Asked by Kim
for what values of r does the function y=e^rx satisfy the differential equation y''-4y'+y=0
Show steps please!
Thank you!
Show steps please!
Thank you!
Answers
Answered by
Damon
y = 1 e^(rx)
y' = r e^(rx)
y" = r^2 e^(rx)
r^2 - 4 r + 1 = 0
r = [4 +/- sqrt (16 -4) ] /2
r = [ 4 +/- 2 sqrt 3 ]2
r = 2 +/- sqrt 3
y' = r e^(rx)
y" = r^2 e^(rx)
r^2 - 4 r + 1 = 0
r = [4 +/- sqrt (16 -4) ] /2
r = [ 4 +/- 2 sqrt 3 ]2
r = 2 +/- sqrt 3