If `(dy)/(dx)=(x-y+2)/(x-y+1),` where `u=x-y`, then

Options

A. `(x-y)/(2)+2x-y=c`
B. `x-y+((2x-y)^(2))/(2)=c`
C. `((x-y)^(2))/(2)+2x-y=c`
D. None of these

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{x - y + 2}{x - y + 1}\) using the substitution \(u = x - y\), we will follow these steps:

### Step 1: Substitute \(u = x - y\)
We start by substituting \(u = x - y\). This implies that \(y = x - u\).

### Step 2: Differentiate \(u\) with respect to \(x\)
Differentiating \(u\) with respect to \(x\), we have:
\[ \frac{du}{dx} = 1 - \frac{dy}{dx} \]
From this, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = 1 - \frac{du}{dx} \]

### Step 3: Substitute \(\frac{dy}{dx}\) into the original equation
Now, we substitute \(\frac{dy}{dx}\) into the original differential equation:
\[ 1 - \frac{du}{dx} = \frac{u + 2}{u + 1} \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ -\frac{du}{dx} = \frac{u + 2}{u + 1} - 1 \]
This simplifies to:
\[ -\frac{du}{dx} = \frac{u + 2 - (u + 1)}{u + 1} = \frac{1}{u + 1} \]
Thus, we have:
\[ \frac{du}{dx} = -\frac{1}{u + 1} \]

### Step 5: Separate variables
We can separate variables:
\[ (u + 1) du = -dx \]

### Step 6: Integrate both sides
Integrating both sides, we get:
\[ \int (u + 1) du = -\int dx \]
This results in:
\[ \frac{u^2}{2} + u = -x + C \]

### Step 7: Substitute back for \(u\)
Substituting back \(u = x - y\):
\[ \frac{(x - y)^2}{2} + (x - y) = -x + C \]

### Step 8: Rearranging the equation
Rearranging gives us:
\[ \frac{(x - y)^2}{2} + 2x - y = C \]

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