`(x+y)(dx-dy)=dx+dy`, where `x+y=u`

Options

A. `x+y=log(x-y)+c`
B. `x+y=log(x+y)+c`
C. `x-y=logu+c`
D. `u=logu+c`

Answer

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

### Step 1: Rewrite the equation
Start with the given equation:
\[ (x+y)(dx-dy) = dx + dy \]
Substituting \(u = x + y\), we can express \(dx\) and \(dy\) in terms of \(u\).

### Step 2: Substitute \(u\)
Using the substitution \(u = x + y\), we can rewrite \(dx\) as:
\[ dx = du - dy \]
Now, substitute \(u\) into the equation:
\[ u(dx - dy) = dx + dy \]

### Step 3: Expand and rearrange
Expanding the left side gives:
\[ udx - u dy = dx + dy \]
Now, rearranging the equation, we have:
\[ udx - dx = dy + u dy \]
Factoring out \(dx\) and \(dy\):
\[ (ud - 1)dx = (1 + u)dy \]

### Step 4: Separate variables
Now, we can separate the variables:
\[ \frac{dx}{1 + u} = \frac{dy}{u - 1} \]

### Step 5: Integrate both sides
Integrate both sides:
\[ \int \frac{dx}{1 + u} = \int \frac{dy}{u - 1} \]
The left side integrates to:
\[ \ln |1 + u| + C_1 \]
And the right side integrates to:
\[ \ln |u - 1| + C_2 \]

### Step 6: Combine the results
Setting the two integrals equal gives:
\[ \ln |1 + u| = \ln |u - 1| + C \]
Where \(C = C_2 - C_1\).

### Step 7: Exponentiate to eliminate the logarithm
Exponentiating both sides results in:
\[ |1 + u| = K|u - 1| \]
Where \(K = e^C\).

### Step 8: Substitute back for \(u\)
Substituting back \(u = x + y\):
\[ |1 + (x + y)| = K|(x + y) - 1| \]

### Final Result
Thus, the solution to the differential equation is:
\[ |1 + x + y| = K|(x + y) - 1| \]