I am trying to reduce the following ODE to Bessel's ODE form and hence 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:


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:


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:


Which is the final solution. I tried to do the same for equation $(6)$, but did not get the solution. Is it plausible to obtain the solution via the Kummer's series method? Any help is appreciated.


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