`(e^(y)+1)cos x dx+e^(y)sin x dy =0`

Options

A. `(e^(y)+1)sin x =c`
B. `(e^(y)+1)=c tan x`
C. `e^(y)sin x + cos x=c`
D. `(e^(x)+1)sin y=c`

Answer

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

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

### Step 2: Simplifying the Equation
Recognizing that \(\frac{\cos x}{\sin x} = \cot x\), we rewrite the equation:
\[ \cot x \, dx = -\frac{e^{y}}{e^{y}+1} \, dy \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \cot x \, dx = -\int \frac{e^{y}}{e^{y}+1} \, dy \]

The integral of \(\cot x\) is:
\[ \int \cot x \, dx = \ln |\sin x| + C_1 \]

For the right-hand side, we can use substitution. Let \(t = e^{y} + 1\), then \(dt = e^{y} \, dy\) or \(dy = \frac{dt}{e^{y}} = \frac{dt}{t-1}\). Thus, we can rewrite the integral:
\[ -\int \frac{e^{y}}{e^{y}+1} \, dy = -\int \frac{t-1}{t} \cdot \frac{dt}{t-1} = -\int \left(1 - \frac{1}{t}\right) dt \]
This simplifies to:
\[ -\left(t - \ln |t|\right) + C_2 = -\left(e^{y} + 1 - \ln |e^{y}+1|\right) + C_2 \]

### Step 4: Combining Results
Combining both integrals, we have:
\[ \ln |\sin x| = -\left(e^{y} + 1 - \ln |e^{y}+1|\right) + C \]

### Step 5: Exponentiating Both Sides
Exponentiating both sides gives us:
\[ |\sin x| = C \cdot (e^{y}+1)e^{-e^{y}} \]

### Final Solution
Thus, we can express the final solution as:
\[ e^{y} + 1 = C \sin x \]