Answer to Question #131543 in Differential Equations for Rouhish Ray

Question

Solve: e^(x)sinydx + (e^x +1)cosydy=0

Answer 1

We separate the variables and receive the following:

$$\frac{cos\,y}{sin\,y}dy+\frac{e^x}{e^x+1}dx=0$$

After integration we receive $ln(sin\,y)+ln(e^x+1)+c=0,\,\,c\in\mathbb{R}$

From this we receive that $ln(sin\,y(e^x+1)e^c)=0$ . Thus, $sin\,y(e^x+1)e^c=1$

The latter implies that $sin\,y=\frac{1}{(e^x+1)e^c}$ . From the latter we find that $y=(-1)^narcsin\left(\frac{1}{(e^x+1)e^c}\right)+\pi n.$

Answer: $y=(-1)^narcsin\left(\frac{1}{(e^x+1)e^c}\right)+\pi n$ , $c\in\mathbb{R}$