Consider $$ \big(1+ t\partial_t\big) \left(\partial^3_x+ {6\over x}\partial^2_x + {6\over x^2}\partial_x\right)A(x,t)+ {t\over (1-x)^3} A(x,t)=0 $$ with the initial condition $A(x,0)=1$. In a small $t$ expansion

$A(x,t)= 1+ a_1(x)t + a_2(x) t^2 + a_3(x) t^3+...$

Additional conditions are $a_i(0)=0$, where $i=1,2,3,..$.

The equation is invariant under the transformations $x \to {x\over x-1},~ t\to -t$, which fix the ratio $(1,6,6)$ in the $x$-derivative part.

This PDE looks like a Fuchsian-type equation, but normally Fuchsian equations assume only one variable – Does anyone know this type of "generalised Fuchsian-type equation" with two variables? I wonder if this was studied in the math literature.

I am interested in the PDE's analytic solutions, especially solutions that are non-perturbative in $t$, and symmetry properties.

Note added:

For a slightly simpler PDE $$ \big(1+ t\partial_t\big) \left(\partial^3_x+ {6\over x}\partial^2_x + {6\over x^2}\partial_x\right)A(x,t)+ t A(x,t)=0 $$ I found that the exact solution is given by a hypergeometric ${_0F_3}$ function: $$ A(x,t)=\;{_0F_3}\left(~;{{4\over 3}, {5\over 3},2}; -{x^3 t\over 27}\right). $$ This I believe corresponds to the small $x$ limit of the exact solution to the first PDE with the factor ${1\over (1-x)^3}$. I adopted an indirect computation to arrive at the above solution so if anyone can re-derive this solution in a few lines I'd be happy to know!

(If one knows the solution only depends on $x^3 t \equiv w$ one can just set $A=A(w)$ then the PDE becomes an OPE which can be solved by Mathematica directly. But a prior it is not obvious why $x^3 t$ is special.)

In any case, I wonder if there are some methods which help obtain the exact solution to the original PDE.