1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving a System of Differential Equations with Complex Eigenvalues

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    http://puu.sh/cSK1u/62e2f1c74d.png [Broken]olve the system:
    x' = [-4, -4
    4, -4]
    with x(0) = [ 2, 3]

    Find x1 and x2 and give your solution in real form.


    2. Relevant equations


    3. The attempt at a solution
    Just a note here, I'm basically forced to self-learn this course because my professor doesn't seem to make sense. So I just tried to imitate this website's way of solving one of these problems.
    http://tutorial.math.lamar.edu/Classes/DE/ComplexEigenvalues.aspx

    So I start by finding my complex eigenvalues, which are [itex] -4 \pm 4i [/itex]

    I only need one of them, so I take -4 + 4i, and I plug it into my matrix to solve for the eigenvectors, which I only need one of. I find:

    v1 = [i, 1]

    Which I then plug into my first equation,
    [itex] x_1 (t) = e^{-4t}(cos(4t) + isin(4t)) [i, 1] [/itex]

    Then I multiply in, separate the imaginary from the real, and I have:
    [itex] [-e^{-4t} sin(4t), e^{-4t} cos(4t)] + i[e^{-4t} cos(4t), e^{-4t}sin(4t)] [/itex]

    Next, as the website suggests, I find the constants using the initial values provided, and I get that C1 = 2 and C2 = 3.

    Plugging in the constants, I end up with the equation:
    [itex] 2 [-e^{-4t} sin(4t), e^{-4t} cos(4t)] + 3[e^{-4t} cos(4t), e^{-4t}sin(4t)] [/itex]

    However, I don't understand what the question means by "Real form". Could anyone explain this to me? thank you in advance.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 16, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You need both in principle. And write the general solution as linear combination of these two independent "basic" solutions, belonging to the different eigenvalues. Fitting the linear combination to the initial conditions, you get a real solution of the differential equation.
     
    Last edited by a moderator: May 7, 2017
  4. Nov 16, 2014 #3
    I tried that and I end up with:

    [itex] x = C_1 e^{-4t} (cos(4t) + isin(4t)) [i,1] + C_2e^(-4t) (cos(4t) - isin(4t)) [i, -1] [/itex]

    Then after fitting it to the initial conditions, I end up with :
    [itex] C_1 = -i - \frac {3}{2} , C_2 = -i + \frac{3} {2} [/itex]

    However I don't see how this solution is real, since the i is still present.
     
  5. Nov 16, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Keeping the complex form, the general solution is [itex] \vec x = C_1 e^{(-4+4i)t} [i,1] + C_2e^{(-4-4i)t} [i, -1] [/itex]
    Fitting to the initial conditions, C1=1.5-i; C2=-1.5-i. Your constants does not seem correct.
    Substitute the constants back into the expression for x and changing the exponential form to the trigonometric ones , you get a real expression for both components of x.
     
  6. Nov 16, 2014 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Inserting ##t = 0## into your expression, I get [3,2] instead of [2,3] ...
     
  7. Nov 16, 2014 #6
    Oh, yeah I think I got the two numbers swapped. I recalculated and got

    C1 = 3/2 - i, C2 = -3/2 - i

    However, even after substituting in the constants and changing the exponential forms to trigonometric, I am still left with:

    ( cos(4t) + isin(4t) ) [e^(-4t) ( (3/2) i + 1), e^(-4t) (3/2 - i) ] - (cos(4t) - isin(4t)) [e^(-4t) (3/2 i - 1), e^(-4t) ( (-3/2) - i )]

    When I collect the terms for x1 and x2, I don't see how I am supposed to get rid of the i that makes the expression imaginary.
     
  8. Nov 16, 2014 #7

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Did you try simply multiplying in the trigonometric factors and adding the two vectors together? (Obviously, ##e^{-4t}## can be factored out of the expression so you do not have to care about it.)
     
  9. Nov 16, 2014 #8
    Oh wow! I didn't expect it to work out like that ._.

    So I got x1 = e^(-4t) (-3 sin(4t) + 2cos(4t))
    x2 = e^(-4t) (3cos(4t) + 2 sin(4t))

    Initial condition checks out, did I make any other mistakes? I didn't want to type the entire solution that I did since I expanded the entire thing and had an equation stretch two lines on my piece of paper.
     
  10. Nov 16, 2014 #9

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The final check is of course just plugging your solution back into your differential equation to see that it works out. Once you have done that, you're home safe.
     
  11. Nov 16, 2014 #10
    Alright, thanks guys, I was really confused about this but I'm starting to understand more. Thank you for all the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted