The general 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`

Answer

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 given equation:
\[ \frac{dy}{dx} = e^{x+y} \]
We can rewrite the right-hand side using the property of exponents:
\[ e^{x+y} = e^x \cdot e^y \]
So, we have:
\[ \frac{dy}{dx} = e^x \cdot e^y \]

### Step 2: Separate the variables
Next, we will separate the variables \(y\) and \(x\). We can rearrange the equation as follows:
\[ e^{-y} dy = e^x dx \]

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

### Step 4: Rearranging the equation
To express the solution in a more standard form, we can multiply through by \(-1\):
\[ e^{-y} = -e^x - C \]
This can also be written as:
\[ e^{-y} + e^x + C = 0 \]

### Step 5: Solve for \(y\)
To solve for \(y\), we take the natural logarithm of both sides:
\[ -y = \ln(-e^x - C) \]
Thus, we have:
\[ y = -\ln(-e^x - C) \]

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