The answers already given are quite complete and say it all, but maybe the OP was in search of a very simple explanation of what is happening. Let me try.

You will agree that it is not necessary to work in the full generality of an n-th order equation; if we can do it for the first order case, it is easy to generalize by induction. Moreover, we can assume that the highest order coefficient is equal to 1 (just divide by it). So essentially the question is: let $u$ be a weak solution of the equation
$$ u' + a(x) u = f(x) $$
on an open interval, with $ a(x) $ and $ f(x) $ analytic. Then we want to prove that $u$ is also analytic. A further reduction is possible: multiply both sides by $\exp (A(x)) $ where $A'=a$ and call $ v = e^A u $, $g=e^A f$. Then we are reduced to proving the same thing for the simple equation
$$ v ' = g. $$
Is $v$ analytic if $g$ is analytic? A further reduction is possible! just call $ w=v-G $,
where $G'=g$. Nice trick eh? we are reduced to the even simpler equation
$$ w' = 0 $$
and our serach will be over if we prove the fundamental fact that any weak solution of $w'=0$ must be a constant function. Notice that the same chain of arguments applies if the coefficients are $C^\infty$ (you get that $u$ is also $C^\infty$) and if $f$ is just $C^k$.

Now, in the greatest possible generality, if $w$ is any distribution on an open interval, with vanishing derivative, then $w$ must be a constant. This is proved in the following way: by definition, we know that $ w(\phi')=0 $ for any test function $\phi$. Fix a test function $\chi$ with $\int\chi=1$. Let $\psi$ be an arbitrary test function, define
$$ \psi_1 = \psi-\chi \cdot \int \psi $$
and notice that $\int\psi_1=0$. This means that $ \psi_1 $ can be written as the derivative of another test function (guess which one?) $\psi_1=\psi_2'$ and hence
$$w(\psi_1) = w(\psi_2') = 0 $$
by assumption. This implies
$$ w(\psi) = w(\chi) \cdot \int \psi $$
which in the language of distributions means precisely that $w$ is equal to the constant $w(\chi)$ (notice that $\chi$ is fixed once and for all). That's it.