- #1
chuy52506
- 77
- 0
I'm trying to find the gen. solution to the equation y''''-8y'=0
I found the characteristic polynomial by plugging in ert as a solution to y.
I got,
r^4-8r=0
I simplified to get
r*(r^3-8)
Thus one root is 0, for the other 3 i must find the cubed root of 8.
I know the answer is 2*e2m*pi*i/3 for m=0,1,2
How do I arrive at that answer?
I tried the following:
Represent 8 as 8=8[cos(2*pi)+i*sin(2*pi)]=8*ei*pi
I found the characteristic polynomial by plugging in ert as a solution to y.
I got,
r^4-8r=0
I simplified to get
r*(r^3-8)
Thus one root is 0, for the other 3 i must find the cubed root of 8.
I know the answer is 2*e2m*pi*i/3 for m=0,1,2
How do I arrive at that answer?
I tried the following:
Represent 8 as 8=8[cos(2*pi)+i*sin(2*pi)]=8*ei*pi