Motivating $y'=y \implies y=Ce^x$

Question

Is there some intuitive reason why one should think that a function which is its own derivative should be of the form $Ca^x$ for some number $a$? Of course I can prove that the unique solution set to $f'(x) = f(x)$ is the set $\{f\mid f(x) = Ce^x\}$, but pretending I don't know this, what would be a good rationale for checking a function of this type?

Answer 1

One way to intuitively see this is to write formally the approximation of the derivative as

$$\frac{dy}{dx}\approx\frac{y(x+\Delta x)-y(x)}{\Delta x} \tag 1$$

Then, using $(1)$ reveals that

$$\frac{y(x+\Delta x)-y(x)}{\Delta x}\approx y(x)\implies y(x+\Delta x)=(1+\Delta x)y(x) \tag 2$$

Setting $x=0$ in $(2)$ and iterating yields

$$\begin{align} y(\Delta x)&\approx (1+\Delta x)y(0) \\\\ y(2\Delta x)&\approx (1+\Delta)^2y(0)\\\\ y(n\Delta x)&\approx (1+\Delta x)^ny(0) \tag 3 \end{align}$$

Then, setting $n\Delta x=x_0$ in $(3)$, we find that

$$y(x_0)\approx \left(1+\frac{x_0}{n}\right)^ny(0) \tag 4$$

Recalling that the limit definition of $e^x$ is $e^x=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)$, taking the limit as $n\to \infty$ of $(4)$, and noting that the approximation in $(4)$ becomes exact as $\Delta x\to 0$ reveals

$$y(x_0)=y(0)e^{x_0}$$

Answer 2

Here is one possibility: start with a difference equation instead of a differential equation, and take this simple example:

$$y(x+1) - y(x) = y(x)$$

or

$$y(x+1) = 2 y(x)$$

Now we have the sequence

$y(1) = 2 y(0)$
$y(2) = 2 y(1) = 2^2 y(0)$
$y(3) = 2 y(2) = 2^2 y(1) = 2^3 y(0)$
$\dots$
$y(k) = 2^k y(0)$

This is exactly the form you were looking for with $a = 2$ and $C = y(0)$.

More generally write

$$y(x+h) - y(x) = h y(x)$$

$$y(x+h) = y(x) (1+h)$$

Then we have as before ($k = 1, 2, 3, ...$)

$$y(k h) = (1+h)^k y(0)$$

New let

$x = k*h$ and inserting $k = x/h$

we have

$$y(x) = y(0) \left((h+1)^{1/h}\right)^x$$

For $h\to 0$ we have

$(h+1)^{1/h} \to e$

so that

$$y(x) = y(0) e^x$$