(adsbygoogle = window.adsbygoogle || []).push({}); Higher Order Homogeneous ODE (IVP) [Solved]

I am having problems with this IVP:

y'''' + y' = 0

y(0) = 5

y'(0) = 2

y''(0) = 4

What I have done so far is:

[tex] \lambda^3 + \lambda = 0 [/tex]

[tex] \lambda(\lambda^2 + 1) = 0 [/tex]

So one roots is [tex] \lambda = 0 [/tex]

(though.. can there be a root that is zero??)

And the other root is:

[tex] \lambda^2 = -1 [/tex]

[tex] \lambda = \pm\sqrt{-1} [/tex]

[tex] \lambda = \pmi [/tex]

Therefore [tex] \omega = 1 [/tex]

[tex] y = ce^(0)x + Acos\omegax + Bsin\omegax [/tex]

[tex] y = c + Acosx + Bsinx [/tex]

[tex] y' = -Asinx + Bcosx [/tex]

[tex] y' = -Acosx - Bsinx [/tex]

So, putting in the initial values..

c + A = 5 (so c = 9)

B = 2

A = -4

and so the answer is:

[tex] y = 9e^x - 4cosx + 2sinx [/tex]

But it seems the answer is wrong..

(though, considering my severe lack of math skills, it doesn't surprise me)

I really need to be able to solve this problem.. can anyone help?

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# Homework Help: Higher Order Homogeneous ODE (IVP)

**Physics Forums | Science Articles, Homework Help, Discussion**