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Solving fourth order differential equation ( )

by sasikanth
Tags: differential, equation, order, solving, urgent
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sasikanth
#1
Apr27-10, 02:11 PM
P: 9
I have two second order differential equation which needs to be solved.

x1''(t) = 8 x2(t)
x2''(t) = 2 x1(t)

I have the initial conditions, x1(0) = 0, x2(0) = 1, and terminal conditions x1(pi/4) = 1, x2(pi/4) = 0.

Can anyone help me solve these equations?? What I did was to write the equations in terms of x1 and x2 respectively, but that gives me a fourth order differential

x1''''(t) = 16 x1(t)
x2''''(t) = 16 x2(t)

and I do not know how to solve for these. Can any one help please??
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Cyosis
#2
Apr27-10, 02:39 PM
HW Helper
P: 1,495
You know that the exponential, sine and cosine return to their original form after differentiating them four times. So try a solution of the form [itex]c_1 e^{a t}+c_2 e^{-at}+c_3 \sin(at)+c_4 \cos(at)[/itex].
sasikanth
#3
Apr27-10, 02:45 PM
P: 9
So the solution would be
x1(t) = c1 e^(16t) + c2 e^-(16t) + c3 sin(16t) + c4 cos(16t)??

Cyosis
#4
Apr27-10, 02:55 PM
HW Helper
P: 1,495
Solving fourth order differential equation ( )

No you found the wrong a. Differentiating e^16t 4 times would give 16^4 e^16t, which is not a solution.
sasikanth
#5
Apr27-10, 02:58 PM
P: 9
would a = 4 be the correct solution??
Cyosis
#6
Apr27-10, 03:21 PM
HW Helper
P: 1,495
I am not sure why you have to ask. Take the derivative of your function with a=4 , four times and you will see that it is not the correct solution.
sasikanth
#7
Apr27-10, 03:24 PM
P: 9
I am sorry, I was taking the second dervivative for some reason. I conclude that a = 2 would be thr right solution. Am I correct??
Cyosis
#8
Apr27-10, 04:00 PM
HW Helper
P: 1,495
Yes that's correct.
sasikanth
#9
Apr27-10, 06:14 PM
P: 9
In order to solve for c1,c2,c3 and c4, I would need the second differential of x1. Would that be x1'' = c1 e^4t +c2 e^-4t ??
matematikawan
#10
Apr27-10, 11:47 PM
P: 333
Quote Quote by sasikanth View Post
I have two second order differential equation which needs to be solved.

x1''(t) = 8 x2(t)
x2''(t) = 2 x1(t)
Your equations are linear with constant coefficients. I would handle either the following ways

1. Apply the Laplace transform to the equations - this will transform the problem to solving algebraic system of equations or

2. Let the solution be x1(t)=q1ert and x2(t)=q2ert . Substitute this assumption and you can determine r, q1 and q2 from eigenvalue-eigenvector equation.
sasikanth
#11
Apr28-10, 07:05 PM
P: 9
I got the solution to the equation using the fourth order differntial, but am stuck wolving for the constants c1,c2,c3,c4. If I wanted the second order differntail for x1, would that be x1'' = c1 e^4t +c2 e^-4t ??


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