Is a once-differentiable curve whose tangent lines pass through origin necessarily a line? Let $\alpha: I \to \mathbb{R}^n$, $n\geq 1$ be a differentiable regular curve such that all of it's tangent lines pass through origin. This means that for every $t\in I$ there exists $k(t) \in \mathbb{R}$ such that $\alpha(t)+k(t)\alpha'(t)=0$. So each coordinate $x$ of $\alpha$ satisfies the differential equation $x+kx'=0$.  If $k$ was integrable and non-zero, the solutions to this equation would be $x=Ce^F$ where $F$ is an antiderivative of $-\frac{1}{k}$. And then it follows that the whole curve is a multiple of a constant and therefore contained in a line. However i don't see how to show if $k$ is necessarily integrable/continuous and non-zero.
Note: I am aware there is an easier way if the curve is twice differentiable. My question is more general.
 A: Partial answer: In the following we assume that $|\alpha'|$ is locally bounded. This is the case when $\alpha$ is $C^1$.
Let $\epsilon >0$ and consider the curve
$$ \alpha_\epsilon =\alpha |_{I_\epsilon}, $$
where $I_\epsilon = \alpha^{-1} \{y\in \mathbb R^n: | y| > \epsilon\}$ (that is, consider only the portion of $\alpha$ which is as least distance $\epsilon$ away from the origin). Then $I_\epsilon$ is open and is a disjoint union of connected components (which are intervals). It suffices to show that $\alpha_\epsilon$ is a straight line when restricted to each such intervals.
The condition gives the equality
$$\tag{1}\alpha + k \alpha ' = 0,$$
-taking dot product with $\alpha'$ in (1) gives
$$k = -\frac{\alpha \cdot \alpha '}{\|\alpha'\|^2}.$$
In particular, since $\alpha'$ is measureable (see here), $k$ is also measurable;
-also, (1) implies
$$|k| =\frac{ \|\alpha'\|}{\|\alpha\|}.$$
Hence on $\alpha_\epsilon$ we have $|1/k|\le \|\alpha'\|\epsilon^{-1}$. Since $\|\alpha'\|$ is assumed to be locally bounded, $1/k$ is locally bounded, thus integrable.
As a result, as you did in the post, one obtains that $\alpha = C e^{F}$ on each connected component of $I_\epsilon$.
