What is the solution of the differential equation (x + y) (dx- dy) = dx + dy ?

Options

A. A) x + y + ln (x + y) = c
B. B) x - y + ln (x + y)= c
C. C) y- x + ln (x + y) = c
D. D) y - x - ln (x - y) = c

Answer

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

### Step 1: Expand the equation
Start by expanding the left side of the equation:
\[ (x + y)(dx - dy) = dx + dy \]
This gives:
\[ x \, dx + y \, dx - x \, dy - y \, dy = dx + dy \]

### Step 2: Rearrange the equation
Rearranging the equation, we can group the terms involving \(dx\) and \(dy\):
\[ x \, dx + y \, dx - dx - y \, dy + x \, dy = 0 \]
This simplifies to:
\[ (x + y - 1) \, dx = (y + 1) \, dy \]

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

### Step 4: Substitute for simplification
Let’s use the substitution \(v = x + y\). Then, differentiating both sides with respect to \(x\):
\[ \frac{dv}{dx} = 1 + \frac{dy}{dx} \]
This implies:
\[ \frac{dy}{dx} = \frac{dv}{dx} - 1 \]

### Step 5: Substitute into the equation
Substituting \(\frac{dy}{dx}\) into our equation gives:
\[ \frac{dv}{dx} - 1 = \frac{v - 1}{v + 1} \]
Rearranging yields:
\[ \frac{dv}{dx} = \frac{v - 1}{v + 1} + 1 = \frac{2v}{v + 1} \]

### Step 6: Separate the variables again
Now we can separate the variables again:
\[ \frac{dv}{2v} = \frac{dx}{v + 1} \]

### Step 7: Integrate both sides
Integrating both sides:
\[ \int \frac{1}{2v} \, dv = \int \frac{1}{v + 1} \, dx \]
This gives:
\[ \frac{1}{2} \ln |v| = \ln |v + 1| + C \]

### Step 8: Solve for \(v\)
Exponentiating both sides to eliminate the logarithm:
\[ |v|^{1/2} = |v + 1| e^{C} \]
Let \(C' = e^{C}\), we can rewrite this as:
\[ v^{1/2} = C'(v + 1) \]

### Step 9: Substitute back for \(v\)
Substituting back \(v = x + y\):
\[ (x + y)^{1/2} = C'(x + y + 1) \]

### Step 10: Rearranging to find the solution
Rearranging gives us the final form:
\[ y - x + \ln |x + y| = C \]

This is the required solution to the differential equation.