Question

Substitute y=e^rx into the given differentialequation to determine all values of the constant r for which y=e^rx is a solution of the equation.3y''+3y'-4y=0

Answer 1

Answer:

$r = \dfrac{-3 - \sqrt{57}}{6},\dfrac{-3 + \sqrt{57}}{6}$

Step-by-step explanation:

We are given the following in the question:

$y = e^{rx}$

It is a solution to the differential equation

$3y''+3y'-4y=0$

$y' = \dfrac{dy}{dx} = \dfrac{d(e^{rx})}{dx} = re^{rx}\\\\y'' = \dfrac{d(y')}{dx} = \dfrac{d(re^{rx})}{dx} = r^2e^{rx}$

Putting the values in the differential equation we get,

$3(r^2e^{rx}) + 3(re^{rx})-4e^{rx} = 0\\(3r^2+3r-4)e^{rx} = 0\\e^{rx}\neq 0\\3r^2+3r-4 = 0\\\\r = \dfrac{-3\pm \sqrt{9-4(3)(-4)}}{6}\\\\r =\dfrac{-3\pm \sqrt{57}}{6}\\\\r =\dfrac{-3 - \sqrt{57}}{6},\dfrac{-3 + \sqrt{57}}{6}$