How do we show that $u(x)=ce^{rx}$ is the only solution to $u'=ru$ in $\mathcal D'(X)$?
I tried to decompose a $\phi\in\mathcal D(X)$ into parts and let $u$ acts on each of them, but I couldn't show that the sum converges in $\mathcal D(X)$. I am lost here, can anyone please help?
More generally, for equation of the form $$ u^{(n)} + c_1 u^{(n-1)} + \dots + c_n u =0, $$ why is it true that classical solutions are all posible solutions. What is the reasoning behind that?
The calculations are the same as in classical real analysis:
In $\mathcal D'(X)$ we can multiply with a $C^\infty$ functions, and since $e^{-rx}$ is never $0$ the equation $u'=ru$ will not change the set of solution when multiplied with $e^{-rx}$. Thus, $u'=ru$ is equivalent with $e^{-rx} u' - r e^{-rx} u = 0$. This can be rewritten as $(e^{-rx} u)' = 0$ which means that $e^{-rx} u = c$, where $c$ is a constant. Thus $u = c e^{-rx}$.
Solution of $u''-u=0$:
The equation can be rewritten as $(u'+u)' = u'+u,$ so by the previous result we have $$u'(x)+u(x) = (u'+u)(x) = c_1 e^x.$$
Just as for classical solutions, and for the same reasons, we first seek one particular solution and a set of solutions to the homogeneous equation. One particular solution is $u_p(x) = \frac12 c_1 e^x$ and the homogeneous solutions are $u_h(x) = c_2 e^{-x}$ by a modification of the previous result. Thus the general solution is $$u(x) = u_p(x) + u_h(x) = \frac12 c_1 e^x + c_2 e^{-x}.$$ (Of course, by adjusting $c_1$ we can drop the factor $\frac12$)
How did I come up with $(u'+u)' = u'+u$?
Start with $u''-u = 0$. Rewrite this as $(D^2-1)u=0$. Factor the operator to get $(D-1)(D+1)u=0$. Thus we have $D(D+1)u = (D+1)u,$ i.e. $(u'+u)' = u'+u.$
Generally, when we have an equation of the form $$u^{(n)} + c_1 u^{(n-1)} + \dots + c_n u =0,$$ where $c_1, \ldots, c_n$ are constants, we can factor the differential operator and get $$(D-r_1)(D-r_2)\cdots(D-r_n)u = 0$$ where $r_1, \ldots, r_n$ are the solutions to the characteristic equation $$r^n + c_1 r^{n-1} + \cdots + c_n = 0.$$ Then we can solve the equation by solving for one "differential factor" at a time through multiplication by $e^{-r_k x}$.