`dy/dx=xy-1+x-y , y(0)=0` then find `y(1)`

Answer

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

### Step 1: Rearranging the equation
We start with the given equation:
\[ \frac{dy}{dx} = xy - 1 + x - y \]
Rearranging the terms, we can write:
\[ \frac{dy}{dx} = xy - y + x - 1 \]
Factoring out \(y\) from the first two terms:
\[ \frac{dy}{dx} = y(x - 1) + (x - 1) \]
This can be factored further:
\[ \frac{dy}{dx} = (x - 1)(y + 1) \]

### Step 2: Separating variables
Next, we separate the variables:
\[ \frac{dy}{y + 1} = (x - 1)dx \]

### Step 3: Integrating both sides
Now we integrate both sides:
\[ \int \frac{dy}{y + 1} = \int (x - 1)dx \]
The left side gives:
\[ \ln |y + 1| = \int (x - 1)dx = \frac{x^2}{2} - x + C \]

### Step 4: Applying the initial condition
We apply the initial condition \(y(0) = 0\):
\[ \ln |0 + 1| = \frac{0^2}{2} - 0 + C \]
This simplifies to:
\[ \ln 1 = 0 + C \implies C = 0 \]
Thus, we have:
\[ \ln |y + 1| = \frac{x^2}{2} - x \]

### Step 5: Solving for \(y\)
Exponentiating both sides gives:
\[ y + 1 = e^{\frac{x^2}{2} - x} \]
So,
\[ y = e^{\frac{x^2}{2} - x} - 1 \]

### Step 6: Finding \(y(1)\)
Now, we substitute \(x = 1\) to find \(y(1)\):
\[ y(1) = e^{\frac{1^2}{2} - 1} - 1 = e^{\frac{1}{2} - 1} - 1 = e^{-\frac{1}{2}} - 1 \]

### Final Answer
Thus, the value of \(y(1)\) is:
\[ y(1) = e^{-\frac{1}{2}} - 1 \]