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Eigenvalues of an equation

  1. Mar 24, 2009 #1
    hi I have the following eigenvalue problem
    -(x2y')'=λy for 1<x<2
    y(1)=y(2)=0


    I tried plugging an equation y=xa
    and you get the equation
    a2+a+λ=0
    so for this I get that λ<1/4 to hava a solution. So does this mean, every λ smaller than 1/4 is an eigenvalue?
    do you know what else I could do?

    thanks
     
  2. jcsd
  3. Mar 25, 2009 #2

    HallsofIvy

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    Staff Emeritus
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    Why should [tex]\lambda[/tex] be less than 1/4? That would make the powers of x real numbers but why would that be necessary? In fact, if the powers of x were real numbers wouldn't that make it impossible to satisfy y(1)= y(2)= 0?

    What do solutions to such an equation look like if the characteristic equation has complex roots? Hint: the change of variable t= ln(x) converts an "Euler-type" equation to an equation with constant coefficients having the same characteristic equation.

    Also, since this is a second order linear equation, it has exactly 2 eigenvalues.
     
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