Question

The general solution of the differential equation $(e^y+1)\cos xdx+e^y\sin xdy=0$ is
(where $c$ is constant of integration)

• A. $\left(e^y\right)=c\sin x$
• B. $\left(e^y\right)\sin x=c$
• C. $(e^y+1)\sin x=c$
• D. $(e^y+1)\cos x=c$

Answer 1

The correct option is C $(e^y+1)\sin x=c$

Given: $(e^y+1)\cos xdx+e^y\sin xdy=0$
$\Rightarrow \cot xdx=-\frac{e^y}{(e^y+1)}dy$
Integrating both sides, we get
$\int \cot xdx=-\int \frac{e^y}{(e^y+1)}dy$
Put $e^y+1=t\Rightarrow e^{y}dy=dt$
$\Rightarrow \int \cot xdx=-\int \frac{1}{t}dt$
$\Rightarrow \ln |\sin x|=-\ln |t|+\ln c$
$\Rightarrow (e^y+1)\sin x=c$