Well you are missing the compatibility of slopes at the middle support:
$$\frac{dw(L^-)}{dx} = \frac{dw(L^+)}{dx}$$
In the case of symmetric geometry & loading, the slope of the beam at the middle support will be zero.
Since the bending moment has no derivative at x=L, you will need to derive the deflections of the two halves separately and 'join' them with compatibility.
Update: derivation of beam deflection formula:
Starting from the Euler-Bernoulli beam equation (assuming constant EI), and taking x from the outer supports towards the center:
$$ q = EI \frac{dw^4}{dx^4} $$
Integrating four times:
$$ EI \frac{d^3 w}{dx^3} = q x + A $$
$$ EI \frac{d^2 w}{dx^2} = q \frac{x^2}{2} + Ax +B $$
$$ EI \frac{d w}{dx} = q \frac{x^3}{6} + A\frac{x^2}{2} + Bx + C $$
$$ EI w = q \frac{x^4}{24} + A\frac{x^3}{6} + B\frac{x^2}{2} +Cx + D$$
Noticing the problem is symmetric the boundary conditions are:
$$ w(0)=w(L) = 0$$
$$ \frac{d w(L)}{dx} = 0 $$
$$ \frac{d^2 w(0)}{dx^2} = 0 $$
Therefore we can immediately see that:
$ B = D = 0$
We now have two equations with two unknowns (A,C). Solving we find:
$$ A = -\frac{3 L q}{8}$$
$$ C = \frac{L^3 q}{48}$$
We can now substitue all the constants back into the equation for w. Simplifying results in:
$$ w = \frac{q x}{48EI} (L-x)^2 (L+2 x) $$
Which is identical to the result referenced here (note their coordinate system has x=0 at the centre). Also notice how this is exactly the same result as a propped cantilever. This is due to symmetry, meaning that the beam slope at the center is zero (which is the same boundary condition as a cantilever support).
You can also substitute into the bending moment equation:
$$ M = EI \frac{d^2 w}{dx^2} = \frac{1}{8} q x (4 x-3 L) $$
