Gaussian integer $\alpha + \beta i$

Calculus Level 4

For positive integers $a, b$ , and $c$ , consider the differential equation

$\large ay'' + by' + cy = 0$

If we want solutions of the form

$y(x) = e^{\alpha x}(C_1\cos{\beta x} + C_2\sin{\beta x})$

where $\alpha$ and $\beta$ are non-zero integers, what is the smallest possible value of $a + b + c$ ?

Notations: $y' = \dfrac{dy}{dx}$ , $y'' = \dfrac{d^2y}{dx^2}$ , and $C_1$ and $C_2$ are constants.

1 solution

Geoff Pilling
Jul 27, 2017

One solution is given by, $a=1, b=2 \text{ and } c=2$ .

This gives roots for the characteristic equation, $a\lambda^2 + b\lambda + c = 0$ :

$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\lambda = \frac{-1 \pm \sqrt{2^2 - 4\cdot 2}}{2}$

$\lambda = -1 \pm i$

i.e. $\alpha = -1$ and $\beta = \pm 1$

$\boxed{a + b + c = 5}$

Now to show that this is the smallest sum we can get...

Suppose $a=1$ . This implies that $b$ must be even (to make $\frac{-b}{2a}$ an integer), or actually $2$ if we want a sum smaller than $5$ .

If $b=2$ , the smallest $c$ for which $b^2 - 4ac > 0$ is $c=2$ . This gives us $(a,b,c) = (1,2,2)$ , our earlier solution.

For $a > 1, b$ must be at least $2a$ for $\frac{-b}{2a}$ to be an integer. And this will put $a+b > 5$ .

Therefore, our solution is: