Find the equation of a curve passing through the point $(0,0)$ and whose differential equation is$y' = {e^x}\sin x$.

.: We have, $y' = {e^x}\sin x \Rightarrow \cfrac{{dy}}{{dx}} = {e^x}\sin x$

$\Rightarrow dy = {e^x}\sin xdx$ …(1)

Integrating (1) both sides,

we get
$\int d y = \int {{e^x}} \sin xdx$

$\Rightarrow y = - {e^x}\cos x + \int {{e^x}} \cos xdx$

(Integrating by parts)$\Rightarrow y = - {e^x}\cos x + {e^x}\sin x - \int {{e^x}} \sin xdx$

$\Rightarrow y = - {e^x}\cos x + {e^x}\sin x - y + C$

$\Rightarrow 2y = - {e^x}\cos x + {e^x}\sin x + C$

As point $(0,0)$ lies on it, i.e., $x = 0,y = 0$

$\therefore$ $0 = - {e^0} + C \Rightarrow C = 1$

$\therefore$

Required equation is $2y = - {e^x}\cos x + {e^x}\sin x + 1$

$\Rightarrow 2y - 1 = {e^x}(\sin x - \cos x)$