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Solving Euler's equation

  1. Dec 7, 2013 #1
    How do I solve the following Euler's equation:

    [tex]r^2 B_n'' + r B_n' - n^2 B_n = 3 \delta_{n1} r^2[/tex]

    Such that the solution is:

    [tex]B_n(r) = \beta_n r^n + \delta_{n1}r^2, \forall n \ge 1[/tex]

    where βn is a free coefficient, δ is the Kronecker delta function, and the solutions unbounded at r=0 are discarded.
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2

    Dick

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    You try to solve the differential equation for different values of n. n=1 is obviously different from all other n. The trick for an equation in this form is to use a trial function of the form ##B_n(r)=Cr^k## and solve for k.
     
    Last edited: Dec 7, 2013
  4. Dec 7, 2013 #3
    Is the [itex]\delta_{n1}r^2[/itex] obtained in the solution by linearity? And why is the coefficient '3' not in front?
     
  5. Dec 7, 2013 #4

    Dick

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    I'm not sure what you are asking. The case n=1 is different from the other values of n because then the right side is 3r^2. If n is not 1 then the right side is 0. That's what the Kronecker delta does. They are two different cases. Solve them separately.
     
    Last edited: Dec 7, 2013
  6. Dec 7, 2013 #5
    I've followed your steps, but when I'm solving the case for n = 1, I get [itex]B_{1} = \beta_{1}r + r^{2}[/itex]. Shouldn't it only be [itex]r^{2}[/itex]?
     
  7. Dec 7, 2013 #6

    Dick

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    Yes, it should. You have to put ##\beta_{1}=0## in that case. The problem statement is sloppy.
     
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