How do you arrive at this solution to the modified Bessel equation using the Frobenius method?

Question

I'm looking to solve the equation $$ x^2y'' + xy' - (x^2+p^2)y = 0 $$ in particular for $p = \frac12$. I've already determined the indicial equation to be $r^2 - p^2 = 0$, and so the roots are $r = \pm p$. I've also already determined a series solution (which seems to be a constant factor off, namely $a_0$, from the modified Bessel function $I_p$): $$ y_1 = a_0x^r\sum_{n=0}^\infty\left(\prod_{i=1}^n i(i+p)\right)^{-1}\left(\frac x2\right)^{2n} $$ Now, for $p = \frac12$, I noticed that $x^{-1/2}e^x$ was a solution (by brute-force). However, I can't for the life of me figure out why plugging $p = \frac12$ into the above series equation should give me something that looks remotely like $x^{-1/2}e^x$.

To go through my steps, I assumed $y = \sum_{n\ge 0}a_nx^{n+r}$, and substituted this into the DE to get $$ (r^2 - p^2)a_0x^r + ((r+1)^2 - p^2)a_1x^{r+1} + \sum_{n=0}^\infty ([(n+r)^2 - p^2]a_n - a_{n-2})x^{n+r} = 0 $$ Clearly $r = \pm p$ in order for $a_0\neq 0$. However, it's the $x^{r+1}$ coefficient which is bugging me. Namely, for $p = \frac12$, taking the root $r = -p = -\frac12$ gives $(r + 1)^2 - p^2 = 0$, which means $a_1\neq 0$ can be anything, meaning the odd powers of $x$ also have coefficients in the series solution. This doesn't effect the solution $y = x^{1/2}\sum(\cdots)$ since that still requires $a_1 = 0$, however the solution $y = x^{-1/2}\sum(\cdots)$ changes, which is why I think $x^{-1/2}e^x$ might still arise from the Frobenius solution.

That said, I simply can't figure out how to get to $x^{-1/2}e^x$. Even if you account for the coefficients of the odd powers the series doesn't look like $e^x$ at all as far as I can tell. Are you simply supposed to arrive at $x^{-1/2}e^x$ from another method?

Answer 1

You are correct that the constant $a_1$ becomes arbitrary when $r=-1/2$. The recurrence relation in this case is $$ a_n = \frac{a_{n-2}}{(n-1/2)^2 - 1/4} = \frac{a_{n-2}}{n(n-1)}. $$ You can check that the odd coefficients $a_{2n+1}$ and the even coefficients $a_{2n}$ satisfy, for any $n\ge 1$, \begin{align*} a_{2n} & = \frac{a_0}{(2n)!} \\ a_{2n+1} & = \frac{a_1}{(2n+1)!}. \end{align*} Thus \begin{align*} y(x) & = x^{-1/2}\sum_{n=0}^\infty a_nx^n \\ & = x^{-1/2}\left(\sum_{n=0}^\infty a_0\frac{x^{2n}}{(2n)!} + \sum_{n=0}^\infty a_1\frac{x^{2n+1}}{(2n+1)!}\right) \\ & = x^{-1/2}\Big[a_0\cosh(x) + a_1\sinh(x)\Big] \\ & = x^{-1/2}\left[\left(\frac{a_0 + a_1}{2}\right)e^x + \left(\frac{a_0 - a_1}{2}\right)e^{-x} \right] \\ & = C_1x^{-1/2}e^x + C_2x^{-1/2}e^{-x}, \end{align*} with $C_1, C_2$ arbitrary.