Ok, why not do it. Here's an example of how this is used.
Assume your circuit is a battery E charging a capacator C thru a resistor R.
Let the voltage across the capacitor be V
The current in the capacitor is i = C*dV/dt
The voltage across the resistor is R*i
If you sum the loop voltages you get,
E = R*i + V
Substituting in the formula for current in a capacitor,
E = RC*dV/dt + V
Rearranging to the normal form,
dV/dt + V/RC = E/RC
The above is a first order linear differential equation. This simplest of examples can be solved by separation of variables, but in real life the equations get harder and you'd want to use Laplace transforms if possible.
dV/dt = E/RC - V/RC = (E-V)/RC
dV/(E-V) = dt/RC
Integrating both sides,
ln(E-V) = t/RC + constant1 E-V = exp(t/RC + constant1) = exp(t/RC) * exp(constant1) E-V = exp(t/RC) * constant2
We have to determine the integration constant. At t=0, the capacitor wouldn't have had time to charge, so it's initial value is zero. Therefore at t=0,
E-0 = exp(0) * constant2 constant2 = E
Therefore
E-V = exp(t/RC) * E
V = E*(1 - exp(-t/RC) )
So the voltage across the capacitor has an exponential rise up to the battery voltage with time constant RC.
There ya go,
dave y.