The complete response of a non-homogeneous linear system due to both
the external input and the initial condition can be found as the sum of
the homogeneous and particular solutions of the non-homogeneous DE with
a non-zero right-hand side for the external input:

Specifically in the RC circuit, we have

We want to solve the DE for as the circuit's complete response to the input as well as the initial condition , after a switch is closed at time moment .

Given an RC circuit shown above where the switch is closed at , we want to find the voltage across and voltage across as a function of time for .

First we consider a constant (DC) input applied to the circuit
at , i.e., a step input

The solution of this inhomogeneous DE is composed of two parts,

*Particular solution (forced response)*due to the DC input ;*Homogeneous solution (natural response)*: due to the initial condition .

The constant can be determined from the initial condition :

and the complete solution is

In particular, if , the solution is

The current through is

The voltage across is

We can further verify that

The plots below show (red) and (green) under different initial conditions (purple) and inputs (blue).

- , , i.e.,
- , , i.e.,

Note that in the first case, after the switch is closed at ,

- takes a negative value if for , although the voltage source is positive (both measured with respective to the bottom wire treated as the ground). This is because right after the switch is closed, the voltage on the left side of drops from to , causing the voltage on its right side to also drop from to , lower than the ground level of .
- Voltages on both sides of go through a discontinuous transition to drop (case 1) or jump (case 2) by , however, the voltage remains the same, as the voltage across does not change instantaneously. drops from for to at , while (left side of ) drops from to .

In general, neither the voltage across a capacitor nor the current through
an inductor can be changed instantaneously as it takes time for them to
build up:

**A Shortcut Method:**

Observing the complete solution obtained obtained above, we see that

- When , is the initial condition
- When , is the steady state response.

We can therefore generalize the complete solution obtained above to all
first-order systems, i.e., their responses to a step input, a constant input
that is turned on at moment , always take the same form:

in terms of three essential components of the system's response:

- The steady state response : as discussed in previous section for steady state response.
- The initial value : Denote the value of immediately before and after the moment by and , respectively. If , then use for ;
- the time constant of the system : When there is only one
resistor in the circuit, the time constant is or .
When there are multiple resistors, the time constant can be found by:
- Remove or so that the rest of the circuit () is a one port network.
- Find the equivalent resistance of the network by turning off all energy sources (short-circuit for voltage source, open-circuit for current source).
- Find time constant or .

In particular, note that

- when , the initial condition;
- when
,
the
*steady state response*; - when
, the difference
between the initial
and the steady state values of the response decays exponentially. This term is
the
*transient response*of the system.

**Example 0:** ,
,
.
Find .

- Find initial value .
- Find steady state value :

- Find equivalent resistance :

- Find time constant

- Find the complete response

**Example 1:**

In the circuit below, , , , , the circuit was in steady state at . Find after the switch is closed at .

- Find initial value . As the circuit has been in steady
state before , . Also, as voltage across
cannot change instantaneously (unless therefore ),
we have
.
At , drops from to , also drops from to .

- Find steady state value :

- Find equivalent resistance :

- Find time constant

- Find the complete response

- Find current through :

- Find voltages and across and ,
respectively:

- Find currents and through and ,
respectively:

- Verify current :

**Example 1a:**

In the same circuit above, find the voltages and across and currents and through and , respectively.

- Find
- Find and . Before , the circuit is in steady state, i.e., . However, after the switch closes at , the voltage at node d drops from 10V to 0V, and the voltage at node c drops from 5V to -5V with respect to node b as ground (voltage across a capacitor cannot change instantaneously), i.e., and ;
- Find (same as before);
- Find and :

We see that . - Find and through and :

- Find current through capacitor :