Let $g: \Bbb R \to \Bbb R^3$ be a parametrized curve of $t$. Assume that the curvature $\kappa$ and the torsion $\tau$ are nonzero constants. Show that $g$ represents a circular helix, assuming that the only solution to the differential equation $y''+k^2y=0$ is $y=c_1 \cos(kt)+c_2 \sin (kt)$.
Here is what I have: we know that $g'$ cannot be zero since $\tau = \frac{\lvert g' \times g'' \rvert}{(g')^3}$. Besides this, I am completely lost on how to do this problem, besides I have these formulas: $\kappa = \lvert \frac{dT}{ds} \rvert$ where $s$ is arclength, that $\frac{dN}{ds} = -\kappa T + \lambda B$ where $T$ is $\frac{dr}{ds}$ and $B$ is $T \times N$ and $\lambda$ is the torrsion, and $\frac{dB}{ds} = -\lambda N$. (you can assume I know that some basic formulas for parametrized curves found in calculus books).
An easy way out is the following: Set up a parametric representation of a helix winding around the $z$-axis, and choose the radius and the pitch such that the desired constant curvature $\kappa$ and torsion $\tau$ result. Now appeal to the uniqueness theorem for initial value problems of ODEs in order to show that this helix is congruent to your curve $g$.