Two metallic spheres attached by a long wire.

Two solid metallic spheres with radii R_1 and R_2 are fixed a large distance
apart (large enough that electric fields due to one have a negligible influence
on the other). A long wire with resistance R is connected between them and after a long time the charge on sphere 1 is Q1 and the charge on sphere 2 is Q2. Show that the charge on sphere 2, q2 as a function of time is given by,
$[br]q_2=Q_2\left(1-e^{\left(-\alpha t\right)}\right)\ \mathrm{with}\ \alpha=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{R_1+R_2}{R_1R_2} \right)\dfrac{1}{R}$

I'm having great difficulty setting up the differential equations, can somebody please help me? (I know this is analogous to discharging and charging capacitors)

A quick question.
I take it you know how to set up the differential equation for the case of a single capacitor discharging through a resistor R, and you know how to get the equation from that of the case of a capacitor charging up? The one with the [1 - e^-t/(CR)]?

I think I'm struggling with this because (If I'm correct) I end up with two variables in my DE and can't seem to find a way to get rid of the other variable.

The two variables being the charges (q1 and q2) as functions of time on the spheres

Well
1. the total charge q1 + q2 is constant.
2. The charge Q on a sphere is related to its capacitance via Q=CV, of course. The capacitance of a solid sphere C is 4πεR
3. After a "long time" the discharge/charge stops when the potential of both spheres is equal.
So at the end the spheres are at the same potential where V=Q/C
You know Q1 + Q2 is constant and C is the total capacitance.

I have also noticed that if you look at the formula you have to derive, the value of α is 1/CR in the single capacitor case. ( e^-t/CR)
In this case you have two capacitors which are effectively in parallel (they are at the same pd) and the formula for adding two capacitors in parallel is C₁ + C₂, which in this case is 4πεR₁ + 4πεR₂
If you sub this formula for the total capacitance into the standard formula, you get the value of α required. (I'm not sure why it works though. )