The particular solution of the differential equation $\frac{dy}{dx}=e^{x^2/2} + xy$, when $x = 0, y = 1$, is

$y = (x+1)e^{x^2/2}$

The correct answer is Option (2) → $y = (x+1)e^{x^2/2}$

Given: $\frac{dy}{dx} - xy = e^{x^2/2}$

Integrating factor: $IF = e^{-\int x dx} = e^{-x^2/2}$

Particular solution using formula $y = \frac{1}{IF} \left( \int IF \cdot Q(x) dx + C \right)$:

$y = e^{x^2/2} \left( \int e^{-x^2/2} e^{x^2/2} dx + C \right) = e^{x^2/2} (x + C)$

Apply $y(0)=1 \Rightarrow C=1$

Answer: $y = (x+1) e^{x^2/2}$