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

Rearranging the given equation into standard linear form:

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

This is a first-order linear differential equation of the form $\dfrac{dy}{dx} + P(x)\,y = Q(x)$, where $P(x) = -x$ and $Q(x) = e^{x^2/2}$.

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Finding the Integrating Factor:

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

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Using the solution formula $y \times \text{I.F.} = \int Q(x) \times \text{I.F.}\,dx + C$, we get:

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

Since $e^{x^2/2} \cdot e^{-x^2/2} = e^0 = 1$, this simplifies to:

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

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

Multiplying both sides by $e^{x^2/2}$:

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

This is the general solution.

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Applying the initial condition $y(0) = 1$:

$1 = (0 + C)\,e^{0}$

$C = 1$

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The particular solution is:

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