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If we look at the Bessel ODE: $$x^2 y'' + xy' + (x^2 - \alpha^2)y = 0$$

Suppose I then put the solution to the above ODE as $J_{\alpha}(x)$ in the RHS, and try to solve the following ODE: $$x^2 y'' + xy' + (x^2 - \alpha^2)y = J_{\alpha}(x)$$ Obviously the solution to this equation is the solution to the homogeneous equation plus a specific solution.

Suppose I want to reiterate this procedure of inserting in the RHS of Bessel ODE the solution to the previous equation iteratively.

What would the solution at the $n$-th step will look like?

This is a pure math exercise I am thinking about, I don't think that someone thought of this before.

You can of course generalize it to any special function you might think of with a suitable ODE that defines it.

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    $\begingroup$ How are you choosing which solution to take at each step? are you setting some initial data for $y,y'$? $\endgroup$
    – Sam Zbarsky
    Commented Nov 21, 2019 at 21:24
  • $\begingroup$ No, I am not setting initial data. By choosing a solution, I assume in each iteration that the homogeneous ode is solved by $J_{\alpha}$. $\endgroup$
    – Alan
    Commented Nov 21, 2019 at 21:33
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    $\begingroup$ That's fine, but how are you choosing what the inhomogenous term is? Any possible solution could be the "specific term" you mention; you need some way to decide which one you are taking. $\endgroup$
    – Sam Zbarsky
    Commented Nov 22, 2019 at 14:55
  • $\begingroup$ @SamZbarsky let's assume something like: $y(-\alpha)=0$ and $y(\alpha)=1$ for all the iterations. $\endgroup$
    – Alan
    Commented Apr 22, 2022 at 21:29

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Using Variation of Parameters, your inhomogeneous equation has a particular solution

$$ \frac{\pi}{2} \left( {{Y}_{\alpha}\left(x\right)} \int \!{\frac { {{J}_{\alpha}\left(x\right)} ^{2}}{ x}}\,{\rm d}x- J_\alpha(x) \int \!{\frac {{{ Y}_{\alpha}\left(x\right)}{{ J} _{\alpha}\left(x\right)}}{x}}\,{\rm d}x \right) $$

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  • $\begingroup$ How would you find the solution to the $n$-th iteration of my problem? $\endgroup$
    – Alan
    Commented Nov 22, 2019 at 5:50
  • $\begingroup$ If $F_n$ is a particular solution to the $n$'th iteration, you can take $$F_{n+1}(x) = \frac{\pi}{2} \left( Y_\alpha(x) \int \frac{J_\alpha(x) F_n(x)}{x}\; dx - J_\alpha(x) \int \frac{Y_\alpha(x) F_n(x)}{x}\; dx \right)$$ You may be able to use integration by parts to reduce iterated integrals to single integrals, but things are going to get ugly... $\endgroup$
    – Robert Israel
    Commented Nov 22, 2019 at 13:26
  • $\begingroup$ by all means let get ugly... :-D $\endgroup$
    – Alan
    Commented Apr 22, 2022 at 21:26

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