Solve : `e^(y)dx+e^(x)dy=0`.

Answer

To solve the differential equation \( e^{y}dx + e^{x}dy = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
We start by rearranging the equation to isolate the differentials. We can express it as:
\[ e^{y}dx = -e^{x}dy \]
This can be rewritten as:
\[ \frac{dx}{dy} = -\frac{e^{x}}{e^{y}} \]

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

### Step 3: Integrate both sides
Now we integrate both sides. The left side becomes:
\[ \int \frac{dx}{e^{x}} = \int e^{-x}dx = -e^{-x} + C_1 \]
And the right side becomes:
\[ \int -\frac{dy}{e^{y}} = -\int e^{-y}dy = e^{-y} + C_2 \]

### Step 4: Combine the constants
We can combine the constants of integration \( C_1 \) and \( C_2 \) into a single constant \( C \):
\[ -e^{-x} = e^{-y} + C \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ e^{-x} + e^{-y} = C \]

### Step 6: Final form
We can rewrite this as:
\[ e^{-x}e^{-y} = C \]
or equivalently,
\[ e^{-x} + e^{-y} = C \]

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