The general solution of differential equation $\frac{dy}{dx} = e^{\frac{x^2}{2}} + xy$ is

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

The correct answer is Option (3) → $y = (x + C)e^{x^2/2}$ ##

Given that, $\frac{dy}{dx} = e^{x^2/2} + xy$

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

Which is a linear differential equation.

On comparing it with $\frac{dy}{dx} + P \cdot y = Q$, we get

$P = -x, Q = e^{x^2/2}$

$∴\text{I.F} = e^{\int -x \, dx} = e^{-x^2/2}$

The general solution is

$y \cdot \text{I.F} = \int Q \cdot \text{I.F} \, dx + C$

$y \cdot e^{-x^2/2} = \int e^{-x^2/2} e^{x^2/2} \, dx + C$

$\Rightarrow y \cdot e^{-x^2/2} = \int 1 \, dx + C$

$\Rightarrow y \cdot e^{-x^2/2} = x + C$

$\Rightarrow y = x e^{x^2/2} + C e^{x^2/2}$

$\Rightarrow y = (x + C) e^{x^2/2}$