Answer
$y=c_{1}e^{\frac{-3x}{5}}+c_{2}e^{x}$
As $x$ approches $\pm \infty$, $y$ approaches either $0$ or $\pm \infty$.
Work Step by Step
$5\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}-3y=0$
$5y''-2y'-3y=0$
Use auxiliary equation
$5r^{2}-2r-3=0$
$5r^{2}-2r-3=0$
$5r^{2}+(-5r+3r)-3=0$
$(5r^{2}-5r)+(3r-3)=0$
$5r(r-1)+3(r-1)=0$
$(5r+3)(r-1)=0$
$r_{1}=1, r_{2}=-\frac{3}{5}$
Formula 8
$y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y=c_{1}e^{\frac{-3x}{5}}+c_{2}e^{x}$
Therefore, $f(x)=e^{\frac{-3x}{5}}$ and $g(x)=e^{x}$ are the basic solutions.
FIRST GRAPH
$y=f(x)$ is the blue curve.
$y=g(x)$ is the red curve.
SECOND GRAPH
$y=f(x)$ is the blue dotted curve.
$y=g(x)$ is the red dotted curve.
$y=f+g$ is the solid black curve.
THIRD GRAPH
$y=f(x)$ is the blue dotted curve.
$y=g(x)$ is the red dotted curve.
$y=f-g$ is the solid black curve.
FOURTH GRAPH
$y=f(x)$ is the blue dotted curve.
$y=g(x)$ is the red dotted curve.
$y=-f+g$ is the solid black curve.
As $x$ approches $\pm \infty$, $y$ approaches either $0$ or $\pm \infty$.
