Answer
$f(x)=e^{-2x}cos4x$ and $g(x)=e^{-2x}sin4x$ are the basic solutions.
See graphs below for visual explanations.
Work Step by Step
$\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+20y=0$
$y''+4y'+20y=0$
Use auxiliary equation is
$r^{2}+4r +20=0$
$r^{2}+4r +20=0$
$r^{2}+4r +4=-16$
$(r+2)^{2}=-16$
$r+2=±4i$
$r_{1}=-2+4i$
$r_{2}=-2-4i$
$α=-2, β=4$
Formula 11
$y=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y=e^{-2x}(c_{1}cos4x+c_{2}sin4x)$
Therefore, $f(x)=e^{-2x}cos4x$ and $g(x)=e^{-2x}sin4x$ are the basic solutions.
(GRAPH 1)
$y=f(x)$ is the blue curve.
$y=g(x)$ is the red curve.
(GRAPH 2)
$y=f(x)$ is the blue dotted curve.
$y=g(x)$ is the red dotted curve.
$y=f+g$ is the black solid curve.
(GRAPH 3)
$y=f(x)$ is the blue dotted curve.
$y=g(x)$ is the red dotted curve.
$y=f-g$ is the black solid curve.
From these graphs, we note that $\lim\limits_{ x\to \infty}f = \lim\limits_{x \to \infty}g=0$.
Therefore, all solutions approach $0$ as $x → \infty$.
