`(2x -2y+3) dx -(x-y+1) dy=0 , " when" x=0, y=1 `

Answer

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

### Step 1: Rearranging the Equation
We start with the given differential equation:

\[ (2x - 2y + 3) dx - (x - y + 1) dy = 0 \]

We can rearrange it to isolate \(dy\):

\[ (2x - 2y + 3) dx = (x - y + 1) dy \]

### Step 2: Expressing dy/dx
Dividing both sides by \((x - y + 1)\) gives:

\[ \frac{dy}{dx} = \frac{2x - 2y + 3}{x - y + 1} \]

### Step 3: Substituting Variables
Let \(u = x - y\). Then, we have:

\[ y = x - u \quad \text{and} \quad \frac{dy}{dx} = 1 - \frac{du}{dx} \]

Substituting these into our equation gives:

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

### Step 4: Simplifying the Equation
This simplifies to:

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

Rearranging gives:

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

### Step 5: Finding a Common Denominator
Finding a common denominator, we have:

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

### Step 6: Separating Variables
We can separate variables:

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

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

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

This results in:

\[ -\ln|-u - 2| = x + C \]

### Step 8: Replacing u
Substituting back for \(u\):

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

### Step 9: Applying Initial Conditions
Using the initial condition \(x = 0\) and \(y = 1\):

\[ -\ln|-(0 - 1) - 2| = 0 + C \]

This simplifies to:

\[ -\ln|1 - 2| = C \Rightarrow C = \ln(1) = 0 \]

### Step 10: Final Equation
Thus, the final equation becomes:

\[ -\ln|-(x - y) - 2| = x \]

This can be rewritten as:

\[ 2 + x - y = 0 \]

### Final Answer
The solution to the differential equation is:

\[ 2 + x - y = 0 \]