ODEs in the space of distributions $\mathcal{D}'(a,b)$: some general solution technique. I am studying ordinary differential equations in the space of distribution $\mathcal{D}'$ where $\mathcal{D}$ is the space of bump (aka test) functions.
The equations we treat are linear, although not always with constant coefficients. What my instructor does to solve this equations is the following


*

*Either resorting directly to the lemma which states that $T'=0\ in\
    \mathcal{D}'\Rightarrow T=constant$, in the simplest cases

*In less trivial cases, such as, say $u'+a(x)u=0$ with
$a(x)\in C^\infty$, she exploits some kind of 'suggestions' from the
classical solution and, given some $\phi\in\mathcal{D}$ she lets the
derivative acts on a term $e^{-A(x)}\phi$,   instead of on $\phi$
alone, where $A(x)$ is a primitive of $a(x)$: hence $\phi$ multiplied by the classical solution of the equation. This leads to a
convenient rewriting of the equation which allows to exploit the
lemma of point 1.   


I can smell the idea of the tecnique, but since this is done without justification, I would like to know if this is just an euristic or a general approach. We pretty soon moved to partial differential equations, where through the fundamental solution of an operator and convolution one does get a general approach so any help, suggestion or explanation would be great.
 A: Of course these techniques have to be justified.
Assume that we have a distributional differential equation $u' + A'u = v,$ where $u$ is an unknown distribution, $A$ is a given $C^\infty$ function, and $v$ is a given distribution.
From the classical theory, we get the idea of multiplying $u' + A'u$ with the integrating factor $e^A$. This part is no problem; also in distribution theory it is valid that $e^A (u' + A'u) = (e^A u)'$.
But will multiplication with $e^A$ preserve the set of solutions? Is $e^A w = 0$ equivalent with $w = 0$ for every distribution $w$? Indeed it is, and the reason for this is that $e^{-A} \in C^\infty$ since $e^A \neq 0$ everywhere:
Let $\varphi \in C_c^\infty$. If $w=0$ then it is clear that $e^A w=0$ since $\langle e^A w, \varphi \rangle = \langle w, e^A \varphi \rangle = 0$. On the other hand, if $e^A w = 0$ then
$\langle w, \varphi \rangle = \langle w, e^A e^{-A} \varphi \rangle = \langle e^A w, e^{-A} \varphi \rangle = 0,$ since $e^{-A} \varphi \in C_c^\infty$.
Thus, multiplying $u'+A'u=v$ with $e^A$ doesn't change the set of solutions. Therefore $e^A(u'+A'u)=e^A v$ is equivalent to the initial equation. But this is equivalent with $(e^A u)' = e^A v$. Now we only have to find all primitive distributions $w$ for $e^A v$ to get $e^A u = w,$ and then multiply with the nonzero factor $e^{-A}$ to get $u = e^{-A} w.$
