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System of second order ODEs

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve:

    [ d^2y1/dx^2 ] = [ a -a ] [ y1 ]
    [ d^2y2/dx^2 ] [ -a a ] [ y2 ]

    A = [ a -a ]
    [ -a a ]

    2. Relevant equations

    Everything required is in (1) above

    3. The attempt at a solution

    Reduce to 1st order system

    M = [ 0 I ]
    [ A 0 ]

    Hence, M =
    [ 0 0 1 0 ]
    [ 0 0 0 1 ]
    [ a -a 0 0 ]
    [ -a a 0 0 ]

    The eigenvalues of M are 0, 0, √2a and -√2a
    The eigenvectors are [ 1 1 ], [ 1 1 ], [ 1 -1] and [ 1 -1]

    Hence the general solution is (for y only):

    y(x) = A*e^(0x)*[ 1 1 ]T + B*e^(0x)*[ 1 1 ]T + C*e^(√2a*x) [ 1 -1 ]T +
    D*e^(-√2a*x) [ 1 -1 ]T

    So:

    y1 = A + B + C*cosh(√(2a)*x) + D*sinh(√(2a)*x)
    y2 = A + B - C*cosh(√(2a)*x) - D*sinh(√(2a)*x)

    Is this correct?
     
  2. jcsd
  3. Mar 16, 2014 #2
    The way to check if thats correct is to plug your answer back into the ODE system.
     
  4. Mar 16, 2014 #3

    pasmith

    User Avatar
    Homework Helper

    Do you not recall that the solution of [itex]u'' = 0[/itex] is [itex]u(x) = Ax + B[/itex]?


     
  5. Mar 16, 2014 #4
    Yes, ok. I see you mean because I have repeated roots, it's Ae^0(x) + Bxe^(0x). Even the above without A and B constants is a solution to the system. However, since I have 4 boundary conditions, I thought I'd keep them in there
     
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