I was wondering whether there is an analytical solution to the ODE \begin{equation} -n\int xy(x)dx + ihy'(x) + (x^2+k)y(x) = 0, \end{equation} where $n=0,1,2,...$, $h \in \mathbb{R}$, and $k=+1,0$ or $-1$.

For $n=0$ this can be solved exactly, and the solution is \begin{equation} y(x) = C \exp\left[ih\left(\frac{x^3}{3}+kx\right)\right] \equiv f(x). \end{equation}

However, I struggled to find a general solution for different $n$. Differentiating the ODE gives \begin{equation} ihy''(x)+(x^2+k)y'(x)+(2-n)xy(x)=0. \end{equation} I have tried the ansatz \begin{equation} y(x) = f(x)^{(n-2)/2} + f(x)^{(n-2)/2}\int f(x)^{1-n} dx, \end{equation} but it only satisfies the equation when $n=2$.

I was able to make some progress by separating the equation into its real and imaginary parts and integrate the coupled equations numerically, but I was hoping for an analytical solution.

The other possible solution is the triconfluent Heun function \begin{equation} y(x) = e^{-\frac{x^{3}+3 kx}{3 h}} C_{2} \text { HeunT }\left[0, \frac{-4+n}{h},-\frac{k}{h}, 0,-\frac{1}{h}, x\right] + C_{1} \text { HeunT }\left[0, \frac{-2+n}{h}, \frac{k}{h}, 0, \frac{1}{h}, x\right]. \end{equation}

My question is whether there is an analytical solution of a simpler form?