Given that $u,v$ are functions of $t$, $R$ constant, solve $\left\{\begin{matrix}u'v''-u''v'=0 \\ R^2u'u''+v'v''=0 \end{matrix}\right.$.
When trying to find geodesic on cylinder, I get this terrible ODE, which I've totally no idea. Since the required geodesic must be helix, circle or straight line, I expect $u,v$ can only be either constant or linear function of $t$. Please help me solving this.
On
Set $U=u'$ and $V=v'$. The first equation says (after division by $U^2$) that $V/U=\operatorname{const}=:\lambda$. Then the second equation says that $(U^2)'=0$; hence both $U$ and $V$ are constants. Thus, $u$ and $v$ are linear functions. I hope you can start from here and consider the exceptional case when $U$ can vanish.
$$u'v''-u''v'=0 \implies \frac{u'v''-u''v'}{(v')^2}=-0=0 \implies (u'/v')'=+0=0$$
Thus $$u'=c v'$$
The rest is straight forward...