Question

The solution of differential equation
$(e^y + 1) cos\, x\,dx + e^y\, sin\, x dy = 0$ is

• A. $(e^y + 1) sin \,x = c$
• B. $e^x \,sin\, x = c$
• C. $(e^x + 1) cos \,x = c$
• D. none of these

Answer 1

Answer: (A)

Given $(e^y + 1) \,cos\, x\, dx + e^y\,sin \,x\, dy = 0$
$\Rightarrow \frac{e^{y}}{1+e^{y}}dy=-\frac{cos\,x}{sin\,x}dx$
$\Rightarrow \left(1-\frac{1}{1+e^{y}}\right)dy=-cot\,x\,dx$
$\Rightarrow \left(1-\frac{\left(-e^{-y}\right)}{1+e^{-y}}\right)dy=-cot\,x\,dx$
On integrating, we get $y+log\left(\frac{1+e^{y}}{e^{y}}\right)=log\left(\frac{c}{sin\,x}\right)$.
$\Rightarrow y=log\left(\frac{c}{sin\,x}\right)+log\left(\frac{e^{y}}{1+e^{y}}\right)$
$\Rightarrow y=log\left(\frac{ce^{y}}{sin\,x\left(1+e^{y}\right)}\right)$
$\Rightarrow e^{y}=\frac{ce^{y}}{sin\,x\left(1+e^{y}\right)}$
$\Rightarrow c=sin\,x\left(1+e^{y}\right)$