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2nd order differential equations

  1. Mar 9, 2008 #1
    1) Assume that p and q are continuous on some open interval I, and that y1 and y2 are solutions of y'' + p(t)y' + q(t)y = 0 on I.
    a) Prove that if {y1, y2} form a fundamental set of solutions on I, then they can't have a common inflection point in I, unless p and q are both 0 at this point.
    b) If 0 E I, prove that y(t)=t3 cannot be a solution of the differential equation on I.


    =====================
    1a) My first question: "A unless B", is this equivalent to "A if and only if not B". That is, do I have to prove both directions for question 1a simply by seeing the word "unless"?

    Now my thoughts about this problem:
    Inflection point at c => f ''(c)=0 or f ''(c) does not exist.
    But in this case, since we're given that y1 and y2 are solutions of the differential equation, this means that y1 and y2 must be twice differentiable, so we can reject the possiblity f ''(c) does not exist.
    Now we have: inflection point at c => f ''(c)=0
    Fundamental set of solutions means Wronskian is nonzero.
    But now I am having a lot of trouble deriving one from the another...


    1b) No clue...need some hints...


    Thanks for your help!:smile:
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2
    For part b, since we are given that p, q are continuous, I think this may be related to existence-uniqueness theorem...but I can't figure out how to use it...
     
  4. Mar 11, 2008 #3
    1b) y(t)=t^3
    =>y'(t)=3t^2
    =>y''(t)=6t

    =>y'(0)=0, y''(0)=0

    But the initial value problem (IVP) y'' + p(t)y' + q(t)y = 0, y'(0)=0, y''(0)=0 has the solution y≡0 by inspection, and by existence-uniqueness theorem, y≡0 must be the only solution to the IVP. So y(t)=t^3 cannot be a solution to the ODE on I if 0 E I.

    I've tried my best. Is this a correct proof?
     
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