How to prove exponential functional identity knowing that it is a solution to a first order ODE and knowing its Taylor expansion

Question

Establish the identity

$$E(ax)E(bx) = E[(a+b)x]$$

knowing that

$y = E(px)$ satisfies $y' - py = 0$ and

$E(px) = \sum_{n=0}^\infty\frac{(px)^n}{n!}$

An additional hint the textbook gives ; "Combine the differential equations satisfied by $E(ax)$ and $E(bx)$."

However, I am unsure as to how to get a product when following the hint:

$E'(ax) - aE(ax) = E'(bx) - bE(bx)$

This is also in the context in which we don't know that $E(px)=e^{px}$, meaning we can't use properties of the exponential function.

Answer 1

if $\frac{d}{dx} E(px) - p E(px)=0$ for all $p$ so as $p=a$ and $p=b$, hence $E'(ax)-aE(ax)=0$ and also $E'(bx)-bE(bx)=0$. the hint didn't exactly tell you to substract them or ..., it just said combine them somehow. you need to find a suitable way to do this: $$ \require{mathtools} \left\{ \begin{array}{l} \frac{d}{dx}E(ax)-aE(ax)=0 \\ \frac{d}{dx}E(bx)-bE(bx)=0 \end{array}\right. \xrightarrow [{\times E(bx)}]{\times E(ax)} \require{mathtools} \left\{ \begin{array}{l} E(bx)[\frac{d}{dx}E(ax)-aE(ax)]=0 \\ E(ax)[\frac{d}{dx}E(bx)-bE(bx)]=0 \end{array}\right. $$ if you sum these two equations: $$ E(ax)\frac{d}{dx}E(bx) + E(bx)\frac{d}{dx}E(ax) - (a+b)E(ax)E(bx)=0 $$ which is : $$ \frac{d}{dx} \bigg(E(ax)E(bx) \bigg) -(a+b)E(ax)E(bx)=0. \tag{1} $$ we also know that for $p=a+b$ we have : $$ \frac{d}{dx} E((a+b)x) - (a+b) E(px)=0. \tag{2} $$ if you substract (1) and (2) and consider $\omega(x):=E(ax)E(bx)-E((a+b)x)$ and $c=a+b$: $$ \frac{d}{dx} \omega - c \omega=0 \tag{3} $$ we also have initial condition for $\omega$, using the series $E(px)=\sum_{n=0}^{\infty} \frac{(px)^n}{n!}$ we have: $$ \omega(0)=E(0)E(0)-E(0)=1\times 1 -1=0 $$ so $\omega$ is a function satisfying ODE (3) and $\omega(0)=0$. solve this ode (now we have initial condition) we can see : $$ \omega(x)=0 \Rightarrow E(ax)E(bx)=E((a+b)x) \quad \forall x ,a,b $$