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kape
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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?
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?
Last edited: