Obtaining a linear recurrence from differential equation

Question

I need some guidance with the following problem.

I have a sequence $L_0,L_1,\ldots$ whose ordinary generating series satisfies $$L(x) = \sum_{n=0}^{\infty} L_n \frac{x^n}{n!} = \frac{1}{2-e^x}.$$

Now the exercise I am doing asks to find a linear differential equation that is satisfied by $L(x)$ and use it to obtain a recurrence for $L_n.$

I am very rusty and don't even know how to proceed. I can see that, for example $$x L'(x) - L(x) = \sum_{n=0}^{\infty} (n-1)L_n \frac{x^n}{n!}$$ but I have no idea how to obtain anything useful from this direction.

So my question is :

how does one obtain a linear recurrence satisfied by $L_n$ this case?

Answer 1

If you differentiate the equation $$ (2-e^x)L(x) = 1 $$ you get the differential equation $$ -e^x L(x) +(2-e^x) L'(x) = 0. $$ Now if $$ M(x) = \sum_{n\ge0} M_n \frac{x^n}{n!} $$ then the coefficient of $x^n/n!$ in $M(x)$ is $\sum_{k=0}^n \binom{n}{k} M_k$. Using this to compute the coefficient of $x^n/n!$ in the left side of the differential equation will lead to a recurrence for $L_n$.

Answer 2

I do not know how much this could help you.

Computing the first $L_n$ terms gives [$1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261$] which correspond to sequence $A000670$ of OEIS which, according to OEIS, correspond to the general relation

$$L_n=\sum _{k=1}^n k! \text { }\mathcal{S}_n^{(k)}$$ where appear the Stirling number of the second kind.

OEIS also refers to the differential equation $y'=-y+2y^2$ given by rlartiga. The solution of this differential equation is $$y=\frac{1}{2+c e^{x}}$$ I hope and wish that this answer gives you some ideas for continuing your work.