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Frobenius Method again!

  1. Mar 6, 2009 #1
    Assume that x=0 is a regular singular for x2y" + xp(x)y' + q(x)y = 0
    and the indicial equation has equal roots [tex]\lambda = \lambda_1 = \lambda_2[/tex]

    The first solution is alway known to be of the form [tex]y_1(x) = x^{\lambda_1}\sum a_n x^n [/tex]

    Although tedious, I know how to obtain the second linearly independent solution using the Lagrange reduction of order, y2(x) =u (x)y1.

    I think it is well known that the second solution will be of the form
    [tex]y_2(x) = y_1(x)lnx + x^{\lambda_1}\sum b_n x^n [/tex]


    The book I'm refering, Schaum Outline Series in Differential Equation (cheap and direct to the method :smile: ) gives the following method to compute y2.

    [tex]y_2(x) = \frac{\partial y_1(x,\lambda)}{\partial\lambda} |_{\lambda=\lambda_1} \\\ (*)[/tex] .

    As you all know the book never proved any of their theorem/method.

    My question is why (*) is a solution for the DE? Any proof for it?
     
  2. jcsd
  3. Mar 7, 2009 #2
    I don't remember the method well enough to do this for you but I think its sortof similar to how you prove that a linear constant coefficient equation with a repeated root of the characteristic polynomial will have a second solution given by t times the first. I know I have seen a very good exposition on this topic in Hildebrand's Advanced Calculus for Engieneers. I would highly recommend that book for a lot of stuff and that section was very clear - but I didn't work it through well enough to internalise it and be able to answer your question from the top of my head.
     
  4. Mar 9, 2009 #3
    Thanks b17m4p.
    There is a copy in our library. I will try to get it. But it looks quite an old edition 1958 ?

    QA303 HIL 1949
    Advanced calculus for engineers
    Hildebrand, Francis Begnaud.
    Englewood Cliffs, N.J. : Prentice-Hall, [1958]


    If only I have the note with a click of mouse! :smile:
    Wikipedia ? MIT courses ?
     
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