I need to find all solutions in $\mathcal {D'}(\mathbb{R})$( distributions $T$ ) of this equation:
$id \cdot T' = 0$
$T'(\varphi ) = - T(\varphi ')$
I've already solved a similar equation $id \cdot T = 0$ and there the solutions were $c \cdot \delta_0 , \ c \in \mathbb{R}$
Could you help me with this one?
Thank you
Let us denote $S= T'$, then you have $$id\cdot S = 0,$$ which, as you already solved, yields $$S = c\delta_0.$$
Now you have $$S=T'= c\delta_0.$$ Apparently, now we need to solve this differential equation.
First of all, you should, by now, be able to guess a particular solution to this problem: if $H$ - the Heavyside function $$H(x)=\begin{cases}1,&x\ge 0,\\0,&x<0,\end{cases}$$ then $H'= \delta_0$. Therefore, we consider the equation
$$(T-cH)'=0.$$
It is an easy exercise to show that the distribution with zero derivative is a constant function (if you don't know how to prove it, ask in comments), hence we obtain that $$T-cH = a,$$ or $$T=a+cH,$$ where $a,c\in \Bbb C$ and $H$ - the Heavyside function.