Question

Y^-1 * dy+y e^cosx * sinx * dx=0

Answer 1

$\dfrac{\mathrm dy}y+ye^{\cos x}\sin x\,\mathrm dx=0$
$\implies\dfrac{\mathrm dy}{y^2}=-e^{\cos x}\sin x\,\mathrm dx$

$\displaystyle\int\frac{\mathrm dy}{y^2}=-\int e^{\cos x}\sin x\,\mathrm dx$
$-\dfrac1y=e^{\cos x}+C$
$y=-\dfrac1{e^{\cos x}+C}$