Question

The solution of differential equation
$(e^y+1)cosxdx+e^{y}sinxdy=0$ is

• A. $(e^y+1)sinx=c$
• B. $e^{x}sinx=c$
• C. $(e^x+1)cosx=c$
• D. none of these

Answer 1

Answer: (A)

Given $(e^y+1)cosxdx+e^{y}sinxdy=0$
$\Rightarrow \frac{e^y}{1+e^y}dy=-\frac{cosx}{sinx}dx$
$\Rightarrow \left(1-\frac{1}{1+e^y}\right)dy=-cotxdx$
$\Rightarrow \left(1-\frac{\left(-e^{-y}\right)}{1+e^{-y}}\right)dy=-cotxdx$
On integrating, we get $y+log\left(\frac{1+e^y}{e^y}\right)=log\left(\frac{c}{sinx}\right)$.
$\Rightarrow y=log\left(\frac{c}{sinx}\right)+log\left(\frac{e^y}{1+e^y}\right)$
$\Rightarrow y=log\left(\frac{c{e}^y}{sinx\left(1+e^y\right)}\right)$
$\Rightarrow e^y=\frac{c{e}^y}{sinx\left(1+e^y\right)}$
$\Rightarrow c=sinx\left(1+e^y\right)$