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A Tricky Problem

  1. Nov 10, 2004 #1
    My friend found this problem from Anton

    Suppose that the auxiliary equation of the equation

    [tex]y'' + py' + qy = 0[/tex]

    has a distinct roots [tex]\mu[/tex] and [tex]m[/tex].

    (a)Show that the function

    [tex]g_\mu(x) = \frac{e^{\mu x} - e^{mx} }{\mu - m}[/tex]

    is a solution of the differential equation

    (b)Use L'Hopital's rule to show that

    [tex]\lim_{\mu\rightarrow\ m} g_\mu(x) = xe^{mx}[/tex]

    I tried to proof this using the D-operator method to find the roots, it doesn't seem to work. There seems to be a simpler way of doing this, but I just can't see it.

    Any help is appreciated.
     
    Last edited: Nov 10, 2004
  2. jcsd
  3. Nov 10, 2004 #2
    Ok i havent given this much thought but try this ,
    (Assuming that p and q are constant)
    Standard results of polynomials,
    p = mu + m
    q = mu*m

    Differentiate the function g(x) and substitute in the original equation to show that it satisfies the equation.

    the answer to b is trivial using L'Hospital,
    Differentiate numerator and denominator w.r.t mu,
    its easy to see that numerator differentiates to xe^(mu*x)
    and the denominator is 1.
    the limit evaluates to the required one easily.....

    -- AI
     
  4. Nov 10, 2004 #3
    actually you dont have to use L'Hopital, just the plain definition of derivative, wich is more elegant i think :P
     
  5. Nov 10, 2004 #4

    dextercioby

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    I's definitely more elegant...
    The auxiliary equation reads
    [tex]\lambda^2+p\lambda+q=0 [/tex]
    If u chose the solution with "+" to be"µ",and the one with "-" to be "m",then its solutions verify identically the equation above,i.e.
    [tex] \mu^2+p\miu+q=0;m^2+pm+q=0 [/tex]
    Making the differentiations of "g" correctly and substituting into the original equation,after separating parts with [tex] \exp{\mux} [/tex] and [tex] \exp{mx} [/tex],will find exactly the 2 equations stated above,which actually will ensure you that g is a solution of th equation.
     
    Last edited: Nov 10, 2004
  6. Nov 10, 2004 #5

    dextercioby

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    Hold your horses for a while,junior...
    1.It's irrelevant "which is which",as long as they are different.
    2.Why would complicate that much???The general solution to the given ODE is a linear superposition of fundamental solutions,which are [tex] \exp{\mux} [/tex] and [tex] \exp{mx} [/tex] with arbitrary (hopefully noninfinite,in this case it applies,sice "µ" and "m"are different) coefficients,call them A and B.Who stops you from chosing
    [tex] A=\frac{1}{\mu-m};B=-A [/tex] or viceversa,to find your solution without making any derivatives of the solution given??????? :eek:
    Bonehead... :rofl: :biggrin:
     
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