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, 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-by-Step Solution:

1. **Rearranging the Given Equation**:
Start with the equation:
\[ xy + k = e^{\frac{(x-1)^2}{2}} \]
Rearranging gives:
\[ xy = e^{\frac{(x-1)^2}{2}} - k \]
Thus, we can express \( y \) as:
\[ y = \frac{e^{\frac{(x-1)^2}{2}} - k}{x} \]

2. **Applying the Initial Condition**:
We know that \( y(1) = 0 \). Substituting \( x = 1 \) into the equation for \( y \):
\[ y(1) = \frac{e^{\frac{(1-1)^2}{2}} - k}{1} = 0 \]
This simplifies to:
\[ e^{0} - k = 0 \]
Therefore:
\[ 1 - k = 0 \implies k = 1 \]

3. **Conclusion**:
The value of \( K \) is:
\[ \boxed{1} \]