Question

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

Answer 1

Given,

$y^{\prime }=e^x\sin x$

integrating on both sides, we get,

$\int y^{\prime }=\int e^x\sin xdx$

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

Since the curve passes through $(0,0)$

$0=\frac{1}{2}{e}^0(\sin 0-\cos 0)+c$

$0=-\frac{1}{2}+c$

$\therefore c=\frac{1}{2}$

Therefore the required equation is

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

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

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

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