Solving fourth order differential equation (URGENT)by sasikanth Tags: differential, equation, order, solving, urgent 

#1
Apr2710, 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?? 



#2
Apr2710, 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].




#3
Apr2710, 02:45 PM

P: 9

So the solution would be
x1(t) = c1 e^(16t) + c2 e^(16t) + c3 sin(16t) + c4 cos(16t)?? 



#4
Apr2710, 02:55 PM

HW Helper
P: 1,495

Solving fourth order differential equation (URGENT)
No you found the wrong a. Differentiating e^16t 4 times would give 16^4 e^16t, which is not a solution.




#5
Apr2710, 02:58 PM

P: 9

would a = 4 be the correct solution??




#6
Apr2710, 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.




#7
Apr2710, 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??




#8
Apr2710, 04:00 PM

HW Helper
P: 1,495

Yes that's correct.




#9
Apr2710, 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 ??




#10
Apr2710, 11:47 PM

P: 330

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 x_{1}(t)=q_{1}e^{rt} and x_{2}(t)=q_{2}e^{rt} . Substitute this assumption and you can determine r, q_{1} and q_{2} from eigenvalueeigenvector equation. 



#11
Apr2810, 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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