If K is constant such that ` xy + k = e ^(((x-1)^2)/(2))` satisfies the differential equation ` x . ( dy )/( dx) = (( x^2 - x-1) y+ (x-1)` and ` y (1) =0` then find the value of K .

Answer

To solve the problem step by step, we need to find the value of the constant \( K \) such that the equation

\[ xy + K = e^{\frac{(x-1)^2}{2}} \]

satisfies the differential equation

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

with the initial condition \( y(1) = 0 \).

### Step 1: Rewrite the Differential Equation

First, we rewrite the differential equation in standard form:

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

### Step 2: Identify the Integrating Factor

The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int \frac{x^2 - x - 1}{x} dx} \]

Calculating the integral:

\[ \int \left( x - 1 - \frac{1}{x} \right) dx = \frac{x^2}{2} - x - \ln|x| + C \]

Thus, the integrating factor is:

\[ \mu(x) = e^{\frac{x^2}{2} - x - \ln|x|} = \frac{e^{\frac{x^2}{2} - x}}{x} \]

### Step 3: Multiply the Equation by the Integrating Factor

Now, we multiply the entire differential equation by the integrating factor:

\[ \frac{e^{\frac{x^2}{2} - x}}{x} \frac{dy}{dx} + \frac{e^{\frac{x^2}{2} - x}}{x} \cdot \frac{x^2 - x - 1}{x} y = \frac{e^{\frac{x^2}{2} - x}}{x} \cdot \frac{x - 1}{x} \]

### Step 4: Simplify the Left Side

The left side can be expressed as a derivative:

\[ \frac{d}{dx} \left( y \cdot \frac{e^{\frac{x^2}{2} - x}}{x} \right) = \frac{e^{\frac{x^2}{2} - x}}{x^2} (x - 1) \]

### Step 5: Integrate Both Sides

Integrating both sides gives:

\[ y \cdot \frac{e^{\frac{x^2}{2} - x}}{x} = \int \frac{e^{\frac{x^2}{2} - x}}{x^2} (x - 1) dx + C \]

### Step 6: Solve for \( y \)

Now we need to solve for \( y \):

\[ y = \frac{x}{e^{\frac{x^2}{2} - x}} \left( \int \frac{e^{\frac{x^2}{2} - x}}{x^2} (x - 1) dx + C \right) \]

### Step 7: Apply the Initial Condition

Using the initial condition \( y(1) = 0 \):

\[ 0 = \frac{1}{e^{\frac{1^2}{2} - 1}} \left( \int \frac{e^{\frac{1^2}{2} - 1}}{1^2} (1 - 1) dx + C \right) \]

This implies \( C = e^{\frac{1}{2}} \).

### Step 8: Substitute Back to Find \( K \)

Now we substitute back into the original equation:

\[ xy + K = e^{\frac{(x-1)^2}{2}} \]

At \( x = 1 \):

\[ 1 \cdot 0 + K = e^{\frac{(1-1)^2}{2}} = e^0 = 1 \]

Thus, \( K = 1 \).

### Final Answer

The value of \( K \) is

\[ \boxed{1} \]