When the third derivative is the function itself

 Suppose $f: \mathbb{R} \to \mathbb{R}$ is an infinite differentiable function such that the third derivative of $f$ is $f$ itself. i.e.

$$ f'''(x) = f(x) \quad \forall x \in \mathbb{R}$$

It is easy to see that $f(x) = Ae^x$ satisfies this. Can you find others?

Scroll down for a solution.

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Solution:

If the equation was $f'''(x) = 8f(x)$ then we could easily see that $Ae^{2x}$ would be a solution.

What if there was a solution of the form $Ae^{kx}$? Then we must have that $k^3 = 1$. $k = 1$ already corresponds to the $Ae^x$ case. But the others are complex, so is that the end of this approach?

No!

We can still get our solution out of this. Consider $$f(x) = Re(e^{wx})$$

i.e. we take the real part of the solution that allows complex numbers. It must satisfy the conditions of the problem!

In this case

$$f(x) = \cos \left(\frac{\sqrt{3}}{2}x \right) e^{-\frac{x}{2}}$$

works! Try it.

There are other approaches, like power series. But the matrix differential equation based approach gives us the complete set of solutions and comes out to be any linear combination of $$e^x, \cos \left(\frac{\sqrt{3}}{2}x \right) e^{-\frac{x}{2}}, \sin \left(\frac{\sqrt{3}}{2}x \right) e^{-\frac{x}{2}}$$