Solve the following differential equations.
` (dy)/(dx) = ( x +y+1)/(x +y-1)`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{x + y + 1}{x + y - 1}\), we can follow these steps:

### Step 1: Substitute \(t = x + y\)
Let \(t = x + y\). Then, we can express \(y\) in terms of \(t\) and \(x\):
\[ y = t - x \]
Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{dt}{dx} - 1 \]

### Step 2: Rewrite the differential equation
Substituting \(t\) into the original differential equation gives:
\[ \frac{dt}{dx} - 1 = \frac{t + 1}{t - 1} \]
Rearranging this, we have:
\[ \frac{dt}{dx} = \frac{t + 1}{t - 1} + 1 \]

### Step 3: Simplify the right-hand side
Combine the terms on the right-hand side:
\[ \frac{dt}{dx} = \frac{t + 1 + (t - 1)}{t - 1} = \frac{2t}{t - 1} \]

### Step 4: Separate variables
We can separate the variables \(t\) and \(x\):
\[ \frac{t - 1}{2t} dt = dx \]

### Step 5: Integrate both sides
Now, we integrate both sides:
\[ \int \left(\frac{1}{2} - \frac{1}{2t}\right) dt = \int dx \]
This gives:
\[ \frac{1}{2}t - \frac{1}{2}\ln |t| = x + C \]

### Step 6: Substitute back for \(t\)
Substituting back \(t = x + y\), we have:
\[ \frac{1}{2}(x + y) - \frac{1}{2}\ln |x + y| = x + C \]

### Step 7: Rearranging the equation
Rearranging gives us the final implicit solution:
\[ \frac{1}{2}\ln |x + y| = \frac{1}{2}(x + y) - x - C \]
or
\[ \frac{1}{2}\ln |x + y| = \frac{1}{2}y - \frac{1}{2}x - C \]

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

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