Question

The general solution of the equation

$\left( e ^ { y } + 1 \right) \cos x d x + e ^ { y } \sin x d y = 0$ is

• A. $\left( e ^ { y } + 1 \right) \cos x = c$
• B. $\left( e ^ { y } - 1 \right) \sin x = c$
• C. $\left( e ^ { y } + 1 \right) \sin x = c$
• D. None of these

Answer 1

Answer: (C)

$\left( e ^ { y } + 1 \right) \sin x = c$

Answer 2

$\left( e ^ { y } + 1 \right) \cos x d x + e ^ { y } \sin x d y = 0$

⇒ $\frac { e ^ { y } d y } { e ^ { y } + 1 } + \frac { \cos x } { \sin x } d x = 0$

On integrating both the functions, we get

$\log \left( e ^ { y } + 1 \right) + \log ( \sin x ) = \log c$ ⇒$\left( e ^ { y } + 1 \right) \sin x = c$.