Consider the linear differential equation defined by $$ L^2y=0,\quad \text{where}\ L:=x^2\frac{d}{dx^2}+x\frac{d}{dx}-(x^2+1)\text{id}. \tag{1}$$ Here $\text{id}$ denotes the identity operator, and $L^2$ denotes the operator $L$ applied twice. Written out explicitly, (1) is of the form $$x^4\frac{d^4 y}{dx^4}+6 x^3\frac{d^3y}{dx^3}+(5x^2-2x^4)\frac{d^2y}{dx^2}-(x+6x^3)\frac{dy}{dx}+(x^2-1)^2y=0. \tag{2}$$ I am interested in the general solution to (1) which is bounded at $x=0$.
Since the solution that we are looking for is bounded at $x=0$, we may write it as $$y(x)=\sum_{n=0}^\infty a_n x^n,$$ near the origin. Substituting the above solution into (2) and equating the coefficient of $x^n$ to be zero, we obtain the following recurrence relation $$ a_2=\frac{2}{9}a_0,\quad a_3=\frac{1}{8}a_1,\quad a_4=\frac{1}{75}a_0,\quad a_5=\frac{1}{192}{a_1},\quad a_6=\frac{4}{11025}a_0,\quad\dots.$$ Setting $(a_0,a_1)=(1,0)$ and $(a_0,a_1)=(0,1)$, we obtain two linearly independent solutions whose linear combination gives the general solution to (1) bounded at $x=0$.
The solution to with $(a_0,a_1)=(0,1)$ can be written as $$ y_1(x)=x+\frac{1}{8}x^3+\frac{1}{192}x^5+\cdots=2I_1(x),$$ where $I_1(x)$ is the modified Bessel function of the first kind, while the solution with $(a_0,a_1)=(1,0)$ can be represented as $$ y_2(x)=1+\frac{2}{9}x^2+\frac{1}{75}x^4+\frac{4}{11025}x^6+\cdots$$ My question is that: can $y_2(x)$ be represented in terms of any special function (like $y_1(x))$?
Update: As pointed out in the comments, the $x^0$ coefficient relation gives $a_0=0$, so the solution $y_2(x)$ doest not exist.
Per a c.a.s. the general solution is \begin{split}y(x) = A I_1(x) + B K_1(x) + C \left[-I_1(x) \int_{x_0}^x \frac{K_1(t)^2 \,dt}{t} + K_1(x) \int_{x_0}^x \frac{I_1(x) K_1(x)}{x}\,dt \right] \qquad \qquad \\+ D \left[\frac{1}{2} K_1(x) \left(I_0(x)^2 - I_1(x)^2\right) - I_1(x) \int_{x_0}^x \frac{I_1(x) K_1(x)}{x}\,dt\right] ,\end{split} where $I_\alpha, K_\alpha$ are the modified Bessel functions of the First and Second Kinds, respectively.
Expanding in a series shows that as $x \searrow 0$, $$y(x) = \frac{1}{2} C \frac{\log x}{x} + o\left(\frac{\log x}{x}\right),$$ so for any solution bounded near $x = 0$ we must have $C = 0$. Expanding $y(x) \vert_{C = 0}$ around $x = 0$ now gives $$y(x) = \left(B + \frac{D}{2}\right) \frac{1}{x} + o\left(\frac{1}{x}\right),$$ so boundedness near $0$ implies that $D = -2 B ,$ and expanding $y(x) \vert_{C = 0, D = -2B}$ gives a $2$-parameter family of solutions bounded near $0$; n.b. all of these functions are odd. Now the series expansion as $x \searrow 0$ is $$\frac{B}{2} x \log x + \left(\frac{A}{2} - \frac{B}{4}\right) x + o(x) ,$$ so the solutions are bounded exactly for such choices of coefficient. Rewriting per the below comment of Maxim gives a simple expression for the general bounded solution: $$y(x) = a I_1(x) + b \left(\frac{I_0(x)}{x} - K_1(x)\right) .$$