Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 54

Prove that the function $y = Ae^{-x} + Bxe^{-x}$ satisfies the differential equation

$y'' + 2y' + y = 0$

Solving for 1st Derivative


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} (Ae^{-x}) + \frac{d}{dx} (Bxe^{-x}), \text{ where $A$ and $B$ is constant}
\\
\\
y' =& \frac{d}{dx} \left( \frac{A}{e^x} \right) + \left[ (Bx) \frac{d}{dx} \left( \frac{1}{e^x} \right) + \left( \frac{1}{e^x} \right) \frac{d}{dx} (Bx) \right]
\\
\\
y' =& \left[ \frac{\displaystyle (e^x) \frac{d}{dx} (A) - (A) \frac{d}{dx} (e^x) }{(e^x)^2} \right] + \left[ (Bx) \frac{d}{dx} \left( \frac{1}{e^x} \right) + \left( \frac{1}{e^x} \right) \frac{d}{dx} (Bx) \right]
\\
\\
y' =& \left[ \frac{(e^x) (0) - Ae^x}{e^{2x}} \right] + \left[ (Bx) \left( \frac{-1}{e^x} \right) + \left( \frac{1}{e^x} \right) (B) \right]
\\
\\
y' =& - \frac{A}{e^x} - \frac{Bx}{e^x} + \frac{B}{e^x}
\\
\\
y' =& \frac{-A + B - Bx}{e^x}

\end{aligned}
\end{equation}
$


Solving for 2nd Derivative


$
\begin{equation}
\begin{aligned}

y'' =& \frac{d}{dx} \left( \frac{-A + B - Bx}{e^x} \right)
\\
\\
y'' =& \frac{\displaystyle e^x \frac{d}{dx} (-A + B - Bx) - (-A + B - Bx) \frac{d}{dx} (e^x) }{(e^x)^2}
\\
\\
y'' =& \frac{e^x (-B) - (-A + B - Bx) (e^x) }{e^{2x}}
\\
\\
y'' =& \frac{e^x (-B + A - B + Bx)}{e^{2x}}
\\
\\
y'' =& \frac{Bx + A - 2B}{e^x}


\end{aligned}
\end{equation}
$


Substituting the equations of $y'', y'$ and $y$ in $y'' + 2y' + y = 0$


$
\begin{equation}
\begin{aligned}

& \frac{Bx + A - 2B}{e^x} + 2 \left( \frac{-A + B - Bx}{e^x} \right) + \frac{A + Bx}{e^x} = 0
\\
\\
& \frac{Bx + A - 2B}{e^x} + \frac{-2A + 2B - 2Bx}{e^x} + \frac{A + Bx}{e^x} = 0
\\
\\
& \frac{Bx + A - 2B - 2A + 2B - 2Bx + A + Bx}{e^x} = 0
\\
\\
& \frac{0}{e^x} = 0
\\
\\
& 0 = 0



\end{aligned}
\end{equation}
$