# 4th order differential equation

by chuy52506
Tags: differential, equation, order
 P: 77 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
Homework
HW Helper
Thanks
P: 9,873
 Quote by chuy52506 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
It should be 8 = 8ei*2πm, for any m, right?
 P: 77 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*e2*t+e-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))] I dont know how they arrive to this
 P: 759 4th order differential equation 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)]
Homework
HW Helper
Thanks
P: 9,873
 Quote by 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*e2*t+e-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))] I dont 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 = eλt: P(D)(D-λ)eλt = P(D)(D(eλt - λeλt) = P(D)(λeλt - λeλt) = 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 treλt is also a solution for r = 1, .. n-1.
 P: 77 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
 HW Helper P: 2,264 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)
Homework
HW Helper
Thanks
P: 9,873
 Quote by lurflurf 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.
P: 233
 Quote by 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
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, $$2exp(i\frac{2\pi}{3})=2(cos(\frac{2\pi}{3})+i\sin({\frac{2\pi}{3}})=2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})$$

 Related Discussions Differential Equations 2 Calculus & Beyond Homework 1 Calculus & Beyond Homework 7 Calculus & Beyond Homework 5