Inspired by Matt DeCross

For any real $k$ , we have the non-zero complex-valued solution $y=e^{\sqrt{-k}x}$ . For $k\leq 0$ , this function is real-values and we are done. For $k>0$ , we can use Euler's Theorem and observe that the real and imaginary parts of $y=e^{\sqrt{-k}x}=e^{i\sqrt{k}x}$ $=\cos(\sqrt{k}x)+i\sin(\sqrt{k}x)$ will be non-zero solutions.

Thanks for the shout-out! This problem brings up the interesting point for my problem that wavefunctions in quantum mechanics are expected to be bounded, even if they are not normalizable (c.f. plane wave eigenstates).

Applying the Laplace Transform, one gets:

$s^2 \cdot Y(s) - s \cdot y(0) - \dot{y}(0) = -k \cdot Y(s)$

$Y(s) = \frac{s \cdot y(0) + \dot{y}(0)}{s^2 + k}$

$Y(s) = \frac{1}{2} \cdot (y(0) + \frac{\dot{y}(0)}{\sqrt{k}}) \cdot \frac{1}{s + \sqrt{k}} + \frac{1}{2} \cdot (y(0) - \frac{\dot{y}(0)}{\sqrt{k}}) \cdot \frac{1}{s - \sqrt{k}}$

Which leads to:

$y(t) = \frac{1}{2} \cdot (y(0) + \frac{\dot{y}(0)}{\sqrt{k}}) \cdot e^{-\sqrt{k}t} + \frac{1}{2} \cdot (y(0) - \frac{\dot{y}(0)}{\sqrt{k}}) \cdot e^{\sqrt{k}t}$

Which, in theory, only leads to positive values of k in order to y(t) be real.

However, if we choose $y(0) = \dot{y}(0) = 0$ , we will get $Y(s) = 0$ and, so, $y(t) = 0$ , which is also real, regardless of k