For any real numbers alpha and beta,let y_{al | vedclass

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(B) The given differential equation is $\frac{dy}{dx} + \alpha y = x e^{\beta x}$. This is a linear differential equation of the form $\frac{dy}{dx} +

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$A, C$

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(B) The given differential equation is $\frac{dy}{dx} + \alpha y = x e^{\beta x}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \alpha$ and $Q(x) = x e^{\beta x}$.The integrating factor is $IF = e^{\int \alpha dx} = e^{\alpha x}$.Multiplying both sides by $IF$,we get $\frac{d}{dx}(y e^{\alpha x}) = x e^{(\alpha+\beta)x}$.Case $I$: If $\alpha + \beta = 0$,then $\beta = -\alpha$. The equation becomes $\frac{d}{dx}(y e^{\alpha x}) = x$. Integrating both sides,$y e^{\alpha x} = \frac{x^2}{2} + C$. Using $y(1) = 1$,we get $1 \cdot e^{\alpha} = \frac{1}{2} + C$,so $C = e^{\alpha} - \frac{1}{2}$. Thus,$y = \frac{x^2}{2} e^{-\alpha x} + (e^{\alpha} - \frac{1}{2}) e^{-\alpha x}$. For $\alpha = 1$,$y = \frac{x^2}{2} e^{-x} + (e - \frac{1}{2}) e^{-x}$,which matches option $(A)$.Case $II$: If $\alpha + \beta 
eq 0$,integrating $\int x e^{(\alpha+\beta)x} dx$ by parts gives $\frac{x e^{(\alpha+\beta)x}}{\alpha+\beta} - \frac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$. Thus,$y e^{\alpha x} = \frac{x e^{(\alpha+\beta)x}}{\alpha+\beta} - \frac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$. Using $y(1) = 1$,$C = e^{\alpha} - \frac{e^{\alpha+\beta}}{\alpha+\beta} + \frac{e^{\alpha+\beta}}{(\alpha+\beta)^2}$. For $\alpha = -1, \beta = 2$,we have $\alpha+\beta = 1$. Substituting these values leads to the form in option $(C)$.Therefore,both $(A)$ and $(C)$ belong to $S$.

For any real numbers $\alpha$ and $\beta$,let $y_{\alpha, \beta}(x), x \in R$,be the solution of the differential equation $\frac{dy}{dx}+\alpha y=x e^{\beta x}, y(1)=1$. Let $S=\{y_{\alpha, \beta}(x): \alpha, \beta \in R\}$. Then which of the following functions belong$(s)$ to the set $S$?

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