Question

If $\displaystyle \int e^{- \frac{x^{2}}{2}}dx=f\left(x\right)$ and the solution of the differential equation $\frac{d y}{d x}=1+xy$ is $y=ke^{\frac{x^{2}}{2}}f\left(x\right)+Ce^{\frac{x^{2}}{2}}$ , then the value of $k$ is equal to (where $C$ is the constant of integration)

Answer 1

$\frac{d y}{d x}-xy=1$
Here, I.F. $=e^{- \displaystyle \int x d x}=e^{- \frac{x^{2}}{2}}$
So, solution is $ye^{- \frac{x^{2}}{2}}=\displaystyle \int e^{- \frac{x^{2}}{2}}dx+C$
$y=e^{\frac{x^{2}}{2}}f\left(x\right)+Ce^{\frac{x^{2}}{2}}$
Hence, $k=1$