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4th order differential equation 
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#1
Nov1912, 10:20 PM

P: 77

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^48r=0 I simplified to get r*(r^38) 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} 


#2
Nov1912, 11:32 PM

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#3
Nov2012, 12:01 AM

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),1i*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 dont know how they arrive to this 


#4
Nov2012, 12:34 AM

P: 761

4th order differential equation
Obviously, the cubic root of 8 is 2.
r^48r = r(r2)(rē+2r+4) = r (r2) [r+1+i sqrt(3)] [r+1i sqrt(3)] 


#5
Nov2012, 02:16 AM

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Try the solution y = e^{λt}: P(D)(Dλ)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 (Dλ)^{n}. It's fairly easy to show that t^{r}e^{λt} is also a solution for r = 1, .. n1. 


#6
Nov2012, 03:20 AM

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),1i*sqrt(3) for m=1,2,3



#7
Nov2012, 04:30 AM

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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) 


#8
Nov2012, 04:37 AM

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exp(2∏n ix)=cos(2∏n x) + i sin(2∏n x) ... and cos(2∏/3) = cos(∏/3) = (√3)/2 etc. 


#9
Nov2112, 01:10 AM

P: 233

To see this understand that [tex]exp(i2m\pi)=1[/tex] for any [tex]m\epsilon\mathbb{Z}[/tex]. So if, [tex]r^3=8\cdot1=8exp(i2m\pi)[/tex] then [tex]r =2exp(\frac{i2m\pi}{3})[/tex] 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: [tex]e^{i\phi}=cos(\phi)+isin(\phi)[/tex] So for example, [tex]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})[/tex] 


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