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The ODE $xy'' + y = 0$ has a real degeneracy. Use The Method Of Frobenius to find a fundamental set of solutions.

Here is the procedure, as I understand it:

1) Plug the guess $y = x^s \sum_{n = 0}^\infty a_n x^n$ into the ODE and do the algebra/calculus to separate out the indicial equation and the recurrence relation.

$xy'' + y = 0$
$x(x^s \sum_{n = 0}^\infty a_n x^n)'' + (x^s \sum_{n = 0}^\infty a_n x^n) = 0$
$\sum_{n = 0}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 0}^\infty a_n x^{n + s} = 0$
$\sum_{n = 0}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 1}^\infty a_{n - 1} x^{n + s - 1} = 0$
$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 1}^\infty a_{n - 1} x^{n + s - 1} = 0$
$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + a_{n - 1} x^{n + s - 1} = 0$
$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty [(n + s)(n + s - 1) a_n + a_{n - 1}] x^{n + s - 1} = 0$

Indicial Equation: $s(s - 1) = 0$
Recurrence Relation: $(n + s)(n + s - 1) a_n + a_{n - 1} = 0$, where $n \geq 1$

2) Solve the recurrence relation, treating $s$ as a constant, but without plugging in any indicial equation solutions.

$(n + s)(n + s - 1) a_n + a_{n - 1} = 0$
$a_n = -\frac{1}{(n + s)(n + s - 1)}a_{n - 1}$
$a_n = a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$

Here I would simplify using a formula I was given, $\Pi_{j = c}^d (j + k) = \frac{(d + k)!}{(c + k - 1)!}$.

$a_n = a_0 [\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}] [\frac{1}{\Pi_{m = 1}^n (m + s - 1)}]$
$a_n = a_0 [(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}]$
$a_n = \frac{(-1)^n s! (s - 1)!}{(n + s)! (n + s - 1)!} a_0$

3) To find one fundamental solution, plug the largest indicial equation solution $s_1$ and its associated recurrence relation solution into the guess, and make the particular choice $a_0 = 1$.

$s(s - 1) = 0$
$s = 0, 1$
$s_1 = 1$
$y = x^1 \sum_{n = 0}^\infty \frac{(-1)^n 1! (1 - 1)!}{(n + 1)! (n + 1 - 1)!} 1 x^n$
$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1}}{n! (n + 1)!}\leftarrow$ FIRST FUNDAMENTAL SOLUTION

4) To find another fundamental solution, plug the other indicial equation solution $s_2$ and its associated recurrence relation solution into $y = [$FIRST FUNDAMENTAL SOLUTION$]ln|x| +\ x^{s_2} \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - s_2)a_n)]_{s = s_2} x^n$, and make the particular choice $a_0 = 1$.

$s_2 = 0$

Here I use the earlier, less simplified expression for $a_n$ which is in $\Pi$ product notation, in anticipation of taking the logarithmic partial derivative.

$y = (\sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1}}{n! (n + 1)!})ln|x| + x^0 \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - 0) a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})]_{s = 0} x^n$
$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty [\frac{\partial}{\partial s} (s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})]_{s = 0} x^n$

I make the substitution $b_n = s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$, appearing in the solution as $y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!}$ $+ \sum_{n = 0}^\infty [\frac{\partial}{\partial s} b_n]_{s = 0} x^n$.

$b_n = s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$
$ln|b_n| = ln|s| + ln|a_0| + \sum_{m = 1}^n ln|\frac{1}{m + s}| + ln|\frac{1}{1 - m - s}|$
$\frac{\partial}{\partial s} ln|b_n| = \frac{\partial}{\partial s} ln|s| + \frac{\partial}{\partial s} ln|a_0| + \sum_{m = 1}^n \frac{\partial}{\partial s} ln|\frac{1}{m + s}| + \frac{\partial}{\partial s} ln|\frac{1}{1 - m - s}|$

I set $\frac{\partial}{\partial s} ln|a_0|$ to $0$ because the chapter stipulates that this procedure only seeks solutions where $a_0$ is nonzero and does not depend on $s$.

$\frac{\frac{\partial b_n}{\partial s}}{b_n} = \frac{1}{s} + 0 + \sum_{m = 1}^n -\frac{1}{m + s} + \frac{1}{1 - m - s}$
$\frac{\partial b_n}{\partial s} = \frac{b_n}{s} + b_n \sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)}$
$\frac{\partial b_n}{\partial s} = \frac{s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}}{s} + s a_0 (\Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

Now I can simplify the $\Pi$ products using the above formula. I also make the choice for $a_0$ here.

$\frac{\partial b_n}{\partial s} = \frac{s [\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}] [\frac{1}{\Pi_{m = 1}^n (m + s - 1)}]}{s} + s ([\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}][\frac{1}{\Pi_{m = 1}^n (m + s - 1)}])(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$
$\frac{\partial b_n}{\partial s} = \frac{s [(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}]}{s} + s ([(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}])(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$
$\frac{\partial b_n}{\partial s} = \frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}$

Unwinding the substitution, I plug the expression for $\frac{\partial b_n}{\partial s}$ back into the fundamental solution.

