Hi,
That is a very general question, and usually you have to be more specific
to be able to find starting conditions, especially since you may want
derivatives or just the function at t=0.
One general method for finding f(0) is to use the initial value theorem, which
states that the initial value in the time domain is the F(s=inf) value in
the frequency domain. In this way, you can use Laplace transforms
and take the limit as s approaches infinity.
This really is a varied subject though, because sometimes we need f'(0)
or something like that. One of your other problems (the parallel RLC with
impluse circuit) required the derivative of iL at t=0, and that can come
from either Laplace or taking a good look at the physical layout and how
the various components (L,C,R) respond to the excitation.
Two very useful ideas are for the L and C, and state:
For an inductor, the current can not change by a finite amount in zero time,
and for a capacitor, the voltage can not change by a finite amount in zero time.
There is one exception though, and that is with an impulse. An impulse can cause
a finite change in a cap or inductor in zero time, but that is the only way this
can happen.
Because of the above, many times v(0+)=v(0-) and i(0+)=i(0-), unless there
is an impulse source present and then you have to look at the physical layout
and try to determine how to go about solving it, or use Laplace.
If there is an impulse, then you have to look at the way the circuit element
reacts to the impulse. A cap subject to an impulse will react by integrating it,
so you can say that the cap reacts to the 'area' of the impulse, for example.
In the parallel RLC + impulse circuit, first the cap integrates and causes a voltage,
then the inductor starts to react to the voltage, but even though the inductor current
is zero at t=0, the derivative is not zero. Thus, we have to solve for the derivative of iL
at t=0.
To get the initial cap voltage, we would get 1/C volts. This is because when we integrate
the impulse we get 1, and 1/C is outside the integrand. Then to solve for di/dt, we use
v=L*di/dt
so we get:
di/dt=v/L
and since v=vc and vc=1/C, we get:
iL'(0)=1/(L*C).
That should show how varied this kind of question can be with various circuits.