Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 52

Prove that the function $y = Ae^{-x} + Bx e^{-x}$ satisfies the differential equation $y'' + 2y' + y = 0$.


$
\begin{equation}
\begin{aligned}

\text{if } y =& Ae^{-x} + Bx e^{-x}, \text{ then by using Product Rule}
\\
\\
y' =& Ae^{-x} (-1) + B [xe^{-x} (-1) + (1) e^{-x}]
\\
\\
y' =& -Ae^{-x} + Be^{-x} (-x + 1)

\end{aligned}
\end{equation}
$


Again, by using Product Rule


$
\begin{equation}
\begin{aligned}

y'' =& -Ae^{-x} (-1) + B [e^{-x} (-1) + e^{-x} (-1) (-x + 1)]
\\
\\
y'' =& Ae^{-x} + Be^{-x} [-1 + x - 1]
\\
\\
y'' =&Ae^{-x} + Be^{-x} [-2 + x]

\end{aligned}
\end{equation}
$


Therefore,


$
\begin{equation}
\begin{aligned}

& y'' + 2y' + y = 0
\\
\\
& [Ae^{-x} + Be^{-x} (-2 + x)] + 2 [-Ae^{-x} + Be^{-x} (-x + 1)] + [Ae^{-x} + Bxe^{-x}] = 0
\\
\\
& Ae^{-x} - 2Be^{-x} + Bxe^{-x} - 2Ae^{-x} - 2Bxe^{-x} + 2Be^{-x} + Ae^{-x} + Bxe^{-x} = 0
\\
\\
& 0 = 0

\end{aligned}
\end{equation}
$