Distributional Solution to y' - k y = 0

Theorem
Let $f \in \map {C^1} \R$ be a continuously differentiable function.

Let $T \in \map {\DD'} \R$ be a distribution.

Let $T_f$ be a distribution associated with $f$.

Let $\mathbf 0 \in \map {\DD'} \R$ be the zero distribution.

Let $T$ be a distributional solution to the following distributional differential equation:


 * $\paren {\dfrac \d {\d x} - k} T = \mathbf 0$

Then $T = T_f$ where $f$ is the classical solution to the ordinary differential equation:


 * $\paren {\dfrac \d {\d x} - k} f = 0$

Proof
In the distributional sense we have:

By Vanishing Distributional Derivative of Distribution implies Distribution is Constant:


 * $\exists c \in \C : \map \exp {-k x} T = T_c$

In other words:

If $f = c \map \exp {k x}$, then:


 * $\paren {\dfrac \d {\d x} - k} f = 0$