Question

Solve the differential equation by variable separable method:$(e^y+1)\cos xdx+e^y\sin xdy=0.$

Answer 1

Consider given the deferential equation,

$(e^y+1)\cos xdx+e^y\sin xdy=0$

$(e^y+1)\cos xdx=-e^y\sin xdy$

$\frac{\cos xdx}{\sin x}=-\frac{e^y}{e^y+1}dy$

$\cot x.dx=-\frac{e^y}{e^y+1}dy$

Taking integration both sides,

$\int \cot x.dx=-\int \frac{e^y}{e^y+1}dy$

$\log \sin x=-\log (e^y+1)+\log C$

$\log \sin x=\log \frac{C}{(e^y+1)}$

$\sin x=\frac{C}{(e^y+1)}$

$(e^y+1)\sin x=C$

Hence, this is the answer.