The simplest ODE $x' = x$ An ODE $x'(t) = x(t)$ is one of the simplest equation.
If $x(t) \in C^1(\mathbb{R})$ is a solution of this ODE, we get $x(t) = C \exp(t)$, where $C$ is constant. This is very elementary fact in ODE theory.
However, if $x \in \mathcal{D}'(\mathbb{R})$ satisfies $x'= x$ in $\mathcal{D}'(\mathbb{R})$, does  $x(t) $ need to be $C \exp(t)$ for some constant $C$ ?
The following theorem is well known.

Let $P$ be a polynomial with $\deg(P) = n$. Set $V := \{f \in C^n(\mathbb{R}) \mid P(D)f=0\}$, where $D = d/dt$. Then, $V$ is a vector space and $\dim V = n$.

Does the above theorem hold when we replace $V$ with $W = \{ f \in \mathcal{D}'(\mathbb{R}) \mid P(D)f = 0\}$ ?
 A: Yes it works, you can in the same way prove that $\frac{d}{dt}(x(t)e^{-t})=0$ so that $x(t)=Ce^t$ because if $f\in\mathcal{D}'(\mathbb{R})$ is such that $f'=0$, then $f$ is constant. One way to prove that is by taking $g$ a test function such that $\int_{\mathbb{R}}g=1$, then for all test function $\varphi$, we have
$$ \varphi=\left(\int_{\mathbb{R}}\varphi\right)g+\psi $$
where $\psi:=\varphi-\left(\int_{\mathbb{R}}\varphi\right)g$. Now, since $\int_{\mathbb{R}}g=1$, we have $ \int_{\mathbb{R}}\psi=0 $ and therefore $\xi(x):=\int_{-\infty}^x\psi(t)dt$ is compactly supported, and $\xi'=\psi$. Thus
$$ \begin{aligned} \langle f,\varphi\rangle &= \left(\int_{\mathbb{R}}\varphi\right)\langle f,g\rangle+\langle f,\psi\rangle \\
&=\left(\int_{\mathbb{R}}\varphi\right)\langle f,g\rangle-\langle f',\xi\rangle \\
&=\left(\int_{\mathbb{R}}\varphi\right)\langle f,g\rangle \\
&= \langle C,\varphi\rangle
\end{aligned} $$
where $C:=\langle f,g\rangle$. Therefore $f=C$.
A: Your theorem, generalized to distribution, is a consequence of the following result :

Let $T\in\mathcal D'(\mathbb R)$ and $n\in\mathbb N$ such that $D^n T = 0$. Then, $T$ is a polynomial of degree $< n$.

Proof :
Let us start with the case $n=1$. Let $\varphi \in \mathcal D(\mathbb R)$. If $\int_{\mathbb R}\varphi = \langle 1,\varphi\rangle = 0$, then there is $\psi \in\mathcal D(R)$ with $\phi = D\psi$. We have :
$$\langle T,\varphi\rangle = \langle T,D\varphi\rangle = -\langle DT,\varphi\rangle = 0$$
Therefore, $T$ is constant.
Now, if $D^n T = 0$ for some $n\geq 2$, we have $D^{n-1} T =\lambda$ for some constant $\lambda\in \mathbb R$. Therefore, $D^{n-1}( T - \lambda x^{n-1}/(n-1)!) = 0$. By induction, $T$ is a polynomial of degree $\leq n-1$.
