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298 CHAPTER 10 Systems of Linear Differential Equations

EXAMPLE 10.2
In Example 10.1,
3t 3t
−2e (1 − 2t)e
1 (t) = 3t and 2 (t) = 3t
e te
for all t. Evaluate these at some convenient point, say t = 0:

−2 1
1 (0) = and 2 (0) = .
1 0
Usethese2− vectors as columns of a 2 × 2 determinant:

−2 1

=−1 = 0.
1 0

Therefore 1 and 2 are linearly independent on the real line. In this case this conclusion is
obvious without the determinant test, but this is not always the case.
Proof of Theorem 10.2 Conclusion (2) follows from (1) by the fact that a determinant is zero
exactly when its columns are linearly dependent.
To prove conclusion (1), let t 0 be in I. Suppose first that 1 ,··· , n are linearly dependent
on I. Then one of these solutions is a linear combination of the others. By relabeling if necessary,
suppose 1 is a linear combination of 2 ,··· , n . Then there are numbers c 2 ,··· ,c n such that
1 (t) = c 2 2 (t) + ··· + c n n (t)
for all t in I. In particular, this holds at t = t 0 , hence the vectors
1 (t 0 ),··· , n (t 0 )

are linearly dependent.
n
Conversely, suppose 1 (t 0 ),··· , n (t 0 ) are linearly dependent in R . Then one of these
vectors is a linear combination of the others. Again, suppose for convenience that the first is a
combination of the others:
1 (t 0 ) = c 2 2 (t 0 ) + ··· + c n n (t 0 ).

Define
(t) = 1 (t) − c 2 2 (t) − ··· − c n n (t)
for t in I. Then (t) is a linear combination of solutions, hence it is a solution of the system.
Furthermore,
0
⎛ ⎞
0
⎜ ⎟
(t 0 ) = ⎜.⎟.
⎜ ⎟
.
⎝ . ⎠
0
Therefore, (t) is a solution of the initial value problem

X = AX;X(0) = O.
But the zero function (t) = O is also a solution of this problem. By the uniqueness of the
solution of this initial value problem (Theorem 10.1),
(t) = (t) = O = 1 (t) − c 2 2 (t) − ··· − c n n (t)

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October 14, 2010 20:32 THM/NEIL Page-298 27410_10_ch10_p295-342