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty [\frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}]_{s = 0} x^n$
$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty \frac{(-1)^n (-1)! x^n + (-1)^{n + 1} x^n \sum_{m = 1}^n \frac{2m - 1}{m^2 - m}}{n! (n - 1)!}$

And now we see that something is broken. $(-1)!$ is undefined, and can't even be rescued by being generalized to the Gamma or Pi function. What am I doing wrong, and how can I use this method to solve this ODE?

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The first mistake is with the book/material you are using, as $$y_2 = y_1\ln|x| + u$$ isn't a valid solution of the differential equation if $u = x^{s_2} \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - s_2)a_n)]_{s = s_2} x^n$ and $a_0 = 1$. Instead, you either want $a_0 = -1$ or $y_2 = y_1\ln|x| - u$. Not a rigorous proof, but you can see evidence to support this by looking at the constant term of $xy_2'' + y_2$, which must be $0$.

The main problem is the hole at $s = 0$. You ended up with $$\frac{\partial b_n}{\partial s} = \frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!} \tag{*}$$ and directly plugged in $s = 0$. But the function is undefined there because of the $(-1)!$ term. Instead, take the limit as $s \to 0$. The limit is $1$ for $n = 0$ and $$\frac{(-1)^{n+1}(1+2nH_{n-1})}{(n!)^2}$$ for $n \ge 1$, where $H_n$ is the harmonic number (found using Mathematica, proof at bottom). This means that the second solution is $$y_2 = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} \ln|x|}{n! (n + 1)!} - \left(1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(1+2nH_{n-1})}{(n!)^2} x^n\right)$$

Edit: Coming back to $(*)$: The problem is to evaluate $$\lim_{s \to 0}\frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}$$ For $n = 0$, it is clearly $0$ since the sum is $0$ and then the numerator is the same as the denominator. For other $n$, we can factor it as $$\lim_{s \to 0}\frac{(-1)^ns!}{(n+s)!(n+s-1)!}\left( (s - 1)! - s! \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}\right)$$ The factors converge separately, so moving the first term to the outside and splitting the sum makes it $$\frac{(-1)^n}{n!(n-1)!}\lim_{s \to 0}\left( (s - 1)! -s! \frac{2s +1}{(1 + s)s}- s! \sum_{m = 2}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}\right)$$ which is equal to $$-\frac{(-1)^n}{n!(n-1)!}\lim_{s \to 0} \left(\frac{s!}{s+1}\right)-\frac{(-1)^n}{n!(n-1)!}\left( \sum_{m = 2}^n \frac{2m - 1}{m(m - 1)}\right)$$ which simplifies to $$\frac{(-1)^{n+1}}{(n!)^2}\left(n+n\sum_{m = 2}^n \left( \frac{1}{m}+\frac{1}{m-1} \right)\right) = \frac{(-1)^{n+1}}{(n!)^2}\left(1+2nH_{n-1}\right)$$

Edit 2: By the way, the solutions (found using Mathematica) are $$y_1 = \sqrt{x}\ J_1(2\sqrt{x}), y_2 = \sqrt{x}\ Y_1(2\sqrt{x})$$ where $J_1(z)$ is the Bessel function of the first kind, and $Y_1(z)$ is the Bessel function of the second kind.

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  • $\begingroup$ Hmmm, I was under the assumption that I was allowed to replace $a_0$ with any constant. How is $a_0$ solved for when solving other similar equations? Is it just always $1$ or $-1$ (depending on the $y_2$ formula you use), or does it need to be computed? $\endgroup$
    – user10478
    Feb 25 '21 at 16:26
  • $\begingroup$ It couldn't be any constant - that would imply that any $y_2$ of the form $y_1\ln|x|+c\cdot u$ would work, which is impossible since the two linearly independent solutions are already given by $y_1$ and $y_1\ln|x|-u$. What does work is $a_0\left(y_1\ln|x| - u\right)$. $\endgroup$ Feb 25 '21 at 17:34
  • $\begingroup$ I forgot to say in my earlier comment, but the first coefficient of the series expansion of both $y_1$ and $u$ should be $1$ (or at least the same constant). This is what makes them "cancel" out when plugging into the ODE. $\endgroup$ Feb 25 '21 at 17:44
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Just a pictorial comment. It's always nice to see what the outcome of a differential equation looks like.

enter image description here

And, by the way, it does not seem much of an effort to solve it numerically, for example as in the Pascal snippet below. The solution looks like a sine with decreasing frequency, as has to be expected. The MUCH parameter has been given two different values for checking convergence.

procedure test(much : integer);
var
  y : array of double;
  x,dx : double;
  k : integer;
begin
  SetLength(y,much+1);
  dx := (xmax-xmin)/much;
  y[0] := 0;
  y[1] := y[0] + dx;
  for k := 1 to much-1 do
  begin
    x := k*dx;
    y[k+1] := 2*y[k]-y[k-1]-sqr(dx)/x*y[k];
  end;
end;
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