The solution of the differential equation `(dy)/(dx)=e^(y+x)+e^(y-x),` is

Options

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

Answer

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

### Step 1: Rewrite the equation
The given equation can be rewritten as:
\[ \frac{dy}{dx} = e^y \cdot e^x + e^y \cdot e^{-x} \]
This simplifies to:
\[ \frac{dy}{dx} = e^y (e^x + e^{-x}) \]

### Step 2: Separate variables
We can separate variables by dividing both sides by \(e^y\) and multiplying both sides by \(dx\):
\[ \frac{dy}{e^y} = (e^x + e^{-x}) dx \]

### Step 3: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{dy}{e^y} = \int (e^x + e^{-x}) dx \]
The left side integrates to:
\[ \int e^{-y} dy = -e^{-y} \]
The right side integrates to:
\[ \int (e^x + e^{-x}) dx = e^x - e^{-x} + C \]

### Step 4: Combine results
Combining the results from the integration gives:
\[ -e^{-y} = e^x - e^{-x} + C \]

### Step 5: Rearranging the equation
To express \(e^{-y}\), we can multiply through by -1:
\[ e^{-y} = -e^x + e^{-x} - C \]

### Step 6: Solve for \(y\)
Taking the natural logarithm of both sides to solve for \(y\):
\[ -y = \ln(-e^x + e^{-x} - C) \]
Thus,
\[ y = -\ln(-e^x + e^{-x} - C) \]

### Final Solution
The solution of the differential equation is:
\[ y = -\ln(-e^x + e^{-x} - C) \]