Differentiable $f$ such that the set of translates of multiples of $f$ is a vector space of dimension two How can we derive all of the differentiable functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $V=\{af_b : a,b \in \mathbb{R}\}$ is a vector space of dimension two, where $f_b\colon \mathbb{R} \to \mathbb{R}$ is defined as the translate $f_b(x)=f(x+b)$?
An example is $f=\sin$. The set $V$ contains the linearly independent functions $\sin$ and $\cos$ (since $\cos(x)=\sin(x+\frac{\pi}{2})$). By elementary trigonometry we have $c\sin(x+\phi) = s\sin x + t\cos x$ for any $c, \phi \in \mathbb{R}$ and some $s,t \in \mathbb{R}$, so it follows that $V$ is a vector space of dimension two spanned by $\sin$ and $\cos$.

It would hold for the solution space of a second order ODE which is translation invariant on $\mathbb{R}$. There are a few out there:
$$
           \frac{d^{2}f}{dx^{2}}+B\frac{df}{dx}+Cf=0,\;\;B, C \mbox{ are constants }.
$$
Your example fits into this category. The solution space is 2-dimensional, and any translation of a solution is again a solution because of translation invariance; therefore, translates of solutions are linear combinations of any fixed basis of solutions. Linear sums of such solutions are also solutions.
