Derivative verification I have a function, $a(x)=\sum_{n=0}^\infty\frac{A_n}{n!}x^n=\frac{x^3/3!}{e^x-1-x-(x^2/2)}$. Taking the derivative directly from the exponential fraction function I was was able to derive the identity,
$$3a(x)^2=(3-x)a(x)-xa'(x)$$
Now, taking the second derivative:
$$\frac{d}{dx}[3a(x)^2]=\frac{d}{dx}[(3-x)a(x)-xa'(x)]$$
$$6a(x)a'(x)=(3-x)a'(x)-a(x)-xa''(x)-a'(x)$$
$$6a(x)a'(x)=-a(x)+(2-x)a'(x)-xa''(x)$$
Multiplying both sides by $x$,
$$6a(x)[xa'(x)]=-xa(x)+x(2-x)a'(x)-x^2a''(x)$$
By rearranging the original first derivative identity and substituting for $xa'(x)$ on the LHS,
$$6a(x)[(3-x)a(x)-3a(x)^2]=-xa(x)+x(2-x)a'(x)-x^2a''(x)$$
$$2(3-x)\cdot3a(x)^2-18a(x)^3=-xa(x)+x(2-x)a'(x)-x^2a''(x)$$
$$-2(3-x)\cdot3a(x)^2+18a(x)^3=xa(x)-x(2-x)a'(x)+x^2a''(x)$$
Substituting in from the original first derivative identity for $3a(x)^2$,
$$-2(3-x)\cdot[(3-x)a(x)-xa'(x)]+18a(x)^3=xa(x)-x(2-x)a'(x)+x^2a''(x)$$
$$18a(x)^3-2(3-x)^2a(x)+2x(3-x)a'(x)=xa(x)-x(2-x)a'(x)+x^2a''(x)$$
Isolating $18a(x)^3$,
$$18a(x)^3=[x+2(3-x)^2]a(x)-[x(2-x)+2x(3-x)]a'(x)+x^2a''(x)$$
$$8a(x)^3=(2x^2-11x+18)a(x)+(3x^2-8x)a'(x)+x^2a''(x)$$
I think this is good.  Can anyone verify it's equality?  I'm looking to have it written in terms of the first and second derivatives of $a(x)$ and a similar problem yielded a similar result...The problems are related, but I've been staring at them for a couple of days now, and outside some minor errors I feel that I'm okay...but any problem looked at for that long needs fresh eyes, so I would really appreciate the verification.
 A: Here's a somewhat more systematic way of getting these kinds of relations, for this kind of function. We'll de-clutter notation a little bit. Let
$$e = e^x, t_n = \frac{x^n}{n!}, e_n = \sum_{k=0}^n t_k, e_n - e_{n-1} = t_n$$
You can write a function like your own as
$$a = \frac{t_{n+1}}{e-e_n}$$
To derive the first order relation we'll note that
$$t_{n+1}' = t_n, e_n' = e_{n-1}, \frac{t_{n+1}}{t_n} =\frac{x}{n+1}$$
Then you can write the equation as
$$a(e-e_n) = t_{n+1}$$
and take derivatives
$$a'(e-e_n) + a(e' - e_n') = t_{n+1}'$$
$$a'(e-e_n) + a(e - e_{n-1}) = t_n$$
From the original equation we know that
$$e - e_n = \frac{t_{n+1}}{a}$$
and
$$e - e_{n-1} = \frac{t_{n+1}}{a} + e_n - e_{n-1} = \frac{t_{n+1}}{a} + t_n$$
Putting these together leads to
$$\frac{t_{n+1}a'}{a}  + a\left(\frac{t_{n+1}}{a}+t_n\right) = t_n$$
Multiplying by a, collecting terms and dividing by t_n leads to
$$a^2 = \left(1-\frac{t_{n+1}}{t_n}\right)a - \frac{t_{n+1}}{t_n}a' = \left(1-\frac{x}{n+1}\right)a - \frac{x}{n+1}a'$$
Multiplying by n+1 leads to
$$(n+1)a^2 = (n+1-x)a - xa'$$
To de-clutter things even more we use $n$ instead
$$na^2 = (n-x)a - xa'$$
where
$$a = \frac{t_n}{e - e_{n-1}}$$
From here on out we have
$$xa' = (n-x)a -na^2$$
Taking the first derivative gives
$$a' + xa'' = (n-1)a + (n-x)a' - 2naa'$$
Collecting a' and multiplying by $x$ gives
$$x^2a''+(1-n)ax = (n-x-1-2na)a'x = (n-x-1-2na)((n-x)a -na^2)$$
Expanding and collecting terms gives
$$x^2a''-2n^2a^3+(n-n^2+2nx-x^2)a+(3n^2-n-3nx)a^2 = 0$$
and replacing $na^2$ with $(n-x)a-xa'$ leaves us with
$$2n^2a^3 = (2x^2-(4n-1)x+2n)a + x(3x-(3n-1))a'+x^2a''$$
One could probably take some more derivatives here to get the next few terms In your case $n = 3$ so we find
$$18a^3 = (2x^2-11x+18)a + x(3x-8)a' + x^2a''$$
which is the equation you have. Your linked problem has n = 2 which will leads to
$$8a^3 = (2x^2-7x+8)a + x(3x-5)a' + x^2 a''$$
The exponential generating function for the Bernoulli numbers has n = 1 which leads to
$$2b^3 = (2x^2-3x+2)a + x(3x-2)a' + x^2 a''$$
so the equations you found are not a coincidence. The relationship between the partial sums of the exponential function, their derivatives and differences leads to the first order differential equation. Then the algebra is a bit messy but we get a second order differential equation that will help you get the 3-rd order convolution. I think this is a very interesting problem.
