Question

Differential equations problem. It says and I quote

"Solve the differential equation. Leave answers in exact form"

y^(6)-27y=0

I'm not sure if my teacher has a typo or what because Idk how to solve that without initial values?

Answer 1

You write down the general solution.

Answer 2

The general form I figure is C1 + C2e^(27^(1/5)

Answer 3

C2e^(27^(1/5)t) **

Answer 4

and that general solution has six terms

Answer 5

Hrm, how do you get the general solution for this? Maybe i just don't know how to factor r^6-27r = 0 correctly lol

Answer 6

Note the characteristic equation is
r^6 - 27 = 0

Answer 7

and remember r can be and is for 5 of the solutions definitely complex

Answer 8

4 of the six are complex

Answer 9

Ohhhh, that's what i was doing wrong lol. I was thinking the characteristic equation was r^6-27r =/

Answer 10

How do you go about factoring r^6 - 27 = 0 completely from there... Would you do a pascals triangle or is there another way to do it that's not super difficult?

Answer 11

What are the six solutions of z^6 - 1 = 0 ?

Answer 12

Uhhh, r = + or - 1

Answer 13

No. exp(2jpi.i/6) = exp(pi.j.i/3) for j = 0, 1, 2 ,3 ,4 ,5