Page 170 - Numerical Analysis Using MATLAB and Excel

Solution of the Homogeneous ODE

s t s t s t s t
y = k e , 1 1 y = k e , 2 2 y = k e , 3 3 … , y = k e n (5.24)
3
1
2
n
n
Case I − Distinct Roots
If the roots of the characteristic equation are distinct (different from each another), the solutions
n
of (5.23) are independent and the most general solution is:
s t s t s t
y = k e 1 + k e 2 + … + k e n (5.25)
2
n
N
1
FOR DISTINCT ROOTS
Case II − Repeated Roots
If two or more roots of the characteristic equation are repeated (same roots), then some of the
terms of (5.24) are not independent and therefore (5.25) does not represent the most general solu-
tion. If, for example, s = s 2 , then,
1
s t s t s t s t s t s t
k e 1 + k e 2 = k e 1 + k e 1 = ( k + k ) 2 e 1 = k e 1
2
3
1
1
2
1
and we see that one term of (5.25) is lost. In this case, we express one of the terms of (5.25), say
s t s t
1
1
k e as k te . These two represent two independent solutions and therefore the most general
2
2
solution has the form:
s t s t s t
)
y = ( k + k t e 1 + k e 3 + … + k e n (5.26)
3
1
2
n
N
If there are equal roots the most general solution has the form:
m
1
2
n
y = ( k + k t + … + k t m – 1 ) e s t + k n – i e s t + … + k e s t (5.27)
2
N
1
m
n
FOR M EQUAL ROOTS
Case III − Complex Roots
If the characteristic equation contains complex roots, these occur as complex conjugate pairs.
β
α
Thus, if one root is s = – α + jβ where and are real numbers, then another root is
1
–
s = – α jβ . Then,
1
s t s t – αt + jβt – αt jβt – αt jβt j – βt
–
2
1
k e + k e = k e + k e = e ( k e + k e )
2
1
1
1
2
2
= e – αt ( k cos βt + jk sin βt + k cos βt jk sin βt )
–
2
2
1
1
– αt
β
(
e (= k + k ) [ 2 cos β t + j k – k ) 2 sin t ]
1
1
)
= e – αt ( k cos βt + k sin βt = e – αt k cos ( βt + ϕ )
4
3
5
FOR TWO COMPLEX CONJUGATE ROOTS
(5.28)
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−9
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