I came across this when trying to solve the following recurrence relation via generating function.
$$(4n+3)a_n+2na_{n-1}+(2n+3)a_{n+1}=0$$
If we define $A(x)=\sum_{n=0}^{\infty}a_nx^n$, we see that it satisfies the ODE
$$A'+\frac{2x+1}{2x(x+1)}A=\frac{a_0}{2x(x+1)^2}$$
This equation can be solved to find two generating functions corresponding to the two independent solutions to the recurrence relation,
$$A(x)=\frac{1}{\sqrt{x(1+x)}},\quad A(x)=\frac{1}{1+x}.$$
The second generating function listed tells us that $a_n=(-1)^n$ is a solution. The first generating function has a branch cut at $x=0$, it doesn't seem like there's a way to find the coefficients $a_n$ expanded near $x=0$ for this function.
Luckily, it is still possible to find the second solution: $a_n=(-1)^n\frac{(2n)!!}{(2n+1)!!}$. My question is, how are these coefficients related to the function $\frac{1}{\sqrt{x(1+x)}}$? In fact, the generating function for this solution is
$$A(x)=\sum_{n=0}^{\infty}(-1)^n\frac{(2n)!!}{(2n+1)!!}x^n=\frac{\text{arcsinh}(\sqrt{x})}{\sqrt{x(1+x)}}$$
The factor $\text{arcsinh}(\sqrt{x})$ appears, and gets rid of the branch point. I am very new to the field of generating functions, this is all very confusing!
Note: I would like to understand this so I may get insight into the related recurrence relation (apologies for the alliteration)
$$(4n+3)a_n+2na_{n-1}+(2n+3)a_{n+1}=\lambda a_n$$
Where the problematic generating function is now
$$A(x)=\frac{1}{\sqrt{x(1+x)}}e^{-\frac{\lambda}{2(1+x)}}$$
Here I am not so lucky as to already have the solution.
Best Answer
It appears that the form of the generating function you derived is incomplete. In fact, careful calculation shows that it satisfies a first order ODE similar to yours with an extra term
$$A'(x)+\frac{2x+1}{2x(x+1)}A(x)=\frac{a_0}{2x(x+1)^2}+3\frac{a_1+a_0}{2(x+1)^2}$$
Multiply by the integrating factor $\sqrt{x(1+x)}$ and rewrite to obtain
$$\sqrt{x(1+x)}A(x)=C+a_0\int dx\frac{1}{2\sqrt{x}(x+1)^{3/2}}+3(a_0+a_1)\int dx\frac{\sqrt{x}}{2(x+1)^{3/2}}$$
which evaluate to the final result
$$A(x)=\frac{C}{\sqrt{x(1+x)}}+a_0\frac{1}{1+x}+3(a_1+a_0)\left(\frac{\text{arcsinh}(\sqrt{x})}{\sqrt{x(1+x})}-\frac{1}{1+x}\right)$$
Evidently, for the choice $a_1=-\frac{2}{3}a_0$ one obtains the solution desired, so there's no contradiction.