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Homework Help: Initial condition problem

  1. Aug 30, 2011 #1
    theorem states that for y'' + p(x)y' + q(x)y = r(x);
    if y1 & y2 are random numbers such that y(m) = y1 and y'(m) = y2 then we can find a unique solution y for above differential equation....


    in y'' - y'/x = 0;
    both y = x^2 and y = 0 and y = k x^2 satisfy above with ..
    y(0) = 0; and y'(0) = 0



    then isn't this contradicts above theorem ,,
    i have read that theorem is always true....

    can someone enlighten me on this...?

    thanks
     
  2. jcsd
  3. Aug 30, 2011 #2

    Mark44

    Staff: Mentor

    Aren't you omitting some of the theorem you're citing? For instance, what are the conditions on the p(x) and q(x) functions?
     
  4. Aug 31, 2011 #3
    hmmm ... p and q must be continuous over [a,b] where m lies in the above interval...

    1/x is not continuous .... on 0
    thanks
     
  5. Aug 31, 2011 #4

    HallsofIvy

    User Avatar
    Science Advisor

    I believe you will find that p' must also be continuous on (a, b).
     
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