}+\text{C}$
$\Rightarrow\text{y}.\text{x}^2=\int\text{x}^2\sin\text{x}\text{dx}+\int\text{x}^2\log\text{x}\text{dx}+\text{C}\ .....(\text{i})$
Now $\int\text{x}^2\sin\text{x}\text{dx}=\text{x}^2(-\cos\text{x})+\int2\cos\text{xdx}$
$=-\text{x}^2\cos\text{x}+[2\text{x}(\sin\text{x})-\int2\sin\text{x}\text{dx}]$
$=-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}+2\cos\text{x}\ ......(\text{ii})$
and $\int\text{x}^2\log\text{xdx}$
$=\log\text{x}.\frac{\text{x}^3}{3}-\int\frac{1}{\text{x}}.\frac{\text{x}^3}{3}\text{dx}$
$=\frac{\text{x}^3}{3}\log\text{x}-\frac{1}{3}\int\text{x}^2\text{dx}$
$=\frac{\text{x}^3}{3}\log\text{x}-\frac{\text{x}^3}{9}\ ......(\text{iii})$
On substituting the values from Eqs. (ii) and (iii) in Eq. (i), we get
$\text{y}.\text{x}^2=-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}+2\cos\text{x}+\frac{\text{x}^3}{3}\log\text{x}-\frac{1}{9}\text{x}^3+\text{C}$
$\therefore\text{y}=-\cos\text{x}+\frac{2\sin\text{x}}{\text{x}}+\frac{2\cos\text{x}}{\text{x}^2}+\frac{\text{x}}{3}\log\text{x}-\frac{\text{x}}{9}+\text{C}\text{x}^{-2}$
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