For any real numbers α and β , let y α , β x , x ∈ R , be the solution of the differential equation d y d x + α y = x e β x , y 1 = 1 Let S = y α , β x   : α , β ∈ R . Then which of the following functions belong(s) to the set S ?

Question

For any real numbers &#945; and &#946; , let y &#945; , &#946; x , x &#8712; R , be the solution of the differential equation d y d x + &#945; y = x e &#946; x , y 1 = 1 Let S = y &#945; , &#946; x &#160; : &#945; , &#946; &#8712; R . Then which of the following functions belong(s) to the set S ?

Aicharya · Accepted Answer

$\frac{d y}{d x}+a y=x e^{\beta x}$ 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 $y e^x=\int x d x$ $\Rightarrow y e^x=\frac{x^2}{2}+C$ $y=\frac{x^2}{2} e^{-x}+C e^{-x}$ $y(1)=1$ $1=\frac{1}{2 e}+\frac{C}{e} \Rightarrow C=e-\frac{1}{2}$ $y=\frac{x^2}{2} e^{-x}+\left(e-\frac{1}{2}\right) e^{-x}$

For any real numbers &#945; and &#946; , let y &#945; , &#946; x , x &#8712; R , be the solution of the…

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For any real numbers α and β , let y α , β x , x ∈ R , be the solution of the…