Question

Find the solution of the higher order differential

y'''' - y'' + y' - y = 0

Answer 1

are you familiar with something called "auxilliary solution" or something like that?

Answer 2

Not at the moment.

Answer 3

auxilliary solution is something like replacing the differential operator with a variable

for example:

y''' - y'' + y' - y = 0

becomes

m^3 - m^2 + m - 1 = 0

does that look familiar?

Answer 4

Oh I made a typo, I mean y''' - y '' + y' - y, the highest power was supposed to be 3.

I know the the characteristic equation is r^3-r^2+r -1

Answer 5

yes that's the auxiliary solution i was referring to

Answer 6

r^3 - r^2 + r - 1 = 0

do you know how to factor it out?

Answer 7

I'm kind of brain dead at the moment, kind of forgot how to factor out higher order polynomials :( It's r^2(r-1)+(r-1)=0 right?

So we have a double root at 1, and a real root at 0?

Answer 8

no...you still have to factor r^2(r-1) + (r-1) further

Answer 9

here's a hint: factor out (r-1) because it's a common factor to both terms...what do you get?

Answer 10

Blaaaahh, can't believe I grossly overlooked that. (r^2+1), r-1

Answer 11

right

Answer 12

so your roots are \[r = 1\]
and \[r = \pm i\]

Answer 13

right so the relating equation would be C1E^t + C2 cos t + C3 sin t ?

Answer 14

yes

Answer 15

i just don't know why you used t

Answer 16

Oh I got used to my professor using t... haha

Answer 17

anyway..that would be the right answer

Answer 18

Thank you.