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 will use a substitution method. Let's follow the steps:

### Step 1: Substitution
Let \( u = x + y \). Then, we differentiate \( u \) with respect to \( x \):

\[ \frac{du}{dx} = 1 + \frac{dy}{dx}. \]

From this, we can express \( \frac{dy}{dx} \) in terms of \( u \):

\[ \frac{dy}{dx} = \frac{du}{dx} - 1. \]

### Step 2: Substitute into the original equation
Now, substitute \( u \) into the original differential equation:

\[ \frac{du}{dx} - 1 = \frac{u + 1}{u - 1}. \]

Rearranging gives:

\[ \frac{du}{dx} = \frac{u + 1}{u - 1} + 1. \]

### Step 3: Simplify the right-hand side
We can simplify the right-hand side:

\[ \frac{du}{dx} = \frac{u + 1 + (u - 1)}{u - 1} = \frac{2u}{u - 1}. \]

### Step 4: Separate variables
Now we separate the variables:

\[ \frac{u - 1}{2u} \, du = dx. \]

### Step 5: Integrate both sides
Integrate both sides:

\[ \int \frac{u - 1}{2u} \, du = \int dx. \]

This can be split into two parts:

\[ \int \left( \frac{1}{2} - \frac{1}{2u} \right) du = x + C. \]

The left side integrates to:

\[ \frac{1}{2}u - \frac{1}{2} \ln |u| = x + C. \]

### Step 6: Substitute back for \( u \)
Now, substitute back \( u = x + y \):

\[ \frac{1}{2}(x + y) - \frac{1}{2} \ln |x + y| = x + C. \]

### Step 7: Rearranging the equation
Multiply through by 2 to eliminate the fraction:

\[ x + y - \ln |x + y| = 2x + 2C. \]

Let \( k = 2C \), then we have:

\[ y - \ln |x + y| = x + k. \]

### Final Result
Thus, the solution to the differential equation is:

\[ y - \ln |x + y| = x + k. \]