Matrices and general solutions

  • Thread starter Thread starter hbomb
  • Start date Start date
  • Tags Tags
    General Matrices
hbomb
Messages
57
Reaction score
0
I'm stuck on this problem that involves drawing a phase portrait but I'm at a stand still.

Find the general solution to the following differential equation:
[1 -12 -14]
x'= [1 2 -3]*x
[1 1 -2]

the eigeinvalues that I found are 1 , +5i, -5i

the general solution that I found is

c1e^e[25] + c2[cos5t+5sin5t] + c2[-5cos5t+sin5t] + c3[2] + c3t[1-5i]
[-7] [cos5t ] [sin5t ] [2] [1 ]
[6 ] [cos5t ] [sin5t ] [1] [1 ]

Am I suppose to have imaginary values in my solution?
I'm supposed to find the solution if x(0)=2i-5j+3k
I just plug 0 in for all the t's and solve for the constants which is no problem for me. The problem that I'm having is the fact that I have an imaginary value in the solution.

I'm also suppose to find the phase portrait.
 
Physics news on Phys.org
No, since this problem has only real entries, you should not have any complex numbers in your solution- and I don't see why you should. If your eigenvalues are 1, 5i, and -5i, then your solutions will involve ex, cos(5x), and sin(5x).

I have trouble understanding what you mean by
c1e^e[25] + c2[cos5t+5sin5t] + c2[-5cos5t+sin5t] + c3[2] + c3t[1-5i]
[-7] [cos5t ] [sin5t ] [2] [1 ]
[6 ] [cos5t ] [sin5t ] [1] [1 ]
Is that the "vector" solution? Is [-7][cos5t][sin5t][2][1] 5 separate functions or is that a product? Each row should be a single function like your first row.
 
It's the vector solution. I just realized that this message board gets rid of the extra spaces. 25, -7, 6 is one column matrix. cos5t+5sin5t, cos5t, cos5t is one column matrix. -5cos55t+sin5t, sin5t, sin5t is one column matrix. 2, 2, 1 is one column matrix. 1-5i, 1, 1 is one column matrix. And that's what I thought...about the imaginary values. How should look in the general solution?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top