$\cfrac{{dy}}{{dx}} + 2y = \sin x$

: The given equation is $\cfrac{{dy}}{{dx}} + 2y = \sin x$

…(1)
which is a linear equation of type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = 2$ and $Q = \sin x$

$\therefore$ $I.F. = {e^{\int {Pdx} }} = {e^{\int {2dx} }} = {e^{2x}}$

The solution is $y \cdot ({\rm{I}}.{\rm{F}}.) = \int Q \cdot ({\rm{I}}{\rm{.F}}{\rm{.}})dx + C$

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

…(2)
Now, $I = \int {{e^{2x}}} \sin xdx = {e^{2x}}( - \cos x) - \int 2 {e^{2x}}( - \cos x)dx$

[Integrating by parts]
$= - {e^{2x}}\cos x + 2\int {{e^{2x}}} \cos xdx$

Again integrating by parts
$= - {e^{2x}}\cos x + 2\left[ {{e^{2x}}\sin x - \int {{e^{2x}}} \cdot 2\sin xdx} \right]$

$= - {e^{2x}}\cos x + 2{e^{2x}}\sin x - 4{\rm{I}}$

$\Rightarrow 5I = {e^{2x}}(2\sin x - \cos x) \Rightarrow I = \cfrac{{{e^{2x}}}}{5}(2\sin x - \cos x)$

Substituting the value of $I$ in (2),

we get
$y \cdot {e^{2x}} = \cfrac{1}{5}{e^{2x}}(2\sin x - \cos x) + C$

$\Rightarrow y = \cfrac{1}{5}(2\sin x - \cos x) + C{e^{ - 2x}}$,

which is the required solution