On reducing complex ODE's to Bessel's form using Kummer's series I am trying to reduce the following ODE to Bessel's ODE form and solve it:
$$x^{2}y''(x)+x(4x^{3}-3)y'(x)+(4x^{8}-5x^{2}+3)y(x)=0\tag{1} \, .$$
I tried to solve it via the standard method, i.e., by comparing it with a generalised ODE form and finding the solution from then on. The general form (as given in Mary L. Boas- Mathematical Methods in Physical Sciences) is:

$$y''(x)+\frac{1-2a}{x}y'(x)+\left((bcx^{c-1})^{2}+\frac{a^{2}-p^{2}c^{2}}{x^{2}}\right)y(x)=0\tag{2} \, ,$$
  and the solution:$$y(x)=x^{a}Z_{p}(bx^{c})\tag{3} \, .$$

But I am unable to get the answer via this method. The solution which is as follows:
$$y(x)=x^{2}e^{-\frac{x^{4}}{2}}[AI_{1}(\sqrt{5}x)+BK_{1}(\sqrt{5}x)]\tag{4}$$
Is obtained using comparison with another standard form which is given as follows:

$$x^{2}y''(x)+x(a+2bx^{p})y'(x)+[c+dx^{2q}+b(a+p-1)x^{p}+b^{2}x^{2p})y(x)=0\tag{5} \, ,$$
  and the solution as:
  $$y(x)=x^{\alpha}e^{-\beta x^{p}}[AJ_{\nu}(\lambda x^{q})+BY_{\nu}(\lambda x^{q})]\tag{6} \, .$$

Where: $\alpha=\frac{1-a}{2}$, $\beta=\frac{b}{p}$, $\lambda=\frac{\sqrt{d}}{q}$, $\nu=\frac{\sqrt{(1-a)^{2}-4c)}}{2q}$
If I divide through the ode by $x^{2}$, I would get the Fuchasian form:
$$y''(x)+f(x)y'(x)+g(x)y(x)=0$$
The terms of $xf(x)$ and $x^{2}g(x)$ are expandable in convergent power series $\sum_{n=0}^{\infty}a_{n}x^{n}$, hence there exists a nonessential singularity at the origin.
But I am unable to solve via the Frobenius method. 
Hence, my question- How is the generalised form of equation $(5)$ arrived at and why can't I use $(2)$ instead?  Rather than bringing this ODE to a non-standard form as given in equation $(5)$, is there a way to derive the equation itself (and deduce the general solution)? Any help is appreciated.
Edit:
I found the following form in a book:

$$x^{2}y''(x)+x(a+2bx^{p})y'(x)+[c+dx^{2q}+fx^{q}+b(a+p-1)x^{p}+b^{2}x^{2p})y(x)=0\tag{7} \, .$$

The only difference between the above and equation $(6)$ is the extra term:$fx^{q}$
Now if I substitute $y=we^{-\frac{bx^{p}}{p}}$ in equation $(7)$, it simplifies to the following linear equation:

$$x^{2}w''(x)+axw'(x)+(dx^{2q}+fx^{q}+c)w(x)=0\tag{8}\,.$$

Now using the transformation $z=x^{q}$, and $y=wz^{k}$, where $k$ is the root of the following quadratic equation: $q^{2}k^{2}+q(a-1)k+c=0$; leads to a further simplified and linear form:

$$q^{2}zy''(z)+[qbz+2kq^{2}+q(q-1+a)]y'(z)+(dz+kqb+f)y(z)=0\tag{9}\,.$$
  This equation has the solution: $y(x)=e^{kx}w(z)$, where $w(z)$ is the solution to the hypergeometric equation as given below

