Using the "prime" notation for differentiation $$f'(x) = \frac{\partial }{\partial x}\{f\}(x)\\f''(x) = \frac{\partial^2 }{\partial x^2}\{f\}(x)\\\vdots\\f^{(k)}(x) = \frac{\partial^k }{\partial x^k}\{f\}(x)$$
Consider the differential equation:
$$\sum_{k=0}^N P_k(t)f^{(k)}(t) - g(t) = 0$$
Where $\{P_0,\cdots,P_N\}$ are polynomials, $f$ is the function we want to solve for.
A general function in the exponential family is:
$$f(x) = P_1(x) e^{P_2(x)}$$
For two given polynomials $P_1,P_2$. Now allow them to not be same as $P_k$ above.
Will we be able to express all functions in the exponential family using this differential equation?
Own work:
I have calculated $$\frac{\partial}{\partial x}\{P_1(x) e^{P_2(x)}\} = {P_1}'(x)e^{P_2(x)} + P_1(x)P_2(x)e^{P(x)} = \\ \underset{P_3(x)}{\underbrace{({P_1}'(x)+P_1(x)P_2(x))}}e^{P_2(x)}=P_3(x)e^{P_2(x)}$$
So we can guarantee that differentiation on such a function keeps us inside the family (closure).
But is this enough? Is it perhaps even overkill?
In fact every function of the form $P_1(t) e^{P_2(t)}$ is a solution of a homogeneous linear differential equation with polynomial coefficients.
Note that if $f(t) = P_1(t) e^{P_2(t)}$, $f'(t) - P_2'(t) f(t) = P_1'(t) e^{P_2(t)}$, and $P_1'$ has degree one less than $P_1$ (if $P_1$ is not constant). Thus if $P_1$ has degree $n$, $f$ is a solution of the differential equation $T^{n+1} f = 0$ where $T$ is the differential operator $f \mapsto f' - P' f$.