I need to find all $u \in \mathcal{D}'$ ( space of distributions) such that
$ e^x (e^{-x} u ) ' = \delta_0 +1$.
For any $\phi \in C_0^\infty(\mathbb{R})$ we have
$\langle e^x (e^{-x} u ) ' , \phi \rangle = \langle \delta_0 +1, \phi\rangle$ and so
$\langle u , e^{-x}(e^x\phi) ' \rangle = \phi(0)+ \langle 1, \phi\rangle$
$\langle u , \phi+\phi ' \rangle = \phi(0)+ \langle 1, \phi\rangle$.
$\langle u , \phi\rangle+\langle u, \phi ' \rangle = \phi(0)+ \langle 1, \phi\rangle$.
How should I continue from here ?
Thanks.
You have the equation $$e^x(e^{-x}u)'=\delta_0+1.$$ First step is to notice that the function $x\to e^{x}$ is $C^\infty$, strictly positive everywhere, hence we can safely divide by it both sides of the equation without producing and/or losing solutions. Thus, we get
$$ (e^{-x}u)'=e^{-x}\delta_0+e^{-x}=\delta_0+e^{-x}.$$
It is easy to take the antiderivative of the right hand side: if $H$ is the Heavyside function and $c\in \Bbb C$, we obtain $$e^{-x}u = H(x)-e^{-x}+c.$$
By the similar argument we can safely divide by $e^{-x}$ to obtain $$ u = H(x)e^{ x}-1+ce^{ x}.$$