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System of 4 equations in 4 unknowns

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data
    0 = A+B+D
    0 = A-B+C
    0 = A exp(k) + B exp(-k) + C sin(k) + D cos(k)
    0 = A exp(k) - B exp(-k) + C cos(k) - D sin(k)

    Solve for A, B, C, D in the above system. (k is a positive real number)

    2. Relevant equations
    N/A

    3. The attempt at a solution
    1st equation=> D = -A-B
    2nd equation=> C = B-A
    Put these into the 3rd and 4th equation, we get:
    0 = A exp(k) + B exp(-k) + (B-A) sin(k) + (-A-B) cos(k)
    0 = A exp(k) - B exp(-k) + (B-A) cos(k) - (-A-B) sin(k)

    How should I continue??

    Just wondering: In a system of 4 equations in 4 unknowns, is it POSSIBLE to have infinitely many solutions? or must the solution be unique?

    Any help is greatly appreciated!
     
  2. jcsd
  3. Oct 27, 2009 #2
    Now you have two equations and two unknowns, so solve one equation for A or B and plug that into the other equation.
     
  4. Oct 27, 2009 #3
    But I don't think we can divide by that stuff involoving sin, cos, and exp becuase it may be zero.

    Also, I am expecting to have infinitely many solutions because this actually comes from an eigenvalue problem: ODE X'''' = λ X with boundary conditions X(0)=X'(0)=X(1)=X'(1)=0
    Here λ=k^4, with k>0
    where k is solution cos(k) cosh(k) = 1.
    To find the eigenfunctions, I got the above system of 4 equations in 4 unknowns. Since any nonzero multiple of an eigenfunction is again an eigenfunciton, I am expecting the solution of the system to have one arbitrary constant.

    Can someone please help??
     
  5. Oct 27, 2009 #4
    If you know linear algebra, do it in a matrix. You should be able to get all unknowns A,B,C, and D. k will remain an arbitrary constant.
     
  6. Oct 27, 2009 #5
    But I think k here is a FIXED (given) positive real number??
     
  7. Oct 28, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's what flatmaster just said: " k will remain an arbitrary constant."
     
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