Starting with:
V = iR + 1/C integral(i.dt)
the integral must be current, not current squared or any other function, just simply current. So it could be oscillatory (sine/cos) or exponential. As we have good reason to think that the current will diminish, an exponential looks promising. Try: A.exp(bt). But this keeps increasing, so A.exp(-bt) is better.
Evaluate at time t=0 when A.exp(-bt) = 1 and the capacitor voltage is 0 V then:
V = A.R so 'A' is just the initial current. Io.
Evaluate after infinite time when the current is zero. The constant of integration (if any) vanishes and:
V = Io/(bC) or b = Io/(VC)
whence b =1/RC
substitute and: i = Io.exp(-t/(RC))
and substitute again for the voltages.

 

The solution to the first equation doesn't involve 'trying' anything or guessing about whether it's oscillatory or exponential. What's wrong with just solving the equation?

 

No problem at all solving the equation. The above is quicker and just seems clearer for people who are not so mathematically minded.