Rearranging the equation:
\frac{dy}{dx} - xy = e^{\frac{x^2}{2}}$ This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)\,y = Q(x)$, where $P(x) = -x$ and $Q(x) = e^{\frac{x^2}{2}}$. <hr> The integrating factor is:\text{I.F.} = e^{\int P(x),dx} = e^{\int (-x),dx} = e^{-\frac{x^2}{2}}<hr>Usingthestandardsolutionformula:y \cdot \text{I.F.} = \int Q(x) \cdot \text{I.F.},dx + Cy \cdot e^{-\frac{x^2}{2}} = \int e^{\frac{x^2}{2}} \cdot e^{-\frac{x^2}{2}},dx + C$
Since e2x2⋅e−2x2=e0=1, this simplifies to:
y⋅e−2x2=∫1dx+C
y \cdot e^{-\frac{x^2}{2}} = x + C$ <hr> Multiplying both sides by $e^{\frac{x^2}{2}}$:\boxed{y = (x + C),e^{\frac{x^2}{2}}}$
where C is an arbitrary constant.