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Solve system of two differential equations

  1. Dec 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the system
    ##\dot{x}=-2x+y+5sint## and ##\dot{y}=x-2y+3## for ##x(0)=1## and ##y(0)=1##


    2. Relevant equations



    3. The attempt at a solution

    Well, I tried expressing x from the second equation, derived it and inserted it in the first equation but this process is long. Just to be clear, I will write a few steps..:
    from second equation ##x=\dot{y}+2y-3## therefore ##\dot{x}=\ddot{y}+2\dot{y}##

    Now inserting these two in the first expression gives me ##\ddot{y}+4\dot{y}+3y=6+5sint##

    Homogeneous part is: ##y_h=Ae^{-t}+Be^{-3t}##

    particular 1 (for the constant polynom 6): Constant C=2

    particular 2 (for sin function): ##y_p=\frac{1}{2}sint-cost##

    All together, the solution is than: ##y=Ae^{-t}+Be^{-3t}+\frac{1}{2}sint-cost+2##

    and x from ##x=\dot{y}+2y-3= Ae^{-t}-Be^{-3t}+\frac{1}{2}cost+2sint-3##

    because ##x(0)=1## and ##y(0)=1##, than ##A=-B=\frac{7}{4}##

    HOWEVER, I do know that this can be done using matrices... It goes somehow like this...:

    ##\begin{bmatrix}
    \dot{x}\\
    \dot{y}
    \end{bmatrix}=\begin{bmatrix}
    -2 & 1\\
    1& -2
    \end{bmatrix}\begin{bmatrix}
    x\\
    y
    \end{bmatrix}+\begin{bmatrix}
    5sint\\
    3
    \end{bmatrix}##

    Eigenvalues ##\lambda _1=-1## and ##\lambda _2=-3## give eigenvectors for matrix ##A=\begin{bmatrix}
    -2 & 1\\
    1& -2
    \end{bmatrix}##:

    ##v_1=(-1,1)## and ##v_2=(1,1)##

    SO, now my question:

    What will the solution look like? If the system would me homogeneous it would be something like ##\vec{r}=Pe^{Dt}\vec{c}## where P is matrix from eigenvectors and D matrix from eigenvalues. C is vector of constants.
    But my system also has 5sint and +3 .... What do I do now?

    Thanks for all the help!
     
  2. jcsd
  3. Dec 22, 2013 #2

    haruspex

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    Not sure if it's what you're thinking of, but here's one way.
    In matrix form, you have ##\dot x = A x + B(t) ## where A is constant. Writing D for the differential operator, (D-A)x = B, x = -(A-D)-1B = A-1(I-A-1D)-1B.
    Expanding binomially, x = A-1Æ©A-nDnB. If you write that out I think you'll see how to sum the series.
     
  4. Dec 22, 2013 #3

    Ray Vickson

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    Homework Helper

    If you know how to compute the matrix exponential ##e^{uA}## for scalar ##u## you can solve the linear system
    [tex] \dot{x} = A x + h(t)[/tex]
    as
    [tex] x(t) = e^{At} x(0) + \int_0^t e^{(t-s)A} h(s) \, ds.[/tex]

    The easiest way to get the matrix exponential is to note that for an nxn matrix with distinct eigenvalues ##r_1, r_2, \ldots, r_n## we have
    [tex] f(A) = \sum_{j=1}^n E_j f(r_j) [/tex]
    for any analytical function ##f(x) = c_0 + c_1 x + c_2 x^2 + \cdots## (with ##f(A)## defined as ## c_0 I + c_1 A + c_2 A^2 \cdots ##). Here, the matrices ##E_1, E_2, \ldots, E_n## are the same for any function f, so we can find them by looking at some simple, convenient functions. In particular,
    [tex] f(x) = 1 = x^0 \Longrightarrow A^0 = I = E_1 r_1^0 + E_2 r_2^0 = E_1 + E_2 \\
    f(x) = x \Longrightarrow A^1 = A = E_1 r_1 + E_2 r_2[/tex]
    for the 2x2 case with two distinct nonzero eigenvalues ##r_1, r_2##. Thus, we get ##E_1,E_2## by solving
    [tex] E_1 + E_2 = I\\
    r_1 E_1 + r_2 E_2 = A [/tex]
    Now we can determine the matrix exponential from
    [tex] f(x) = e^{ux} \Longrightarrow e^{uA} = E_1 e^{u r_1} + E_2 e^{u r_2}.[/tex]
     
  5. Dec 22, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Here's how I would do it. The matrix [itex]\begin{pmatrix}-2 & 1 \\ 1 & -2 \end{pmatrix}[/itex]
    has eigenvalues -1 and -3 with corresponding eigenvectors [itex]\begin{pmatrix}1 \\ 1\end{pmatrix}[/itex] and [itex]\begin{pmatrix}1 \\ -1\end{pmatrix}[/itex]. That tells us that
    [tex]\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}\begin{pmatrix}-2 & 1 \\ 1 & -2\end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -3\end{pmatrix}[/tex]

    ([itex]\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}[/itex] is the inverse matrix to [itex]\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}[/itex], the matrix having the eigenvectors as columns.)

    Multiplying both sides of [itex]\begin{pmatrix}x' \\ y'\end{pmatrix}= \begin{pmatrix}-2 & 1 \\ 1 & -2\end{pmatrix}+ \begin{pmatrix}5 sin(t) \\ 3\end{pmatrix}[/itex] by [itex]\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}[/itex] gives
    [tex]\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}x' \\ y'\end{pmatrix}= \begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}-2 & 1 \\ 1 & -2\end{pmatrix}\left(\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}\right)\begin{pmatrix}x \\ y\end{pmatrix}+ \begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}5 sin(t) \\ 3\end{pmatrix}[tex]
    where I have used the fact that [tex]\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}= \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}[/tex] to be able to write that product in the middle.

    That reduces to
    [tex]\begin{pmatrix}(1/2)x'+ (1/2)y' \\ (1/2)x'- y'\end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -3\end{pmatrix}\begin{pmatrix}(1/2)x+ (1/2)y \\ (1/2)x- (1/2)y\end{pmatrix}+ \begin{pmatrix}(5/2)sin(t)+ 3/2 \\ (5/2)sin(t)- 3/2\end{pmatrix}[/tex]

    Now, let U= (1/2)x+ (1/2)y and V= (1/2)x- (1/2)y and that becomes
    [tex]\begin{pmatrix}U' \\ V' \end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -3\end{pmatrix}\begin{pmatrix}U \\ V \end{pmatrix}+ \begin{pmatrix}((1/2) sin(t)+ 3/2 \\ (1/2) sin(t)- 3/2\end{pmatrix}[/tex]
    That is, we have uncoupled the two equations. We can now write them as [itex]U'= -U+ (1/2)sin(t)+ 3/2[/itex] and [itex]V'= -3V+ (1/2)sin(t)- 3/2[/itex].

    Once you have solved those, of course
    [tex]\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}U \\ V\end{pmatrix}[/tex]
     
  6. Dec 23, 2013 #5
    Big thanks to everybody!

    I was able to get the same result. :)
     
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