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  $S=\{y_{\alpha ,\beta }\left(x\right):\alpha ,\beta \in R\}$ . Then which of the following functions belong(s) the set S?

 $\frac{dy}{dx}+\alpha y=x{e}^{\beta x}$
I.F :  $e^{\int \alpha dx}=e^{\alpha x}$
 $d(e^{\alpha x}.y)=x{e}^{(\alpha +\beta )x}$
Case–I:   $\alpha +\beta \ne 0$
 $\Rightarrow e^{\alpha x}.y=\frac{x{e}^{(\alpha +\beta )x}}{\alpha +\beta }-\frac{e^{(\alpha +\beta )x}}{{(\alpha +\beta )}^2}+C$
 $\Rightarrow y=\frac{x{e}^{\beta x}}{\alpha +\beta }-\frac{e^{\beta x}}{{(\alpha +\beta )}^2}+c{e}^{-\alpha x}$
 $\alpha =2,\beta =1\Rightarrow y=\frac{x{e}^x}{3}-\frac{e^x}{9}+c{e}^{-2x}$
  $y\left(1\right)=1\Rightarrow 1=\frac{e}{3}-\frac{e}{9}+c{e}^{-2}\Rightarrow c=e^2(1-\frac{2e}{9})$
 $y=\frac{x{e}^x}{3}-\frac{e^x}{9}+e^2{(1-\frac{2e}{9})}^{e^{-2x}}$
=  $\frac{e^x}{3}(x-\frac{1}{3})+(e^2-\frac{2e^3}{9})e^{-2x}$
Case–II:  $\alpha +\beta =0$
 $$\begin{gathered} \Rightarrow y{e}^{\alpha x}=\int xdx=\frac{x^2}{2}+C \\ \Rightarrow y=\frac{x^2}{2}{e}^{-\alpha x}+c{e}^{-\alpha x} \end{gathered}$$
If   $\alpha =2y=\frac{x^2}{2}{e}^{-2x}+c{e}^{-2x}$
 $y\left(1\right)=1\Rightarrow 1=\frac{1}{2}{e}^{-2}+c{e}^{-2}$
 $\Rightarrow (e^2-\frac{1}{2})=C$
 $y=\frac{x^2}{2}{e}^{-2x}+(e^2-\frac{1}{2})e^{-2x}$