What is the general solution of `(1 + e^(x)) y dy = e^(x) ` dx ?

Options

A. `y^(2)` = ln [`C^(2) (e^(x) + 1)^(2)]`
B. dy = ln `(e^(x) + 1)`
C. `y^(2)` = ln `[ C (e^(x) + 1)]`
D. None of these

Answer

To find the general solution of the differential equation \( (1 + e^{x}) y \, dy = e^{x} \, dx \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (1 + e^{x}) y \, dy = e^{x} \, dx \]
We can rearrange this to isolate \( y \, dy \) on one side:
\[ y \, dy = \frac{e^{x}}{1 + e^{x}} \, dx \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side integrates to:
\[ \int y \, dy = \frac{y^2}{2} + C_1 \]
For the right side, we need to integrate:
\[ \int \frac{e^{x}}{1 + e^{x}} \, dx \]
To solve this integral, we can use the substitution \( u = 1 + e^{x} \), which gives \( du = e^{x} \, dx \). Thus, the integral becomes:
\[ \int \frac{1}{u} \, du = \ln |u| + C_2 = \ln |1 + e^{x}| + C_2 \]

### Step 3: Equating the Results
Now we equate the results of the integrals:
\[ \frac{y^2}{2} = \ln |1 + e^{x}| + C \]
where \( C = C_2 - C_1 \).

### Step 4: Solving for \( y^2 \)
To express \( y^2 \) in terms of \( x \), we multiply both sides by 2:
\[ y^2 = 2 \ln |1 + e^{x}| + 2C \]

### Step 5: Final Form of the General Solution
We can rewrite \( 2C \) as a new constant \( C' \):
\[ y^2 = 2 \ln |1 + e^{x}| + C' \]
This is the general solution of the given differential equation.