This is an exercise in a book.
So we need to find the most general function $g$ such that this curve from $\mathbb{R}^3$
$f(t) = (a\cos(t), a\sin(t), g(t))$
lies in a plane.
It's not mentioned but there, but I guess we can assume that
- $g$ is e.g. 3 times differentiable
- $a \ne 0$
With these assumptions in mind, I know that for the curve to be planar we must have that the torsion $\tau(t) = 0$ for any $t$.
The formula for the torsion is well-known. From it I got that this requirement translates to $g'(t) + g'''(t) = 0$.
I don't know differential equations so I cannot continue further from here.
But the book gives this answer:
$g(t) = b\cos(t) + c\sin(t) + d$
I could easily verify that this function satisfies the equation I got. OK... So my question is: does that mean that the only solution to the differential equation I got is the function given in the book's answer? Where can I loop up some basic theory about this differential equation?
Setting $g'(t)=u(t)$ your equation becomes: $$ u''(t)=-u(t) $$ whose general solution is $$ u(t)=A\cos t+B\sin t, $$ with $A$ and $B$ arbitrary constants. Then integrate $u(t)$ to get $g(t)$.
EDIT.
The same result can be reached without calculus. The projection of the curve on the $xy$ plane is a circle. The curve could lie on any plane not perpendicular to $xy$ plane, whose general equation can be written as: $$ z=bx+cy+d. $$ Subsitute here $x=a\cos t$ and $y=a\sin t$ to get $z=g(t)$.