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 $A(x,0)=1$.

This PDE 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
$$
\big(1+ t\partial_t\big) \left(\partial^3_x+ {6\over x}\partial^2_x + {6\over x^2}\partial_x\right)A(x,t)+ t A(x,t)=0
$$ 
with $A(x,0)=1$, 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 helps obtain the exact solution to the original PDE.