Page 169 - Numerical Analysis Using MATLAB and Excel

Chapter 5 Differential Equations, State Variables, and State Equations

solution, will be referred to as the natural response, and will be denoted as y () or simply y N . The
t
N
particular solution of a non-homogeneous ODE will be referred to as the forced response, and will
be denoted as y t() or simply y F . Accordingly, we express the total solution of the non-homoge-
F
neous ODE of (5.12) as:

yt() = y Natural + y Forced = y + y F (5.18)
N
Response Response
The natural response y N contains arbitrary constants and these can be evaluated from the given

initial conditions. The forced response y F , however, contains no arbitrary constants. It is impera-
tive to remember that the arbitrary constants of the natural response must be evaluated from the
total response.

5.4 Solution of the Homogeneous ODE

Let the solutions of the homogeneous ODE

n
d y d n – 1 y dy
a --------- + a n – 1 ---------------- + … + a ------ + a y = 0 (5.19)
1
0
n
dt n dt n – 1 dt
be of the form
y = ke st (5.20)
Then, by substitution of (5.20) into (5.19) we obtain

–
st
n st
e +
a ks e + a n – 1 ks n1 st … + a kse + a ke st = 0
n
1
0
or
n
( a s + a n – 1 s n – 1 + … + a s + a ) 0 ke st = 0 (5.21)
1
n
We observe that (5.21) can be satisfied when
n
( a s + a n – 1 s n – 1 + … + a s + a ) 0 = 0 or k = 0 or s = – ∞ (5.22)
1
n
but the only meaningful solution is the quantity enclosed in parentheses since the latter two yield
trivial (meaningless) solutions. We, therefore, accept the expression inside the parentheses as the
only meaningful solution and this is referred to as the characteristic (auxiliary) equation, that is,
n
a s + a n1 s n – 1 + … + a s + a = 0
n
0
–
1
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (5.23)
Characteristic Equation
Since the characteristic equation is an algebraic equation of an nth-power polynomial, its solutions
are s s s … s, 1 2 , 3 , , n , and thus the solutions of the homogeneous ODE are:

5−8 Numerical Analysis Using MATLAB® and Excel®, Third Edition

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