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

\(\frac{d y}{d x}+a y=x e^{\beta x}\)
Integrating factor (I.F.) \(=\mathrm{e}^{\mathrm{adx}}=\mathrm{e}^{\mathrm{ax}}\)
So, the solution is \(y \cdot e^{a x}=\int x e^{\beta x} \cdot e^{a x} d x\)
\(\Rightarrow y e^{a x}=\int x e^{(\alpha+\beta) x} d x\)
If \(\alpha+\beta \neq 0\)
\(\Rightarrow y e^{a x}=x \frac{e^{(\alpha+\beta) x}}{(\alpha+\beta)}-\frac{e^{(\alpha+\beta) x}}{(\alpha+\beta)^2}+C\)
\(\Rightarrow y=\frac{x e^{\beta x}}{(\alpha+\beta)}-\frac{e^{\beta x}}{(\alpha+\beta)^2}+C e^{-a x}\)
\(\Rightarrow y=\frac{e^{\beta x}}{(\alpha+\beta)}\left(x-\frac{1}{a+\beta}\right)+C e^{-\alpha x}\)
Put \(\alpha=\beta=1\) in (1)
\(y=\frac{e^x}{2}\left(x-\frac{1}{2}\right)+C e^{-x}\)
\(y(1)=1\)
\(1=\frac{e}{2} \times \frac{1}{2}+\frac{C}{e} \Rightarrow c=e-\frac{e^2}{4}\)
So, \(y=\frac{e^x}{2}\left(x-\frac{1}{2}\right)+\left(e-\frac{e^2}{4}\right) e^{-x}\)
If \(\alpha+\beta=0 \& \alpha=1\)
\(\begin{aligned}
&\frac{d y}{d x}+y=x e^{-x} \\
&\text { I.F. }=e^x
\end{aligned}\)
Solution is \(y e^x=\int x d x\)
\(\Rightarrow y e^x=\frac{x^2}{2}+C\)
\(y=\frac{x^2}{2} e^{-x}+C e^{-x}\)
\(y(1)=1\)
\(1=\frac{1}{2 e}+\frac{C}{e} \Rightarrow C=e-\frac{1}{2}\)
\(y=\frac{x^2}{2} e^{-x}+\left(e-\frac{1}{2}\right) e^{-x}\)