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=ke^(x-y)`
D. `x-y=ke^(x+y)`

Answer

To solve the differential equation \((x+y)(dx-dy) = dx + dy\), we will follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the given equation:
\[ (x+y)(dx - dy) = dx + dy \]
Expanding the left side gives:
\[ (x+y)dx - (x+y)dy = dx + dy \]

### Step 2: Collecting Like Terms
Next, we can collect the terms involving \(dx\) and \(dy\):
\[ (x+y)dx - dx = (x+y)dy + dy \]
This simplifies to:
\[ [(x+y) - 1]dx = [(x+y) + 1]dy \]

### Step 3: Separating Variables
Now, we can separate the variables:
\[ \frac{dx}{(x+y) + 1} = \frac{dy}{(x+y) - 1} \]

### Step 4: Integrating Both Sides
We will integrate both sides:
\[ \int \frac{dx}{(x+y) + 1} = \int \frac{dy}{(x+y) - 1} \]
Let \(z = x + y\). Then \(dx = dz - dy\). Substituting this into the integral gives:
\[ \int \frac{dz - dy}{z + 1} = \int \frac{dy}{z - 1} \]

### Step 5: Solving the Integrals
Now we can solve the integrals:
1. The left side becomes:
\[ \int \frac{dz}{z + 1} - \int \frac{dy}{z + 1} \]
which gives:
\[ \log|z + 1| - \log|z + 1| = \log|z + 1| - \log|z - 1| \]

2. The right side becomes:
\[ \log|z - 1| \]

### Step 6: Combining the Results
Combining the results from both sides gives:
\[ \log|z + 1| = \log|z - 1| + C \]
where \(C\) is the constant of integration.

### Step 7: Exponentiating Both Sides
Exponentiating both sides leads to:
\[ \frac{z + 1}{z - 1} = e^C \]
Let \(k = e^C\), then we have:
\[ z + 1 = k(z - 1) \]

### Step 8: Substituting Back for \(z\)
Substituting back \(z = x + y\):
\[ x + y + 1 = k(x + y - 1) \]

### Step 9: Rearranging to Find the Solution
Rearranging gives:
\[ x + y + 1 = kx + ky - k \]
This can be rearranged to find the relationship between \(x\) and \(y\).

### Final Solution
The final solution can be simplified to:
\[ \frac{x + y + 1}{x + y - 1} = k \]