One very simple way to solve this problem is by using an electrical circuit analogy where your rigid container with a compressed fluid would be represented by a charged capacitor (voltage equivalent to pressure) and the hole in the container as a resistor that's connected to the capacitor with the other end at a voltage equivalent to the pressure on the outside of the container.
Since you're willing to assume ideal gas in an isothermal process as gas expands through the orifice, the relationship between container pressure and the volume of fluid molecules in the container is linear: $P_1 = V/C$. Here $V$ is not $V_o$ ,the container volume, but rather the volume of gas molecules at $p_r$ that were used to charge the pressure in the container to pressure $P_1$. The pneumatic equivalence of capacitance is 'compliance' and that can be determined using the container volume: $C = V_o/p_r$.
With the two elements you have two equations,
The pressure in the container
$$p_1(t)={1\over C} \int Q(t)dt$$
and the pressure drop across the orifice
$$p_1(t) - p_r = Q(t)^2R$$
where $Q$ is the volumetric flow through the orifice, which is equivalent to current in the electrical circuit.
and the volume that leaves the container at pressure $p_r$ is
$$V(t) = \int Q(t)dt$$
Solving the second of these equations for $Q$ and substituting into the first, then solving for $p_1$ gives you the differential equation you need.
For an incompressible (or essentially incompressible) fluid, pressure is determined only to within an arbitrary constant. The only way to establish an absolute pressure is to match a boundary condition, which is not available in this problem specification.
Best Answer
Though the answer to the respective principles behind compressible fluid flow through a pipe still interests me, it transpires that one can treat air as incompressible at speeds below the velocity of sound.
As such continuity (mass conservation), momentum and Bernoulli equations are all applicable in this situation.