The ODE y'=y Characterizes Exponentials

Solutions of y'=y satisfy the law of exponents.

Here’s a self-contained proof that the solution to $y^\prime = y$ with $y(0)=1$ is an exponential function. This is one possible means of proving the existence of exponentials, when defined functionally:

Exponential Functions

An exponential function is a continuous nonconstant function $E\colon\mathbb{R}\to\mathbb{R}$ satisfying the law of exponents $E(x+y)=E(x)E(y)$ for all $x,y\in\mathbb{R}$ .

First a uniqueness result we’ll need:

Let $f,g$ be two solutions to the differential equation $y^\prime =y$ . Then they are constant multiples of one another.

Consider the function $h(x)=\tfrac{f(x)}{g(x)}$ . Differentiating with the quotient rule,

\begin{align} h^\prime(x)&=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^2}\\ &= \frac{f(x)g(x)-f(x)g(x)}{g(x)^2}\\ &=\frac{0}{g(x)^2}\\ &=0 \end{align}

Thus $h^\prime(x)=0$ for all $x$ , which implies $h=f/g$ is a constant function, and $g$ is a constant multiple of $f$ as claimed.

Now we’re ready for the main theorem:

Let $g$ be any differentiable function which solves $g^\prime = g$ and has $g(0)=1$ . Then $g$ is an exponential.

Let $g\colon\RR\to\RR$ solve $Y^\prime = Y$ and satisfy $g(0)=1$ . We wish to show that $g(x+y)=g(x)g(y)$ for all $x,y\in\RR$ .

So, fix an arbitrary $y$ , and consider each of these separately, defining functions $L(x)=g(x+y)$ and $R(x)=g(x)g(y)$ .
Differentiating,

\begin{align*} L^\prime(x)&=\left(g(x+y)\right)^\prime\\ &=g(x+y)(x+y)^\prime\\ &=g(x+y)\\ &=L(x) \end{align*}

\begin{align*} R^\prime(x)&=\left(g(x)g(y)\right)^\prime\\ &=(g(x))^\prime g(y)\\ &=g(x)g(y)\\ &=R(x) \end{align*}

Thus, both $L$ and $R$ satisfy the differential equation $Y^\prime=Y$ . Our previous proposition implies they are constant multiples of one another,

$\frac{L(x)}{R(x)}=k\hspace{1cm} \forall x\in\RR$

To find this constant we evaluate at $x=0$ where (using $g(0)=1$ ) we have $L(0)=g(0+y)=g(y)$ $R(0)=g(0)g(y)=g(y)$

They are equal at $0$ so the constant is $1$ :

$\frac{L(x)}{R(x)}=\frac{L(0)}{R(0)}=\frac{g(y)}{g(y)}=1$ $\implies L=R$

But these two functions are precisely the left and right side of the law of exponents for $g$ . Thus their equality is equivalent to $g$ sayisfying the law of exponents for this fixed value of $y$ :

$\forall x,\,\, L(x)=g(x+y)=g(x)g(y)=R(x)$

As $y$ was arbitrary, this holds for all $y$ , and $g$ is an exponential.

Note this proof does not establish the existence of a solution to this ODE, it only says if you have a solution then its an exponential. But this provides a much easier means of rigorously constructing an exponential: if we can develop techniques to build such a function (eg power series, and term by term differentiation) we will immediately know the result satisfies the law of exponents.