Question

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

• A. $f(x)=\frac{x^{2}}{2} e^{-x}+\left(e-\frac{1}{2}\right) e^{-x}$
• B. $f(x)=-\frac{x^{2}}{2} e^{-x}+\left(e+\frac{1}{2}\right) e^{-x}$
• C. $f(x)=\frac{e^{x}}{2}\left(x-\frac{1}{2}\right)+\left(e-\frac{e^{2}}{4}\right) e^{-x}$
• D. $f(x)=\frac{e^{x}}{2}\left(\frac{1}{2}-x\right)+\left(e+\frac{e^{2}}{4}\right) e^{-x}$

Answer 1

Answer: (A), (C)

$\frac{d y}{d x}+\alpha y=x e^{\beta x}$
$d\left(e^{\alpha x} \cdot y\right) x e^{\alpha x} \cdot e^{\beta x}$
$d\left(e^{\alpha x} \cdot y\right)=x e^{(\alpha+\beta) x}$ ... (i)
Case-I : $\alpha+\beta \neq 0$
$d \left( e ^{\alpha x} \cdot y \right)= xe ^{(\alpha+\beta) x}$
$e ^{\alpha x} \cdot y =\frac{ xe ^{(\alpha+\beta) x}}{(\alpha+\beta)}-\frac{ e ^{(\alpha+\beta) x}}{(\alpha+\beta)^{2}}+ C$
$y =\frac{ xe ^{\beta x}}{(\alpha+\beta)}-\frac{ e ^{\beta x}}{(\alpha+\beta)^{2}}+ Ce ^{-\alpha x}$
$\alpha=1, \beta=1 \Rightarrow y =\frac{ xe ^{x}}{2}-\frac{ e ^{x}}{4}+ Ce ^{- x }$
as $y (1)=1 \Rightarrow C = e \left(1-\frac{ e }{4}\right)$
$y ( x )=\frac{x e ^{x}}{2}-\frac{ e ^{x}}{4}+\left( e -\frac{ e ^{2}}{4}\right) e ^{- x }$
Case-II : $\alpha+\beta=0$
$\Rightarrow \frac{d y}{d x}-\beta y=x e^{\beta x}$
$d\left(e^{-\beta x} y\right)=x$
$e^{-\beta x} \cdot y=\frac{x^{2}}{2}+C$
$y=\frac{e^{\beta x} x^{2}}{2}+C e^{\beta x}$
$y(1)=1$
$\Rightarrow C=\left(1-\frac{e}{2}\right) \frac{1}{e} $
$\Rightarrow y=e^{\beta x} \cdot \frac{x^{2}}{2}+\left(1-\frac{e}{2}\right) \frac{1}{e} \cdot e^{\beta x}$
Take $\beta=-1$
$ \Rightarrow y=\frac{x^{2}}{2} e^{-x}+\left(1-\frac{e}{2}\right) e^{-x}$