Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fundamental set of solutions

  1. Feb 9, 2009 #1


    User Avatar

    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


    User Avatar
    Science Advisor

    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


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook