The differential equation ` (e^(x)+1)y dy = (y+1) e^(x) dx ` has the solution

Options

A. `(y-1)(e^(x)-1) = Ce^(y)`
B. ` (y-1)(e^(x)+1) = Ce^(y)`
C. ` (y+1) (e^(x)-1)=Ce^(y)`
D. ` (y+1)(e^(x)+1)= Ce^(y)`

Explanation

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

### Step 1: Rearranging the Equation
We start by rearranging the given differential equation:

\[ (e^{x}+1)y \, dy = (y+1) e^{x} \, dx \]

This can be rewritten as:

\[ \frac{y \, dy}{y + 1} = \frac{e^{x} \, dx}{e^{x} + 1} \]

### Step 2: Integrating Both Sides
Next, we integrate both sides.

**Left Side:**

\[ \int \frac{y \, dy}{y + 1} \]

To integrate this, we can use the method of partial fractions or rewrite it as:

\[ \int \left( 1 - \frac{1}{y + 1} \right) dy = \int dy - \int \frac{1}{y + 1} dy \]

This gives us:

\[ y - \log|y + 1| \]

**Right Side:**

\[ \int \frac{e^{x} \, dx}{e^{x} + 1} \]

Let \( t = e^{x} + 1 \), then \( dt = e^{x} \, dx \) which gives:

\[ \int \frac{dt}{t} = \log|t| = \log|e^{x} + 1| \]

### Step 3: Equating the Integrals
Now we equate the results from both integrations:

\[ y - \log|y + 1| = \log|e^{x} + 1| + C \]

### Step 4: Rearranging the Equation
To express the solution clearly, we can rearrange this equation:

\[ y - \log|y + 1| - \log|e^{x} + 1| = C \]

### Step 5: Final Form of the Solution
We can express this in a more standard form:

\[ y + 1 = e^{C} (e^{x} + 1)^{k} \]

Where \( k \) is a constant that can be determined based on initial conditions.

### Summary of the Solution
The solution to the differential equation is given by:

\[ y + 1 = K (e^{x} + 1) \]

where \( K \) is a constant determined by initial conditions.