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

$\dfrac{dy}{dx}+\alpha y = x e^{\beta x}$

$y(1)=1,\quad \alpha,\beta\in\mathbb{R}$

$S = \{y_{\alpha,\beta}\}$

$\text{find the general form of } y_{\alpha,\beta}$

$\text{then test each option's structure and } y(1)=1$

$1.\ \text{integrating factor } e^{\alpha x}$

$2.\ \text{general form of } y$

$3.\ \text{test each option}$

$\mu=e^{\alpha x}:\quad \left(ye^{\alpha x}\right)' = x e^{(\alpha+\beta)x}$

$ye^{\alpha x} = \int x e^{(\alpha+\beta)x}\,dx + C$

$k=\alpha+\beta\ne0:\quad y=e^{\beta x}\!\left(\dfrac{x}{k}-\dfrac{1}{k^2}\right)+Ce^{-\alpha x}$

$k=0\ (\beta=-\alpha):\quad y=e^{\beta x}\!\left(\dfrac{x^2}{2}+C\right)$

$\text{so } y=e^{\beta x}(Ax+B)+Ce^{-\alpha x},\quad B=-A^2$

$\text{or (when } \beta=-\alpha\text{)}:\ y=e^{\beta x}\!\left(\tfrac{x^2}{2}+C\right)$

$f_3(1) = \dfrac{e}{2}\cdot\dfrac12+\left(e-\dfrac{e^2}{2}\right)e^{-1} = 1-\dfrac{e}{4}$

$f_4(1) = 1+\dfrac{e}{4}$

$\ne 1 \Rightarrow (3),(4)\notin S$

$f_1 = e^{-x}\!\left(\dfrac{x^2}{2}+e-\dfrac12\right):\ \beta=-1,\ \alpha=1,\ C=e-\dfrac12$

$f_1'+f_1 = x e^{-x},\quad f_1(1)=1 \Rightarrow (1)\in S$

$f_2 = e^{-x}\!\left(-\dfrac{x^2}{2}+e+\dfrac12\right)$

$k=0 \text{ needs } +\dfrac{x^2}{2};\quad k\ne0 \text{ gives a linear term} \Rightarrow (2)\notin S$

$\text{only } (1)\in S$

$\boxed{\text{option }(1)}$