Now, let a function $\Omega(b,a;x)$ be an arbitrary solution to the degenerate hypergeometric equation: 

$$xy''(x)+(a-x)y'(x)-by(x)=0\tag{10}\,.$$

And $Z_{\nu}(x)$ be an arbitrary solution of the Bessel equation.
Now in equation $(10)$, if $b\neq0,-1,-2,-3,...$, the solution is given by the Kummer's series as:
$$\Phi(b,a;x)=1+\sum_{k=1}^{\infty}\frac{(b)_{k}x^{k}}{(a)_{k}k!}$$
Where:$(b)_{k}=b(b+1)...(b+k-1)$
When $a$ is not an integer, the solution can be written as:
$$y=C_{1}\Phi(b,a;x)+C_{2}x^{1-a}\Phi(b-a+1,2-a;x)$$
Make the following replacements:
$b=2n$ and $a=n$
Now the series becomes:

$$\Phi(n,2n;x)=\Gamma\left(n+\frac{1}{2}\right)e^{\frac{x}{2}}\left(\frac{x}{4}\right)^{(-n+\frac{1}{2})}I_{n-\frac{1}{2}}(\frac{x}{2})$$
  And
  $$\Phi(-n,-2n;x)=\frac{1}{\sqrt{\pi}}e^{\frac{x}{2}}\left(x\right)^{(-n+\frac{1}{2})}K_{n-\frac{1}{2}}(x)$$

Substituting the above in the solution of equation $(9)$, the general solution becomes:
$$y=e^{x(k+\frac{1}{2})}\left[C_{1}\Gamma\left(n+\frac{1}{2}\right)\left(\frac{x}{4}\right)^{(-n+\frac{1}{2})}I_{n+\frac{1}{2}}(x)+C_{2}\frac{1}{\sqrt{\pi}}\left(x\right)^{(-n+\frac{1}{2})}K_{n+\frac{1}{2}}(x)\right]$$
Which simplifies to:

$$y(x)=\left(x\right)^{(-n+\frac{1}{2})}e^{x(k+\frac{1}{2})}\left[C_{1}\Gamma\left(n+\frac{1}{2}\right)\left(\frac{1}{4}\right)^{(-n+\frac{1}{2})}I_{n+\frac{1}{2}}(x)+C_{2}\frac{1}{\sqrt{\pi}}K_{n+\frac{1}{2}}(x)\right]$$

Which is the final solution. I tried to do the same for equation $(6)$, but did not get the solution. Any help is appreciated. 
 A: Working in Mathematica, I find that (3) is the general solution to the differential equation $$x^{2}y''(x)+x(4x^{4}-3)y'(x)+(4x^{8}-5x^{2}+3)y(x)=0.$$ 
This almost entirely matches (1), but differs in that the linear term contains $x^4$ rather than $x^3$. So you should check the source of the problem to see if (1) was transcribed improperly: If it was, then that fixes things; if the book indeed had $x^3$, though, then it's a typo and the prof should be informed.
To determine the ODE which the general solution satisfies, I defined
y1[x_] := x^2 Exp[-x^4/2] BesselI[1, x Sqrt[5]]
as one of the two linearly-independent  solutions to the ODE. Assuming that the relevant desired ODE is of the form $$x^2y''(x)+x A(x) y'(x)+B(x)y(x)=0$$ for some appropriate $A,B$, I plugged the above solution into the left-hand side and simplify:
x^2 y1''[x] + x A[x] y1'[x] + B[x] y1[x] // FullSimplify

==> E^(-(x^4/2)) x^2 (Sqrt[5]x (3 - 4 x^4 + A[x]) BesselI[0,Sqrt[5] x] 
      + (5 x^2 - 10 x^4 + 4 x^8 + A[x] - 2 x^4 A[x] + B[x]) BesselI[1, Sqrt[5] x])`

From this we can read off that both terms will vanish identically when 
\begin{align}
A(x)&=4x^4-3\\\\
B(x)&=(2 x^4 - 1) A(x) + 10 x^4 - 4 x^8 - 5 x^2\\&=4x^8 - 5 x^2 + 3
\end{align}
which is the result cited above. This can be further checked by asking Mathematica to solve the 
ODE directly:
 DSolve[x^2 y''[x]+x (4x^4-3)y'[x]+(3-5 x^2+4 x^8) y[x]==0,y[x],x]

 ==> {{y[x]->E^(-(x^4/2)) x^2 (-I BesselI[1,Sqrt[5] x] C[1]
               +BesselY[1,-I Sqrt[5] x] C[2])}}

Trading Bessel functions $J,Y$ for modified Bessel functions $I,K$ then gives the indicated general solution up to integration constants.
