Solution of the differential equation `((x+y-1)/(x+y-2))(dy)/(dx)=((x+y+1)/(x+y+1)),` given that `y = 1` when `x = 1`, is

Options

A. `log|((x-y)^(2)-2)/(2)|=2(x+y)`
B. `log|((x-y)^(2)+2)/(2)|=2(x-y)`
C. `log|((x-y)^(2)+2)/(2)|=2(x-y)`
D. none of these

Explanation

To solve the differential equation

\[ \frac{(x+y-1)}{(x+y-2)} \frac{dy}{dx} = \frac{(x+y+1)}{(x+y+1)}, \]

given the initial condition \(y = 1\) when \(x = 1\), we will follow these steps:

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

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

This implies:

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

### Step 2: Rewrite the Differential Equation
Substituting \(u\) into the differential equation, we get:

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

### Step 3: Simplifying the Equation
Expanding this, we have:

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

Rearranging gives:

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

### Step 4: Combine Terms
Combining the terms on the right side:

\[ \frac{(u-1)}{(u-2)} \frac{du}{dx} = \frac{(u-2) + (u-1)}{(u-2)} = \frac{2u - 3}{(u-2)}. \]

### Step 5: Cross-Multiplication
Cross-multiplying gives:

\[ (u-1) du = (2u - 3) dx. \]

### Step 6: Separate Variables
Rearranging gives:

\[ \frac{(u-1)}{(2u - 3)} du = dx. \]

### Step 7: Integrate Both Sides
Integrating both sides:

\[ \int \frac{(u-1)}{(2u - 3)} du = \int dx. \]

### Step 8: Solve the Integral
To solve the left integral, we can use partial fraction decomposition. After integrating, we will have:

\[ \frac{1}{2} \ln |2u - 3| + C = x + C_1. \]

### Step 9: Substitute Back for \(u\)
Substituting back \(u = x + y\):

\[ \frac{1}{2} \ln |2(x + y) - 3| = x + C_1. \]

### Step 10: Solve for \(y\)
To express \(y\) in terms of \(x\), we rearrange:

\[ \ln |2(x + y) - 3| = 2x + C. \]

Exponentiating both sides gives:

\[ |2(x + y) - 3| = e^{2x + C}. \]

### Step 11: Apply Initial Condition
Using the initial condition \(y(1) = 1\):

\[ |2(1 + 1) - 3| = e^{2(1) + C} \implies |4 - 3| = e^{2 + C} \implies 1 = e^{2 + C}. \]

Thus, \(C = -2\).

### Final Solution
Substituting back, we get:

\[ |2(x + y) - 3| = e^{2x - 2}. \]

This leads to the final form of the solution:

\[ 2(x + y) - 3 = \pm e^{2x - 2}. \]