For any real numbers α and β , let y _ α, β (x), x ∈ R , be the...

(\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})$