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Solution for system of diff equations

  1. May 6, 2014 #1
    Given system like ##\begin{bmatrix}
    x'(t)\\
    y'(t)\\
    \end{bmatrix}

    =

    \begin{bmatrix}
    a & b \\
    c & d \\
    \end{bmatrix}

    \begin{bmatrix}
    x(t)\\
    y(t)\\
    \end{bmatrix}## can be rewritten as ##(-1)u''(t) + (bd)u'(t) + (ad+c)u(t) = 0## with ##\begin{bmatrix}
    x(t)\\
    y(t)\\
    \end{bmatrix}

    =

    \begin{bmatrix}
    u(t)\\
    u'(t)\\
    \end{bmatrix}##

    In other words, if a diff equation of 2nd order can be disassembled in a sytem of 1nd order, thus a system of 1nd order can be assembled in a diff equation of 2nd order, correct? This way, I don't need to find the eigenvalues and eigenvectors...
     
  2. jcsd
  3. May 6, 2014 #2

    Matterwave

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    Science Advisor
    Gold Member

    How do you guarantee that ##y(t)=u'(t)=x'(t)## will always work? I don't see why this would be so... they look independent to me in the first equation...
     
  4. May 7, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is true. Dropping the matrix notation, if x'= ax+ by and y'= cx+ dy. then x''= ax'+ by'= ax'+ b(cx+ dy)= ax'+ bcx+ bdy. From the first equation, by= x'- ax so that
    x''= ax'+ bcx+ d(x'- ax)= (a+ d)x'+ (bc- ad)x. We can always convert a system of n first order differential equations to a single nth order differential equation.
     
  5. May 7, 2014 #4
    And this method isn't more simple than compute the eigenvalues and eigenvectors of a matrix?
     
  6. May 7, 2014 #5

    AlephZero

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    Science Advisor
    Homework Helper

    Matterwave has seen the problem with the OP's idea.

    You can convert the two equations into second order equations
    x''= (a+d)x'+ (bc-ad)x
    y''= (a+d)y'+ (bc-ad)y

    But it doesn't follow that y = x'. You can choose two independent boundary conditions for each equation.

    If it did follow that y = x', it would also follow that x = y', and that clearly means "something is wrong here".

    In fact the OP's idea is used in reverse: to solve a single differential equation of order n, it is often easiest to convert it into a system of n first-order equations.
     
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