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Re: differential equations help

  1. Jul 29, 2008 #1
    I'm having a couple problems with my diff eq assignment if anyone can help me. The problem is 1. (a) show that the diff eq. for y=c1u1(x) + c2u2(x), where u1(x) and u2(x) are functions which are at least twice differentiable, may be written in determinant form as
    | y u1(x) u2(x) |
    | y u1'(x) u2'(x) | = 0
    | y u1''(x) u2''(x)|

    b) what happens in the case where

    w = | u1(x) u2(x) | = 0
    | u1'(x) u2'(x)|
     
  2. jcsd
  3. Jul 30, 2008 #2

    HallsofIvy

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    In that case, the two function, u1(x) and u2(x), are not independent. You need two independent solutions to write the general solution to a second order equation.
     
  4. Jul 30, 2008 #3
    I understand that if the determinant is zero, the equations are linearly dependent, but I don't know how to go about doing the proof required to answer the problem.
     
  5. Jul 30, 2008 #4

    Defennder

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  6. Jul 30, 2008 #5
    That's part B... part A though asks for proof of the determinant of the three equations equaling zero...that's the part I'm really having issues with.
     
  7. Jul 30, 2008 #6

    Defennder

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    Well you didn't provide the DE, so I assume it's a 2nd order ODE. Is the first column of the matrix supposed to read y,y',y'' from top to bottom instead? HallsOfIvy has said you need 2 linearly independent solutions for a second order ODE, so what can you say if the determinant of that is zero?
     
  8. Jul 30, 2008 #7
    yes it is supposed to read y, y', y"... the problem is as it is written in the book... it didn't say that it was a 2nd order ODE.
    Clearly the original differential eq. and it's two derivatives are linearly dependent since the determinant is zero, but I don't understand what I'm exactly supposed to be showing
     
  9. Jul 30, 2008 #8

    Defennder

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    This can be verified directly. You are given y=c1u1+c2u2. Find y' and y'' and evaluate the determinant. No proof is needed because none is asked for.
     
  10. Jul 30, 2008 #9

    HallsofIvy

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    Yes, Defennder is right. I had overlooked A and answered B.

    If y= c1u1+ c2u2, then
    [tex]\left|\begin{array}{ccc} y & u & v \\ y' & u' & v' \\ y" & u" & v" \end{array}\right|= \left|\begin{array}{ccc} c1u+ c2v & u & v \\ c1u1'+ c2u2' & u' & v' \\ c1u1"+ c2u2" & u" & v" \end{array}\right|[/tex]

    If you know a little linear algebra, and properties of determinants, the fact that the first column is a linear combination of the second and third columns, that tells you all you need. If you don't recognise that, go ahead and do the calculations.
     
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