Vtgyjfy

I have a question regarding an alternate form of the Bessel equation and how that alternate form translates to the modified Bessel equation and its solution. The modified form is from:
http://mathworld.wolfram.com/BesselDifferentialEquation.html

and looks like this:
$$ frac{d^2 y}{dx^2}+frac{1-2alpha}{x}frac{dy}{dx}+left(beta^2gamma^2x^{2gamma-2}+frac{alpha^2-n^2gamma^2}{x^2}right)y=0 $$
and has the following solutions:
$$
y=
begin{cases}
x^alphaleft[AJ_n(beta x^gamma)+BY_n(beta x^gamma)right] &text{ for integer }n \
\
x^alphaleft[AJ_n(beta x^gamma)+BJ_{-n}(beta x^gamma)right] &text{ for noninteger }n
end{cases}
$$
For the moment, I am only concerned with the case where $gamma=1$ and $alpha=n$ so the equation simplifies to:
$$ frac{d^2 y}{dx^2}+frac{1-2alpha}{x}frac{dy}{dx}+beta^2y=0 $$
with the following solutions:
$$
y=
begin{cases}
x^alphaleft[AJ_alpha(beta x)+BY_alpha(beta x)right] &text{ for integer }alpha \
\
x^alphaleft[AJ_alpha(beta x)+BJ_{-alpha}(beta x)right] &text{ for noninteger }alpha
end{cases}
$$
What I'd like to know is if instead of being the unmodified Bessel equation, the equation was in the form of the modified Bessel equation as shown below, what would the solutions be?
$$ frac{d^2 y}{dx^2}+frac{1-2alpha}{x}frac{dy}{dx}-beta^2y=0 $$
A quick search on wolframalpha tells me they might look like this:
$$
y=
begin{cases}
x^alphaleft[AJ_alpha(-ibeta x)+BY_alpha(-ibeta x)right] &text{ for integer }alpha \
\
x^alphaleft[AJ_alpha(-ibeta x)+BJ_{-alpha}(-ibeta x)right] &text{ for noninteger }alpha
end{cases}
$$
Could these then be translated to the modified Bessel functions so:
$$
y=
begin{cases}
x^alphaleft[CI_alpha(beta x)+DK_alpha(beta x)right] &text{ for integer }alpha \
\
x^alphaleft[CI_alpha(beta x)+DI_{-alpha}(beta x)right] &text{ for noninteger }alpha
end{cases}
$$

Any insights you could provide would be greatly appreciated, thanks!

Yes. If we start from
begin{equation}
frac{d^2y}{dx^2} + frac{1-2alpha}{x} frac{dy}{dx} - beta^2 y = 0,
end{equation}
we write $beta^{alpha} y = xi^{alpha} z(xi)$ with a new independent variable $xi = beta x$ and with a new dependent variable $z$. Computing the first two derivatives of $y$ by the product and chain rules yields
begin{eqnarray}
beta^{alpha} frac{dy}{dx} &=& alpha xi^{alpha-1} beta z + xi^{alpha} beta frac{dz}{dxi},\
beta^{alpha} frac{d^2y}{dx^2} &=& alpha(alpha-1) xi^{alpha-2} beta^2 z + 2 alpha xi^{alpha-1} beta^2 frac{dz}{dxi} + xi^{alpha} beta^2 frac{d^2z}{dxi^2}.
end{eqnarray}
We now obtain
begin{eqnarray}
0 &=& beta^{alpha} left( frac{d^2y}{dx^2} + frac{1-2alpha}{x} frac{dy}{dx} - beta^2 y right) = beta^{alpha} frac{d^2y}{dx^2} + frac{1-2alpha}{xi} beta beta^{alpha} frac{dy}{dx} - beta^2 beta^{alpha} y\
&=& dots = xi^{alpha-2} beta^2 left( xi^2 frac{d^2z}{dxi^2} + xi frac{dz}{dxi} - (xi^2 + alpha^2) z right).
end{eqnarray}
So now we have the modified Bessel's equation $xi^2 frac{d^2z}{dxi^2} + xi frac{dz}{dxi} - (xi^2 + alpha^2) z = 0$ for $z$, and a fundamental system of solutions is given by ${I_{alpha}(xi),K_{alpha}(xi)}$ for any value of $alpha$, or by ${I_{alpha}(xi),I_{-alpha}(xi)}$, if $alpha not in mathbb{Z}$.

Transforming back we obtain $y(x) = beta^{-alpha}(beta x)^{alpha} z(beta x) = x^{alpha} z(beta x)$.

The only thing that I don't like about the Wolfram notation is that it might be misinterpreted in the way that ${I_{alpha}(xi),K_{alpha}(xi)}$ is a fundamental system of solutions only if $alpha in mathbb{Z}$, which is not true.