Answer:
[tex]r = \dfrac{-3 - \sqrt{57}}{6},\dfrac{-3 + \sqrt{57}}{6}[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]y = e^{rx}[/tex]
It is a solution to the differential equation
[tex]3y''+3y'-4y=0[/tex]
[tex]y' = \dfrac{dy}{dx} = \dfrac{d(e^{rx})}{dx} = re^{rx}\\\\y'' = \dfrac{d(y')}{dx} = \dfrac{d(re^{rx})}{dx} = r^2e^{rx}[/tex]
Putting the values in the differential equation we get,
[tex]3(r^2e^{rx}) + 3(re^{rx})-4e^{rx} = 0\\(3r^2+3r-4)e^{rx} = 0\\e^{rx}\neq 0\\3r^2+3r-4 = 0\\\\r = \dfrac{-3\pm \sqrt{9-4(3)(-4)}}{6}\\\\r =\dfrac{-3\pm \sqrt{57}}{6}\\\\r =\dfrac{-3 - \sqrt{57}}{6},\dfrac{-3 + \sqrt{57}}{6}[/tex]
