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

Options

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

Explanation

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{dy}{dx} = e^{x+y} \]
We can rewrite the right-hand side using the property of exponents:
\[ \frac{dy}{dx} = e^x \cdot e^y \]

### Step 2: Separate the variables
Next, we separate the variables \(y\) and \(x\):
\[ \frac{dy}{e^y} = e^x \, dx \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side requires the integral of \(\frac{1}{e^y}\):
\[ \int \frac{dy}{e^y} = \int e^x \, dx \]
The left side integrates to:
\[ -\frac{1}{e^y} = e^x + C \]
where \(C\) is the constant of integration.

### Step 4: Rearranging the equation
Now, we can rearrange the equation:
\[ -\frac{1}{e^y} = e^x + C \]
Multiplying both sides by \(-1\):
\[ \frac{1}{e^y} = -e^x - C \]

### Step 5: Take the reciprocal
Taking the reciprocal gives us:
\[ e^y = \frac{1}{-e^x - C} \]

### Step 6: Solve for \(y\)
Now, we take the natural logarithm of both sides to solve for \(y\):
\[ y = \ln\left(\frac{1}{-e^x - C}\right) \]
This can be simplified to:
\[ y = -\ln(-e^x - C) \]

### Final Solution
Thus, the solution of the differential equation is:
\[ y = -\ln(-e^x - C) \]