Can’t see that an ODE is equivalent to a Bessel equation I can solve the following differential equation without any trouble using the method of Frobenius
$$
x^2 y’’ - (2 + 3x) y = 0.
$$
When I put the differential equation in Mathematica, it gives me the solutions in terms of modified Bessel functions of order 3
$$
y(x) = A \sqrt{x} I_3\big(2\sqrt{3x}\big) + B \sqrt{x} K_3\big(2\sqrt{3x}\big).
$$
I cannot for the life of me see how to put the given ODE into the form of a modified Bessel equation. Can anyone point me in the right direction?
Some Added Information
Generally, the equation
$$
\frac{d}{dx}\left(x^a \frac{dy}{dx}\right) + b x^c y = 0,
$$
can be transformed to a Bessel equation with solution
$$
y(x) = x^{\nu/\alpha} Z_\nu \left(\alpha\sqrt{|b|} x^{1/\alpha}\right),
$$
where $Z_\nu$ is any Bessel function solution of the transformed equation, if we choose
$$
\alpha = \frac{2}{c-a+2} \quad \text{and} \quad \nu = \frac{1-a}{c-a+2},
$$
Considering the equations for $\alpha$ and $\nu$, I can see from Mathematica’s solution that $\nu = 3$ and $\alpha = 2$, so I can solve them to find $a = -2$ and $c = -3$. Putting those into the ODE above, we get
$$
\frac{d}{dx}\left(x^{-2} \frac{dy}{dx}\right) + b x^{-3} y = 0.
$$
Expanding this out, I get
$$
\frac{1}{x^2}y’’ - \frac{2}{x^3} y’ - \frac{b}{x^3} y = 0 \quad \text{or} \quad x^2 y’’ - 2x y’ - 3xy = 0.
$$
This is my ODE if the $y’$ were instead a $y$. :-( So, I am stuck.
 A: Your equation is indeed a transformed modified Bessel equation. To see that, you need to take the modified Bessel equation
$$
\xi^2 \frac{d^2\eta}{d\xi^2} + \xi \frac{d\eta}{d\xi} - (\xi^2+n^2)\eta=0
$$
and employ following transformation
$$
\eta=\frac{y}{x^\alpha}, \quad \xi=\beta x^\gamma,
$$
you should arrive at
$$
\frac{d^2y}{dx^2} - \frac{(2\alpha-1)}{x}\frac{dy}{dx} - (\beta^2 \gamma^2 x^{2\gamma-2}+\frac{n^2\gamma^2-\alpha^2}{x^2})y=0.
$$
To see the steps, I recommend to look into the book "Bowman: Introduction to Bessel functions, p.117", where it is done for standard Bessel equation.
Since $\eta$ satisfies the modified Bessel equation it follows that $\eta(\xi)=AK_n(\xi)+B I_n(\xi)$, $A,B$ being constants. Employing the transformations above, we have that the transformed Bessel equation has a general solution
$$
y(x)=x^\alpha[A K_n(\beta x^\gamma)+B I_n(\beta x^\gamma)].
$$
Your specific equation is the case $\alpha=\frac12,\gamma=\frac12,\beta=2\sqrt{3},n=3$.
Note, that for non-integer $n$ we usually take the solution in a fom
$$
y(x)=x^\alpha[A I_n(\beta x^\gamma)+B I_{-n}(\beta x^\gamma)].
$$
A: The following Wolfram code
y[x_] := Sqrt[x] K[2 Sqrt[3 x]];
u = (x^2 D[y[x], {x,2}] - (2 + 3 x) y[x]) 8 Sqrt[3]/t;
u/. x -> (t^2/12) // PowerExpand // Simplify

evaluates to
-(9 + t^2) K[t] + t (K'[t] + t K''[t])

which goes a long way to establishing the Bessel connection.
