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

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?

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.
• I see, I suppose I got caught up in the Gamma function expression of the coefficients for arbitrary $p$, and I didn't see that it reduces to the expression you wrote at the top of your answer. Thanks a lot for the help! – user3002473 Jun 9 '19 at 21:05