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Dif.eq.system; complex eigenvalues

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Given system:
    dx/dt=-x-5y
    dy/dt=x+y


    2. Relevant equations



    3. The attempt at a solution
    So I calculated that [itex]\lambda[/itex]_1=-2i and [itex]\lambda[/itex]_2=2i
    Generaly [itex]\lambda[/itex]=+-qi

    next i know that general solution is in form:
    x=C1cos(qt)+C2sin(qt)
    y=C*1cos(qt)+C*2sin(qt)

    So to calculate constsnts C*_1 and C*_2 I can choose C_1 and C_2 how i want.
    Then from sytems second eq.(dy/dt=x+y) i calculate C*_1 and C*_2 .


    My question is- which q value I put in solution,is it
    x=C1cos(2t)+C2sin(-2t)
    or
    x=C1cos(2t)+C2sin(2t)
    ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2013 #2

    vanhees71

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    2016 Award

    You have two different complex eigenvalues [itex]\lambda_1, \lambda_2[/itex] and thus also two linearly independent eigenvectors [itex]\vec{e}_1,\vec{e}_2[/itex]. Then the general solution of the linear ODE system reads
    [tex]\vec{r}(t)=[\vec{x}(t),\vec{y}(t)]^t=A_1 \vec{e}_1 \exp(\lambda_1 t) + A_2 \vec{e}_2 \exp(\lambda_2 t).[/tex]
    Since your eigenvalues are conjugate complex and the original matrix is real, you have [itex]\vec{e}_2=\vec{e}_1^*[/itex]. In order to get the general real solutions you have to set [itex]A_2=A_1^*[/itex].
     
  4. Nov 3, 2013 #3

    D H

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    Staff Emeritus
    Science Advisor

    Have you learned about the matrix exponential? If you have, this problem is nicely formulated to have a very simple matrix exponential solution.


    If you haven't, you'll have to use the techniques you have been taught.
    Does it make any difference? That's a rhetorical question. Since C2 is an arbitrary constant and since sin(-x) = -sin(x), your two forms are equivalent.
     
  5. Nov 3, 2013 #4
    OK,I did not see sin(-x) = -sin(x). So it means that i can put either one of q values and it does not make a difference?

    My goal for this system to draw ( i realy dont know how its called in english meiby integral lines) integral lines, which should look like ellipse (so the form of elipse is dependent from q)
     
  6. Nov 4, 2013 #5

    ehild

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    Homework Helper
    Gold Member

    It is easy to solve the system of equations in the OP by transforming them into a second order equation:
    d^2y/dt^2=dx/dt+dy/dt--> d^2y/dt^2+4y=0. The solutions are y1=sin(2t ) and y2=cos (2t ), and the corresponding x-s are obtained as x1(t)=y1-dy1/dt, x2(t)=y2-dy2/dt. The general solution is x=c1x1+c2x2, y=c1y1+c2y2, or it can be written as Y=Asin(2t+θ), X=2Acos(2t+ θ)-Asin(2t+θ)
    You get the ellipses iin the x,y plane by eliminating t.

    ehild
     
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