I tried to solve the equations $(x-a)T=\delta_b$ and $(x-a)T=\delta_a$, but I have some questions.
I have already established that $ (x-a)T=0$ $\iff$ $T=c\delta_a $
Does this mean that the solution to $(x-a)T=S$ is $c\delta_a + T_0$ with $T_0$ being a particular solution ? (Like we do for xT=S)
For $(x-a)T=\delta_b$, I used the conclusion above and determined that the solution is $c\delta_a + \frac{1}{b-a}\delta_b$, since $\frac{1}{b-a}\delta_b$ is a particular solution. Is the a correct way ?
For $(x-a)T=\delta_a$, I'm struggling to find any particular solution. Any pointers ?
(T and S are distributions, a,b are scalars and c $\in$ $\mathbb{C}$ )
EDIT: for 3), I have realised that $T_0=-\delta_a'$ is a valid particular solution. So I only need to make sure that $c\delta_a + T0$ being the solution is correct.
Thanks !