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Solve the differential system

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a solution of the system ##{x}'''=2x+y## and ##{y}'''=x+2y## for which ##x(0)={x}'(0)=0## and ##{x}''(0)=1## also for ##y(0)={y}'(0)=0## and ##{y}''(0)=1##.


    2. Relevant equations



    3. The attempt at a solution

    I must be doing something wrong:

    ##\begin{bmatrix}
    {x}'''\\
    {y}'''
    \end{bmatrix}=\begin{bmatrix}
    2 & 1\\
    1&2
    \end{bmatrix}\begin{bmatrix}
    x\\
    y
    \end{bmatrix}##

    Eigenvalues are ##\lambda _1=1## and ##\lambda _2=3## so eigenvectors ##v_1=(-1,1)## and ##v_2=(1,1)##.

    Therefore matrix ##D=\begin{bmatrix}
    1 & 0\\
    0& 3
    \end{bmatrix}## and matrix ##P=\begin{bmatrix}
    -1 & 1\\
    1&1
    \end{bmatrix}##

    So general solution should be ##\begin{bmatrix}
    x\\
    y
    \end{bmatrix}=Pe^{Dx}\vec{c}=\begin{bmatrix}
    -1 & 1\\
    1&1
    \end{bmatrix}\begin{bmatrix}
    e^x & 0\\
    0& e^{3x}
    \end{bmatrix}\begin{bmatrix}
    A\\
    B
    \end{bmatrix}##

    But all these conditions ##x(0)={x}'(0)=0## and ##{x}''(0)=1## also for ##y(0)={y}'(0)=0## and ##{y}''(0)=1##.... What do I do? :/

    Thank you for your help!
     
  2. jcsd
  3. Dec 27, 2013 #2

    ehild

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    Gold Member

    λ2 is not 3, but third root of 3. Both x and y are functions of the same variable, say t. And you tried to find the solutions as x(t)=aeλt and y(t)=beλt. You cannot write exponents of x into the solution.

    What does the last matrix equation mean for x(t) and y(t)? ( Expand the product.) You have to find the constants A and B so as the initial conditions are fulfilled.

    ehild
     
    Last edited: Dec 27, 2013
  4. Dec 27, 2013 #3
    ##det\begin{bmatrix}
    2-\lambda &1 \\
    1 & 2-\lambda
    \end{bmatrix}=(2-\lambda )^2-1=\lambda ^2-4\lambda +3=(\lambda -1)(\lambda -3)=0## ????

    am... it means that ##\begin{bmatrix}
    x\\
    y
    \end{bmatrix}=\begin{bmatrix}
    -Ae^x+Be^{3x}\\
    Ae^x+Be^{3x}
    \end{bmatrix}## but i have 6 conditions for only two constants... o_O
     
  5. Dec 27, 2013 #4

    ehild

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    It is not a first order but third order system of equations. And your matrix equation means that x= -Aex+Be3x. Do you think it true?


    ehild
     
    Last edited: Dec 27, 2013
  6. Dec 27, 2013 #5
    Am, I can see the problem but I have to admit that I don't know what to do? Should I calculate the eigen values of matrix ##A^3## ?

    Uf, good point, I guess the ##'## mean derivation by t and not by x.
     
  7. Dec 27, 2013 #6

    ehild

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    Homework Helper
    Gold Member

    No, why ? Still, yo can assume the solutions of form x=aeλt, y=beλt). Substitute into the original equations. You get values 1 and 3 for λ3. But λ is a complex number, and you have the complex third roots, 6 values altogether.

    Or you can introduce dummy variables u=dx/dt, v=du/dt and p=dy/dt, q=dp/dt to transform the original equations into first order ones. Then you have a system of six first order equations and you can solve it with the standard method.

    ehild
     
  8. Dec 27, 2013 #7
    Aaa, ok, i get it now... Is this a standard procedure for systems that are second orders or higher?

    So I will get a polynomial ##\lambda ^6-4\lambda ^3+3=(\lambda -1)(\lambda ^3-3)(\lambda ^2+\lambda +1)##

    This gives me ##\lambda _1=1## and ##\lambda _{1,2}=\frac{1}{2}(-1\pm i\sqrt{3})## and ##\lambda _3= \sqrt[3]{3}##

    How exactly do I get the other two? o_O
     
  9. Dec 27, 2013 #8

    ehild

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    λ6-4λ3+3=0 is a quadratic equation in λ3, with solutions λ3=1 and λ3=3.

    So the roots are 1, (-1/2±√3/2 i) and 31/3, 31/3(-1/2±√3/2 i)

    (You can factorize λ3-3 in the same way you did with λ3. If a=31/3, λ3-3=(λ-a)(λ2+aλ+a2).)

    ehild
     
  10. Dec 27, 2013 #9
    Thank you, now I know how to finish this!
     
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