Higher Order Homogeneous ODE (IVP)

1. Apr 18, 2006

kape

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:

$$\lambda^3 + \lambda = 0$$

$$\lambda(\lambda^2 + 1) = 0$$

So one roots is $$\lambda = 0$$

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

And the other root is:

$$\lambda^2 = -1$$

$$\lambda = \pm\sqrt{-1}$$

$$\lambda = \pmi$$

Therefore $$\omega = 1$$

$$y = ce^(0)x + Acos\omegax + Bsin\omegax$$

$$y = c + Acosx + Bsinx$$

$$y' = -Asinx + Bcosx$$

$$y' = -Acosx - Bsinx$$

So, putting in the initial values..

c + A = 5 (so c = 9)
B = 2
A = -4

$$y = 9e^x - 4cosx + 2sinx$$

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: Apr 18, 2006
2. Apr 18, 2006

dextercioby

1.You put one prime more than was necessary in the first equation.

2.Make the substitution

$$y'(x)=f(x)$$

and then integrate the resulting ODE. Then you'll know what to do.

Daniel.

3. Apr 18, 2006

HallsofIvy

Staff Emeritus
Yes, it is certainly possible for 0 to satisfy an equation!

Well, yes, of course, that's wrong! You started by saying (correctly) that the solution must be of the form
$$y = ce^{(0)x} + Acos \omega x + Bsin \omega x$$
$$y = c+ A cos x+ B sin x$$

But after solving (correctly) for c=9, A= -4, and B= 2, you give the solution as
$$y = 9e^x - 4cosx + 2sinx$$

Do you see the difference?

Last edited: Apr 18, 2006
4. Apr 18, 2006

kape

oh!!! thank you!! i just understood! :D