Solution of the differential equation `(2x-2y+3)dx-(x-y+1)dy=0,` when y(0)=1 is

Options

A. `2x-y+log|x-y+2|+1=0`
B. `2x+y-log|x-y+2|+1=0`
C. `2x-y-log|x-y+2|+1=0`
D. `2x+y+log|x-y+2|+1=0`

Answer

To solve the differential equation \((2x - 2y + 3)dx - (x - y + 1)dy = 0\) with the initial condition \(y(0) = 1\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ (2x - 2y + 3)dx - (x - y + 1)dy = 0 \]
Rearranging gives us:
\[ (2x - 2y + 3)dx = (x - y + 1)dy \]
Dividing both sides by \(dx\) and \(dy\) leads to:
\[ \frac{dy}{dx} = \frac{2x - 2y + 3}{x - y + 1} \]

### Step 2: Simplifying the Equation
We can simplify the right-hand side:
\[ \frac{dy}{dx} = \frac{2(x - y) + 3}{x - y + 1} \]
Let \(T = x - y\). Then, we have:
\[ \frac{dy}{dx} = 2T + 3 \]
Now, differentiating \(T\) gives:
\[ \frac{dT}{dx} = 1 - \frac{dy}{dx} \]
Substituting \(\frac{dy}{dx}\) into this equation:
\[ \frac{dT}{dx} = 1 - (2T + 3) = -2T - 2 \]

### Step 3: Separating Variables
Now we can separate variables:
\[ \frac{dT}{-2T - 2} = dx \]
This can be rewritten as:
\[ \frac{dT}{-2(T + 1)} = dx \]

### Step 4: Integrating Both Sides
Integrating both sides:
\[ \int \frac{1}{-2(T + 1)} dT = \int dx \]
This gives us:
\[ -\frac{1}{2} \ln |T + 1| = x + C \]

### Step 5: Solving for \(T\)
Multiplying through by -2:
\[ \ln |T + 1| = -2x - 2C \]
Exponentiating both sides:
\[ |T + 1| = e^{-2x - 2C} = Ke^{-2x} \quad \text{(where \(K = e^{-2C}\))} \]
Thus, we have:
\[ T + 1 = \pm Ke^{-2x} \]
Substituting back \(T = x - y\):
\[ x - y + 1 = \pm Ke^{-2x} \]
This leads to:
\[ y = x + 1 \mp Ke^{-2x} \]

### Step 6: Applying Initial Condition
Using the initial condition \(y(0) = 1\):
\[ 1 = 0 + 1 \mp K \implies 0 = \mp K \implies K = 0 \]
Thus, we have:
\[ y = x + 1 \]

### Final Solution
The solution to the differential equation is:
\[ \boxed{y = x + 1} \]