Question

find the general solution for the given differential

y^(6) + y = 0 (6th power differential)

The homogeneous equation is r^6 + 1 = 0 so there are 6 roots. r = [-1]^(1/6) so there are at least 4 complex roots. Can anyone help from here?

Answer 1

So since there are 6 roots can we just divide the unit circle into 6ths and use those as reference points? r_k = e^ i ([pi+2k pi]/3) ?

Answer 2

the roots are when y(x)=0

Answer 3

Wolfram has a solution but , but the method is long,
http://www.wolframalpha.com/input/?i=d%5E6y%2Fdx%5E6+%2By%3D0
perhaps there is a simpler method

Answer 4

I was looking at another example (y''''+y = 0) and they found the roots using the unit circle. r = [-1]^1/4 -> (-1 = cos pi + i sin pi)

This eventually led down to r_k = e^ i ([pi+2k pi]/4)

Answer 5

Assuming that's a valid method r_0 = e^ i(pi/3) = cos (pi/3) + i sin (pi/3) = \[\sqrt{3}/2 + i 1/2\] So this would be one of the possible roots complex conjugate roots.

Answer 6

oooooo I got it now, \[\alpha = \sqrt{3}/2, \beta = \frac{ 1 }{ 2 }\]

So this would eventually lead down to y_1 = e^alpha t ( C1 cos 1/2 t + C2 sin 1/2 t) as one part of the solution where the complex conjugate of y_1 would be the other one and another solution using r = \[\pm i\].

Answer 7

great work

Answer 8

Thanks haha.