4th order differential equation

[chuy52506]

I'm trying to find the gen. solution to the equation y''''-8y'=0
I found the characteristic polynomial by plugging in e^rt 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*e^2m*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*e^i*pi

 

[haruspex]

I'm trying to find the gen. solution to the equation y''''-8y'=0
I found the characteristic polynomial by plugging in e^rt 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*e^2m*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*e^i*pi

It should be 8 = 8e^i*2πm, for any m, right?

 

[chuy52506]

The roots are r=0,2*e^(2*m*pi*i)/3
then it says this is equivalent to
r=0,2,-1+i*sqrt(3),-1-i*sqrt(3)

Then the gen solution is
y=c1+c2*e^2*t+e^-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))]

I don't know how they arrive to this

 

[JJacquelin]

Obviously, the cubic root of 8 is 2.
r^4-8r = r(r-2)(r²+2r+4)
= r (r-2) [r+1+i sqrt(3)] [r+1-i sqrt(3)]

 

[haruspex]

The roots are r=0,2*e^(2*m*pi*i)/3
then it says this is equivalent to
r=0,2,-1+i*sqrt(3),-1-i*sqrt(3)

Then the gen solution is
y=c1+c2*e^2*t+e^-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))]

I don't know how they arrive to this

Suppose λ is a root of the polynomial. That is, the differential equation has a factor (D-λ), where D = d/dt, so the equation can be written P(D)(D-λ)y = 0, for some polynomial P.
Try the solution y = e^λt: P(D)(D-λ)e^λt = P(D)(D(e^λt - λe^λt) = P(D)(λe^λt - λe^λt) = 0. So the general solution is a linear combination of such terms.
A complication arises when there is a repeated root, i.e. a factor (D-λ)ⁿ. It's fairly easy to show that t^re^λt is also a solution for r = 1, .. n-1.

 

[chuy52506]

But where is the root e^(2*m*pi*i)/3 derived from? and how is this equivalent to the roots:2,-1+i*sqrt(3),-1-i*sqrt(3) for m=1,2,3

 

[lurflurf]

exp(2∏n i)=1
so
(exp(2∏n i/3))^3=1
also
exp(2∏n ix)=cos(2∏n x)+i sin(2∏n x)
so
exp(2∏n i/3)=cos(2∏n /3)+i sin(2∏n /3)

 

[haruspex]

exp(2∏n i)=1
so
(exp(2∏n i/3))^3=1
also
exp(2∏n ix)=cos(2∏n ix)+i sin(2∏n ix)
so
exp(2∏n i/3)=cos(2∏n i/3)+i sin(2∏n i/3)

Small correction (too many i's):

exp(2∏n ix)=cos(2∏n x) + i sin(2∏n x)
so
exp(2∏n i/3)=cos(2∏n/3) + i sin(2∏n/3)
​

... and cos(2∏/3) = -cos(∏/3) = -(√3)/2 etc.

 

[Jim Kata]

But where is the root e^(2*m*pi*i)/3 derived from? and how is this equivalent to the roots:2,-1+i*sqrt(3),-1-i*sqrt(3) for m=1,2,3

What you are getting confused about is the field of the reals vs. the field of complex numbers. Over the field of the reals there is only one solution to the equation r^3-8=0 namely r=2, but over the field of complex numbers there are 3 distinct solutions to the equation r^3-8=0.

To see this understand that exp(i2m\pi)=1 for any m\epsilon\mathbb{Z}. So if, r^3=8\cdot1=8exp(i2m\pi) then
r =2exp(\frac{i2m\pi}{3}) but you can see the only distinct ones are for m= 0,1,2 since for m beyond or below that you repeat your answers. To answer your other question you need to know about Euler's formula which says that: e^{i\phi}=cos(\phi)+isin(\phi)

So for example, 2exp(i\frac{2\pi}{3})=2(cos(\frac{2\pi}{3})+i\sin({\frac{2\pi}{3}})=2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})