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

Options

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

Explanation

To solve the differential equation \((x+y)(dx-dy)=dx+dy\), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rearrange the Equation**:
Start with the given equation:
\[ (x+y)(dx-dy) = dx + dy \]
We can rewrite this as:
\[ dx - dy = \frac{dx + dy}{x + y} \]

2. **Express in Terms of Differentials**:
Notice that the left side can be expressed as:
\[ d(x-y) = dx - dy \]
and the right side can be expressed as:
\[ d(x+y) = dx + dy \]
Thus, we have:
\[ d(x-y) = \frac{d(x+y)}{x+y} \]

3. **Integrate Both Sides**:
Now, we will integrate both sides:
\[ \int d(x-y) = \int \frac{d(x+y)}{x+y} \]
The left side integrates to:
\[ x - y \]
For the right side, we can use the substitution \(t = x + y\), which gives \(dt = d(x+y)\). Therefore, we have:
\[ \int \frac{dt}{t} = \ln |t| + C = \ln |x+y| + C \]

4. **Combine the Results**:
Setting the results from both integrals equal gives:
\[ x - y = \ln |x+y| + C \]

5. **Exponentiate to Solve for \(x+y\)**:
To eliminate the logarithm, we exponentiate both sides:
\[ e^{x - y} = e^{\ln |x+y| + C} \]
This simplifies to:
\[ e^{x - y} = |x+y| \cdot e^C \]
Let \(k = e^C\) (where \(k\) is a constant), we have:
\[ |x+y| = k e^{x - y} \]

6. **Final Form**:
We can drop the absolute value for simplicity (assuming \(x+y\) is positive):
\[ x + y = k e^{x - y} \]

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