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Finding a solution to this equation using Frobenius method 
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#1
Sep1911, 04:05 AM

P: 17

Hi, I have this equation to solve.
y'' + (1/x)y' + [(x^2) + k + (m^2 / x^2)]y = 0 now, I've tried to solve this using frobenius method but cannot formulate a solution. I have that a_(n+4) = [ka_(n+2)  a_(n)] / [n^2 +/ 2inm] is my recurrence relation, but now I'm stuck and don't know how to proceed, any help greatly appreciated. thanks. 


#2
Sep1911, 03:46 PM

P: 1,666

Using the standard convention of letting [itex]a_0=1[/itex], I get:
[tex]a_1=0[/tex] [tex]a_2=\frac{h a_0}{(2+c)(1+c)+1+m^2}[/tex] with the remaining odd a_n=0 and even a_n equal to: [tex]a_n=\frac{a_{n2}+a_{n4}}{(n+c)(n+c+1)+1+m^2}[/tex] with a similar relation for the other root expressed in b_n and so I can write the solution as: [tex]y(x)=K_1 \sum_{n=0}^{\infty} a_n x^{n+mi}+K_2 \sum_{n=0}^{\infty} b_n x^{nmi},\quad x>0[/tex] and I believe because of the conjugates in the solution, for real initial conditions [itex]y(x_0)=y_0[/itex], and [itex]y'(x_0)=y_1[/itex], the imaginary component is annihilated leaving the desired real solution. If you into it, here's some Mathematica code to check the solution. I believe it's correct but not as close as I would have expected. May be some convergence issues though.



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