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Second order differential equation problem

  1. Dec 7, 2005 #1
    Consider the second order linear equation

    z" + c(t)z = 0
    Where c(t) is a continuous real-valued function of a real variable.

    (a) Show that every (nontrivial) solution of this equation is non-oscillatory if c(t) < (1 - epsilon)/(4t^2) for t>=1, where epsilon > 0 is a real number.

    (b)Show that every (nontrivial) solution of this equation is oscillatory if c(t) > (1+epsilon)/(4t^2) for t>=1, where epsilon > 0 is a real number.
  2. jcsd
  3. Dec 8, 2005 #2


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    ODE, not sure of your intentions about this interesting problem (my view anyway), but if I were to be confronted with such a task I would frist, before doing anything analytically, just solve a few of them numerically (cus it's easy that way in Mathematica esp. with 'EquationTrekker') just to see what's going on qualitatively with this, then I suppose I'd look into the criteria for underdamping, overdamping, critial damping, then start to approach it analytically.
  4. Dec 27, 2005 #3


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    Just for closure I'd like to complete this problem since I hate not being able to solve something I feel I'm capable of (hope you guys don't mind):

    For starters damping does not enter into the equation since the first derivative is absent. Ideally then one seeks to find an analytical expression for the solution and then analyze it for the various restrictions indicated above for c(t). It took me a while but after an hour or two (I'm never gonna' get good at this stuf), I realized the equation is in Euler-Cauchy form:

    [tex]4x^2 y^{''}+0xy^{'}+a(1+\epsilon)y=0[/tex]

    Note that I'm using the parameter 'a' to distinguish the two cases above:

    [tex] \frac{a(1+\epsilon)}{4t^2}<\frac{1+\epsilon}{4t^2}\:\text{when}\:a<1\:\text{and}\:\epsilon<0[/tex]

    [tex] \frac{a(1+\epsilon)}{4t^2}>\frac{1+\epsilon}{4t^2}\:\text{when}\:a>1\:\text{and}\:\epsilon>0[/tex]



    and substituting into the ODE and solving for the auxiliary equation:




    The solution then becomes:


    Thus becoming clear that the two cases hinge upon the quantity inside the radicals. That is, when:

    [tex]1\geq a(1+\epsilon)[/tex]

    the solution is non-oscillatory and exponential in form as indicated above. This of course will be the case when a<1 and [itex]\epsilon<0[/itex].

    However if:


    then complex roots enter the solution. Using Euler's expansion of such and noting that:


    the solution becomes:



    Which of course is oscillatory.

    Ok, I'm done.:smile:
    Last edited: Dec 27, 2005
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