Solve each of the following differential equations
`cos y "" dx +(1+2e^(-x)) sin y "" dy=0`.

Answer

To solve the differential equation \( \cos y \, dx + (1 + 2e^{-x}) \sin y \, dy = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables \( x \) and \( y \):
\[ \cos y \, dx + (1 + 2e^{-x}) \sin y \, dy = 0 \]
This can be rewritten as:
\[ \sin y \, dy = -\frac{\cos y}{1 + 2e^{-x}} \, dx \]

### Step 2: Separating Variables
Now, we can separate the variables:
\[ \frac{\sin y}{\cos y} \, dy = -\frac{1}{1 + 2e^{-x}} \, dx \]
This simplifies to:
\[ \tan y \, dy = -\frac{1}{1 + 2e^{-x}} \, dx \]

### Step 3: Integrating Both Sides
Next, we integrate both sides:
\[ \int \tan y \, dy = -\int \frac{1}{1 + 2e^{-x}} \, dx \]
The left side integrates to:
\[ -\ln |\cos y| + C_1 \]
For the right side, we can use substitution. Let \( t = 1 + 2e^{-x} \), then \( dt = -2e^{-x} \, dx \) or \( dx = -\frac{dt}{2e^{-x}} \). We can express \( e^{-x} \) in terms of \( t \):
\[ e^{-x} = \frac{t - 1}{2} \]
Thus, we have:
\[ dx = -\frac{dt}{t - 1} \]
Now, substituting this into the integral gives:
\[ -\int \frac{1}{t} \left(-\frac{dt}{t - 1}\right) = \int \frac{dt}{t(t - 1)} \]
Using partial fractions, we can write:
\[ \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \]
Solving for \( A \) and \( B \), we find:
\[ 1 = A(t - 1) + Bt \]
Setting \( t = 1 \) gives \( A = 1 \), and setting \( t = 0 \) gives \( B = -1 \). Thus:
\[ \int \frac{dt}{t(t - 1)} = \ln |t| - \ln |t - 1| + C_2 \]

### Step 4: Combining Results
Combining both sides, we have:
\[ -\ln |\cos y| = \ln |t| - \ln |t - 1| + C \]
This simplifies to:
\[ \ln |\cos y| = -\ln |t| + \ln |t - 1| + C \]
Exponentiating both sides gives:
\[ |\cos y| = \frac{C(t - 1)}{t} \]

### Step 5: Substituting Back for \( t \)
Now substituting back \( t = 1 + 2e^{-x} \):
\[ |\cos y| = \frac{C(2e^{-x})}{1 + 2e^{-x}} \]

### Final Solution
Thus, the solution to the differential equation can be expressed as:
\[ \sec y = \frac{2e^{-x}}{C(1 + 2e^{-x})} \]
or rearranging gives:
\[ C = \frac{2e^{-x}}{\sec y (1 + 2e^{-x})} \]