Page 317 - Advanced engineering mathematics

10.1 Linear Systems 297

EXAMPLE 10.1
Consider the system

1 −4

X = X.
1 5
It is routine to verify by substitution that
3t 3t
−2e (1 − 2t)e
1 (t) = 3t and 2 (t) = 3t
e te
are two solutions. These are linearly independent on the entire real line, since neither is a constant
multiple of the other, for all t.
A third solution is
3t
(−5 − 6t)e
3 (t) = 3t .
(4 + 3t)e
However, these three solutions are linearly dependent, since, for all real numbers t,

3 (t) = 4 1 (t) + 3 2 (t).
There is a test for linear independence of n solutions of an n × n homogeneous system
X = AX.

THEOREM 10.2 Test for Independence of Solutions

Suppose that

⎛ ⎞ ⎛ ⎞ ⎛ ⎞
ϕ 11 (t) ϕ 12 (t) ϕ 1n (t)
ϕ 21 (t) ϕ 22 (t) ϕ 2n (t)
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
. . .
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
. ⎟, 2 (t) = ⎜ . ⎟,··· , n (t) = ⎜ . ⎟
1 (t) = ⎜
⎝ . ⎠ ⎝ . ⎠ ⎝ . ⎠
ϕ n1 (t) ϕ n2 (t) ϕ nn (t)
are n solutions of X = AX on an open interval I.Let t 0 be any number in I. Then

1. 1 , 2 ,··· , n are linearly independent on I if and only if 1 (t 0 ), 2 (t 0 ),··· , n (t 0 ) are
n
linearly independent, when considered as vectors in R .
2. 1 , 2 ,··· , n are linearly independent on I if and only if

ϕ 11 (t 0 ) ϕ 12 (t 0 ) ··· ϕ 1n (t 0 )

ϕ 21 (t 0 ) ϕ 22 (t 0 ) ··· ϕ 2n (t 0 )

. . .

= 0.
. . .

. . ··· .

ϕ n1 (t 0 )ϕ n2 (t 0 ) ··· ϕ nn (t 0 )

Conclusion (2) is an effective test for linear independence of solutions of the homoge-
neous system on an interval. Evaluate each solution at any number t 0 in the interval and
form the n × n determinant having j (t 0 ) as column j. We may choose t 0 in the interval to
suit our convenience (to make this determinant as easy as possible to evaluate). If this deter-
minant is nonzero, then the solutions are linearly independent; otherwise the solutions are
linearly dependent. This is similar to the Wronskian test for second order linear differential
equations.
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October 14, 2010 20:32 THM/NEIL Page-297 27410_10_ch10_p295-342