Page 331 - Advanced engineering mathematics

10.2 Solution of X = AX for Constant A 311

This is the system
⎛ ⎞ ⎛ ⎞
0 −1 −5 −11
⎝ 25 −5 E 3 =
0 ⎠ ⎝ −4 ⎠
0 1 5 1
with general solution
⎛ ⎞
(1 − 25α)/25
E 3 = ⎝ 1 − 5α ⎠ .
α
Let α = 1 to get
⎛ ⎞
−24/25
E 3 = ⎝ −4 ⎠ .
1

A third solution is
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
−1 −1 −24/25
1
2 −2t
3 (t) = ⎝ −5 ⎠ t e + −4 ⎠ te −2t + ⎝ −4 ⎠ e −2t
⎝
2
1 1 1
⎛ 2 ⎞
−24/25 − t − t /2
2 −2t
= ⎝ −4 − 4t − 5t /2 ⎠ e .
2
1 + t + t /2
We now have three linearly independent solutions and can use these as columns of the
fundamental matrix
⎛ ⎞
−e −2t (−1 − t)e −2t (−24/25 − t − t /2)e −2t
2
2
(t) = −5e −2t (−4 − 5t)e −2t (−4 − 4t − 5t /2)e −2t ⎠ .
⎝
2
e −2t (1 + t)e −2t (1 + t + t /2)e −2t
The general solution is X(t) = (t)C.
These examples suggest a procedure to follow. Suppose we know the eigenvalues of A.If
these are all distinct, the corresponding eigenvectors are linearly independent and we can write
the general solution.
Thus, suppose an eigenvalue λ has multiplicity k > 1. If there are k linearly inde-
pendent solutions associated with λ, then we can produce k linearly independent solutions
corresponding to λ.
If λ only has r linearly independent associated eigenvectors and r < k, we need from λ a
total of r − k more solutions linearly independent from the others.
If r − k = 1, we need one more solution, which can be obtained as in Example 10.10.
If r − k = 2, proceed as in Example 10.11 to find another linearly independent solution.
If r − k = 3, follow the pattern of the previous cases, trying
1 1
3 λt
λt
2 λt
4 (t) = E 1 t e + E 2 t e + E 3 te + E 4 e λt
3! 2
where E 1 ,E 2 , and E 3 were found in generating preceding solutions.
If r − k = 4, try
1 1 1
2 λt
λt
λt
3 λt
4 λt
5 (t) = E 1 t e + E 2 t e + E 3 t e + E 4 te + E 5 e .
4! 3! 2
This process must be continued until k linearly independent solutions have been obtained
associated with the eigenvalue λ.
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October 14, 2010 20:32 THM/NEIL Page-311 27410_10_ch10_p295-342