Physics question: why Q=Q0exp(-t/RC) is the solution of the equation dQ/dt=-Q/CR?

@Henk, this is very fine, but all that Tung needs to do is take the derivative of Q(t) and see if he gets the RHS of the differential equation.

The equation dQ/dt = -Q/(CR) is a differential equation in the unknown function Q.
So 1/(CR) can be treated as a constant k = 1/(CR) :
dQ/dt = - k Q
=> dQ/dt + kQ = 0
This is one of the simplest forms of diff. eq. , a linear first order diff. eq. with constant coefficients. We use the interesting property of exp(x) that it's derivative is also exp(x). So we try Q(t) = A exp(B x) as solution
=> dQ/dt = AB exp( B x)
=> dQ/dt + k Q = AB exp( B x) + k A exp(B x) = 0
dividing by A exp( B x ) <> 0 yields
B + k = 0 => B = -k
So we have as solution
Q(t) = A exp( -k t)
As k = 1/(CR), this is
Q(t) = A exp( -t/ (RC))
If we call the beginvalue of Q(t) for t=0, Q0, we get
Q(0) = A exp( 0 ) = A = Q0
=> Q(t) = Q0 exp( -t / (RC) )