Question

For any real numbers $\alpha$ and $\beta$, let $y_{\alpha ,\beta }(x),x\in R$, be the solution of the differential equation $\frac{dy}{dx}+\alpha y=x{e}^{\beta x},y(1)=1$.
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)

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