am I guaranteed a certain number of weak solutions (i.e. for a second order equation, there are 2 weak solutions and 2 strong solutions...?)
No, you can have more or fewer. For nice ODE, every weak solution is a classical solution: you get nothing new by allowing weak solutions. Let's take $$y'=2y\tag1$$ as a simple example. Its
weak form is $$\int (y\phi'+2y\phi)=0\tag2$$ for every test function $\phi$. Rewrite it as
$$\int e^{-2x}y(e^{2x}\phi)'=0\tag3$$
Given any test function $\psi$, we can let $\phi=e^{-2x}\psi$ and conclude from (3) that
$$\int e^{-2x}y \psi'=0\tag4$$
for all test functions $\psi$. From (4) you can conclude that $e^{-2x}y$ is in fact a constant function. Indeed,
every test function that integrates to $0$ over $\mathbb R$ is of the form $\psi'$ for some test function $\psi$.
Therefore, the integral $\int e^{-2x}y \varphi$ depends only on the value of $\int \varphi$. By linearity
of $\int e^{-2x}y \varphi$ there is $c\in\mathbb R$ such that $\int e^{-2x}y \varphi=c\int \varphi$ for
all test functions $\varphi$. But this means precisely that $e^{-2x}y=c$ as distributions.
More generally, for any constant-coefficient linear ODE every weak solution is a classical solution. One
way to see this is to take the Fourier transform (I ignore the existence issue): then the equation
$\mathcal L(y)=0$ becomes $p\hat y=0$ where $p$ is a polynomial
(the symbol of $L$). Therefore, $\hat y$ is supported on the zero set of $p$, in particular it's compactly
supported. Therefore, $y$ is real-analytic.
For linear ODE with variable smooth coefficients everything works about the same
as long as the coefficient of the highest derivative does not vanish. This is because we can rewrite
the ODE as $y'=F(x,y)$ with vector-valued $y$, where $F$ is smooth in $x$ and linear in $y$. Instead of
giving a proof, I'll just say that $F(x,y)$ is as regular as $y$ is, which makes $y=\int F(x,y)$ more regular than that: it self-improves ad infinitum.
But your question is motivated by an ODE where the coefficient of the highest derivative (call it $\alpha(x)$) vanishes. What can we do then? The set $\{x:\alpha(x)\ne 0\}$ is a union of open intervals. On each such intervals the above applies: weak solutions are classical solutions. The question is how/if they can be glued together at the zeros of $\alpha$, called the singular points of the equation. This is where we have to look at the particular form of equation, because a lot of different things can happen at singular points.
Let's generalize your example to $\alpha(x)y'=0$ where $\alpha$ is a smooth function with $n$ zeros on the real line, denoted $b_1< \dots <b_n$. On each of $n+1$ open intervals $(-\infty,b_1)$, $(b_1,b_2)$, ... ,
$(b_n,\infty)$ the function $y$ must be constant. This may be an $(n+1)$-dimensional space of weak
solutions, but we must check that they are indeed weak solutions. It turns out that they are:
the weak form of equation is $$\int (\alpha y\phi' +\alpha'y \phi)=0$$ which holds because on each interval we
can move the constant $y$ out of the integral, leaving
$$\int_{b_k}^{b_{k+1}} (\alpha \phi'+\alpha '\phi) = (\alpha \phi)\bigg|_{b_k}^{b_{k+1}}=0 \tag5$$
where the last step is based on $\alpha \phi$ being zero at $b_1,\dots,b_n$.
Thus, we have a first-order equation with $(n+1)$-dimensional space of weak solutions. In your example with
$\alpha(x)=x$ there is only one zero, and therefore the solution space is $2$-dimensional.
On the other extreme, if $\alpha(x)=\sin x$, the solution space becomes infinite-dimensional.
Let's consider two more examples.
$$xy'=2y \tag6$$
Outside of $x=0$, the solutions are $y=cx^2$. The constant can be different for $x<0$ and for $x>0$.
The solutions nicely patch up at $0$, forming a $2$-dimensional space of classical solutions.
$$xy'=-2y \tag7$$
Outside of $x=0$, the solutions are $y=cx^{-2}$. The constant can be different for $x<0$ and for $x>0$.
But they only solution that is locally integrable near $0$ is $y\equiv 0$. Thus, we have only one weak solution
on $\mathbb R$.
Upshot: a lot of things can happen at singular points, where the leading coefficient vanishes. Power series
methods are pretty good at finding the asymptotic behavior of solutions there: see Regular singular point, Frobenius method.
Take $\phi \in C_c^\infty ((-1,1))$ with $\phi(0) = 1$. Let $\phi_n(x) = \phi(nx)$ (where we consider $\phi$ defined on all of $\mathbb{R}$ by trivial extension; since the support of $\phi$ is a compact subset of $(-1,1)$, the trivial extension is still smooth). Consider
$$\int_{-1}^1 g(x)\phi_n(x)\,dx.$$
On the one hand, it should be $\phi_n(0) = \phi(n\cdot 0) = \phi(0) = 1$ for all $n$. On the other hand, what can you say about $\lVert\phi_n\rVert_{L^2((-1,1))}$?
Best Answer
Yes, there are more immediate proofs than invoking Dominated Convergence, and maybe you're intuiting some such thing in saying "it's obvious". But, still, if a thing is obvious, it should be possible to explain it persuasively...
For this example, I'd convince myself of the truth of the conclusion that $\lim_{\delta\to0^+}\int_{-\delta}^\delta |g(x)|\;dx=0$ for every locally absolutely integrable $g$, as follows: by countable-disjoint additivity, $$ \sum_{n=1}^\infty \int_{{1\over n+1}}^{1\over n} |g(x)|\;dx \;=\; \int_0^1 |g(x)|\;dx \;<\; \infty $$ Thus, the infinite sum of integrals converges, so the infinite tails $\int_0^{1/n}$ must go to $0$.