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 these steps:

### Step 1: Rearranging the Equation
Start by rearranging the given equation:
\[ (x+y)(dx - dy) = dx + dy \]
Expanding the left-hand side:
\[ (x+y)dx - (x+y)dy = dx + dy \]
Now, we can move all terms involving \(dx\) and \(dy\) to one side:
\[ (x+y)dx - dx - (x+y)dy - dy = 0 \]
This simplifies to:
\[ [(x+y) - 1]dx - [(x+y) + 1]dy = 0 \]

### Step 2: Separating Variables
We can separate the variables by rearranging the equation:
\[ \frac{dx}{dy} = \frac{(x+y) + 1}{(x+y) - 1} \]
This allows us to express the equation in a form suitable for integration.

### Step 3: Integrating Both Sides
Now, we will integrate both sides. We can rewrite the equation as:
\[ \int \frac{(x+y) + 1}{(x+y) - 1} dy = \int dx \]
Let \(z = x + y\), then \(dz = dx + dy\):
\[ \int \frac{z + 1}{z - 1} dz = \int dz \]
This integral can be solved using partial fractions or substitution.

### Step 4: Solving the Integral
The left-hand side can be simplified:
\[ \int \left(1 + \frac{2}{z - 1}\right) dz = z + 2 \ln |z - 1| + C \]
The right-hand side integrates to:
\[ x + C_1 \]

### Step 5: Combining Results
Now, we equate both integrals:
\[ z + 2 \ln |z - 1| = x + C_1 \]
Substituting back \(z = x + y\):
\[ x + y + 2 \ln |(x + y) - 1| = x + C_1 \]
This simplifies to:
\[ y + 2 \ln |(x + y) - 1| = C \]
where \(C\) is a constant.

### Final Solution
Thus, the solution of the differential equation is:
\[ y + 2 \ln |(x + y) - 1| = C \]