I am trying to see if the following ODE of $g(x)$ admits a series solution the coefficient of which does not have to be expressed recursively but has a direct and closed expression.
$$g'-\Big(\frac1{x^2}+\frac1x+\frac1{1-x}\Big)g=\Big(\frac x{1-x}\Big)^2-\frac{a_0}{1-x}+(2a_0-a_1)\frac x{1-x}$$ where $x\in(0,1)$, and $a_0$ and $a_1$ are constants.
I multiplied both sides by $e^h=\frac{1-x}xe^{\frac1x}$ where $h(x):=\frac1x+\ln\frac{1-x}x$ and get $$\frac{d}{dx}\Big(\frac{1-x}xe^{\frac1x}g\Big)=e^{\frac1x}\Big(\frac x{1-x}-\frac{a_0}x+(2a_0-a_1)\Big).$$ The difficulty comes from the factor $e^\frac1x$ which seems to call for special functions for the integration with respect to $x$. Is there a power series expression for $g(x)$ where the coefficients of it is some "closed" expression, special functions such as hypergeometric functions included?