The Solution of differential equation $\frac{{dy}}{{dx}} + \frac{y}{x} = \sin x$ is

Given differential equation is

$\frac{{dy}}{{dx}} + y\frac{1}{x} = \sin x$

which is linear differential equation.
Here,

$P = \frac{1}{x}$

and $Q = \sin x$

$\therefore$ ${\rm{IF}} = {e^{\int {\frac{1}{x}} dx}} = {e^{\log x}} = x$

The general Solution is

$y \cdot x = \int x \cdot \sin xdx + C$
……(i)
Take $I = \int x \sin xdx$

$- x\cos x - \int - \cos xdx$

$= - x\cos x + \sin x$

Put the value of $l$ in Eq. (i),

we get
$xy = - x\cos x + \sin x + C$

$\Rightarrow$ $x(y + \cos x) = \sin x + C$