Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 100

The constants C 1 and C 2 are found by applying the initial conditions

y(0) ¼ C 1 þ C 2 ¼ 1

y 0 (0) ¼C 1 2C 2 ¼ 4

The solutions are C 1 ¼ 2 and C 2 ¼ 3. Therefore,

2e t 3e 2t

y(t) ¼ for t 0

Next, we consider the homogenous solution. The homogenous solution is the
result of driving the system with an input with zero initial conditions. The homo-
genous solution is typically similar to the input signal; for example, if the input
signal is an exponential function, the response will be exponential and if the input
signal is a sinusoid, the output is also sinusoid with different amplitude and phase.
This is due to the properties of linear systems. A simple example that shows the
process of obtaining the homogenous solution is given below.

Example 3.2

Solve the following second-order DE with initial conditions y(0) ¼ 1 and

dy(t)
dt

j t¼0 ¼ 4:

d 2 y(t)
dt 2

3 dy(t) ¼ e 4t u(t) þ

dt þ

2y(t) (3 : 7)

where u(t) is the unit step function.

S OLUTION
Since the input is an exponential function, the homogenous solution will be
another exponential, that is,

¼ Ae 4t y h (t) (3 : 8)

Substituting Equation 3.8 into Equation 3.7 yields

16Ae 4t 12Ae 4t 2Ae 4t ¼ e 4t þ

Solving for A yields

1
6

A ¼

Therefore, the total solution is the sum of the zero-input response and the homo-
geneous solution

¼ C 1 e t þ C 2 e 2t

1

6 e 4t

y(t) þ