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

The Correct Option is A, C

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