Answer
$y=c_{1}e^{-\frac{x}{3}}+c_{2}xe^{-\frac{x}{3}}$
Therefore, all solutions approach $0$ as $x→\infty$ and $\pm \infty$ as $x$ approaches $-\infty$.
Work Step by Step
$9\frac{d^{2}y}{dx^{2}}+6\frac{dy}{dx}+y=0$
$9y''+6y'+y=0$
Use auxiliary equation
$9r^{2}+6r+1=0$
$(3r)^{2}+2 \times 3 \times r +1^{2}=0$
$(3r+1)^{2}=0$
$3r+1=0$
$r=-\frac{1}{3}$
Formula 10
$y=c_{1}e^{rx}+c_{2}xe^{rx}$
$y=c_{1}e^{-\frac{x}{3}}+c_{2}xe^{-\frac{x}{3}}$
Therefore, $f(x)=e^{-\frac{x}{3}}$ and $g(x)=xe^{-\frac{x}{3}}$ 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 solid black curve.
GRAPH 3
$y=f(x)$ is the blue dotted curve.
$y=g(x)$ is the red dotted curve.
$y=f-g$ is the solid black curve.
All solutions are linear combinations of the basic solutions.
From the graph, we see that $\lim\limits_{x \to \infty}f = \lim\limits_{x \to \infty}g=0$.
Therefore, all solutions approach $0$ as $x→\infty$ and $\pm \infty$ as $x$ approaches $-\infty$.
