Higher Derivatives - Acceleration?

Question

The function y=e^(rx) satisfies the equation
y"-8.3 y'+16.66 y=0

if r =___________ or if r = __________ .

Can anyone help me with this question

I don't get it..

Thank you

Answer 1

Well since we have y=e^rx, substituting we get:
16.66 e^rx - 8.3 e^rx + r^2 e^rx = 0 and so
(r-3.4) (r-4.9) = 0 since e^rx is non-zero everywhere.

So , answers are:


You do the rest.

Answer 2

y=e^(rx)
y'= re^rx
y" = r^2e^rx
r^2e^rx -8.3re^rx + 16.66e^rx = 0
r^2 -8.3 r +16.66 = 0
r = [8.3 +/- sqrt((8.3)^2 -4(1)(16.66))]/(2*1)
r = [8.3 +/- sqrt(2.25)]/2
r= 4.15 +/- 1.5/2
r = 4.15 +/- .75
r = 4.9 or r = 3.4