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 is:
\[ \frac{dy}{dx} = e^{y+x} - e^{y-x} \]
We can factor out \(e^y\):
\[ \frac{dy}{dx} = e^y(e^x - e^{-x}) \]

### Step 2: Separate the variables
Now, we will separate the variables \(y\) and \(x\):
\[ \frac{1}{e^y} dy = (e^x - e^{-x}) dx \]
This can be rewritten as:
\[ e^{-y} dy = (e^x - e^{-x}) dx \]

### Step 3: Integrate both sides
Now, we will integrate both sides:
\[ \int e^{-y} dy = \int (e^x - e^{-x}) dx \]
The left side integrates to:
\[ -e^{-y} + C_1 \]
The right side integrates to:
\[ e^x + e^{-x} + C_2 \]
Thus, we have:
\[ -e^{-y} = e^x + e^{-x} + C \]
where \(C = C_2 - C_1\).

### Step 4: Rearranging the equation
Now, we can rearrange the equation:
\[ e^{-y} = -e^x - e^{-x} - C \]

### Step 5: Expressing \(y\)
Taking the reciprocal gives:
\[ e^y = \frac{1}{-e^x - e^{-x} - C} \]
Taking the natural logarithm:
\[ y = \ln\left(\frac{1}{-e^x - e^{-x} - C}\right) \]
This can be simplified to:
\[ y = -\ln(-e^x - e^{-x} - C) \]

### Final Solution
Thus, the solution to the differential equation is:
\[ e^{-y} + e^{-x} + C = 0 \]