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Fundamental set of solutions

  1. Feb 9, 2009 #1

    d_b

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    For a second order linear differential homogeneous equation, if the two solution y1 and y2 is a multiple of one another. It means that it is linearly dependent which mean they can not form a fundamental set of solutions to second order differential homogeneous equation.

    Am I correct?? or could it be any cases where y1 and y2 is a mulitple of one another and still can form a fundamental set of solutions.

    Also if y1 and y2 are L.D is that mean wronskian equals zero????
     
  2. jcsd
  3. Feb 10, 2009 #2
    You are correct, they cannot be multiples, they cannot be linearly dependent, as that means they are the same solution. (u=au1+bu1=cu1.)
     
  4. Feb 10, 2009 #3

    HallsofIvy

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    Staff Emeritus
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    And yes, if y1 and y2 are multiples of each other, then their Wronskian is equal to 0:
    Specifically, if y1(t)= ay2(t) for some number a, then it is also true that y1'= ay2' so the Wronskian is
    [tex]\left|\begin{array}{cc}y1 & y2 \\ y1' & y2'\end{array}\right|= \left|\begin{array}{cc} y1 & ay1 \\ y1' & ay1'\end{array}\right|= a(y1)(y1')- a(y1)(y1')= 0[/tex]
     
  5. Feb 10, 2009 #4

    d_b

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    what about y2=U(t)y1 ? Isn't y1 and y2 is the set of fundamental solution?? why is wronskian is not equal to zero then???
     
  6. Feb 10, 2009 #5
    those are not linearly dependent.If U was a constant instead of U(t), they would be.
     
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