Answer
$r=-2, 3$
Work Step by Step
$y''-y'-6y=0$ _____($\alpha$)
$y=e^{rx}$ _____(1)
Differentiate (1) with respect to $x$
$y'=re^{rx}$ _____(2)
Differentiate (2) with respect to $x$
$y''=r^2e^{rx}$ _____(3)
Given that $y=e^{rx}$ is solution of ($\alpha$) so it satisfies ($\alpha$)
$r^2e^{rx}-re^{rx}-6e^{rx}=0$
$e^{rx}[r^2-r-6]=0$
$e^{rx}\neq 0$
Therefore, $r^2-r-6=0$
$r^2-3r+2r-6=0$
$r(r-3)+2(r-3)=0$
$(r+2)(r-3)=0$
$r=-2$ and $r=3$
Hence, $r=-2, 3$
