Higher Order Homogeneous ODE (IVP)

[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

and so the answer is:

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

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?

 

[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.

 

[HallsofIvy]

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??)

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

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

and so the answer is:

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

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?

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?

 

[kape]

oh! thank you! i just understood! :D