Page 334 - Advanced engineering mathematics

314 CHAPTER 10 Systems of Linear Differential Equations

The general solution of the nonhomogeneous system is
−t 6t
5e −2e
X(t) = (t)C + (t)U(t) = −t 6t C
e e
−t 6t t
5e −2e (t + 1)e /7
+ −t 6t −6t −6t
e e (−29/252)e + (1/42)te
−t 6t
5e −2e 1 17/6 + (49/7)t
= −t 6t C + .
e e 3 1/12 + t/2
Although in this example the coefficient matrix A was constant, this is not required to apply
the method of variation of parameters.

10.3.2 Solution by Diagonalizing A
If A is a diagonalizable matrix of real numbers, then we can solve the system X = AX + G by

the change of variables X = PZ, where P diagonalizes A.

EXAMPLE 10.13
We will solve the system

33 8

X = X + 3t .
15 4e
The eigenvalues of A are 2,6, with eigenvectors, respectively,

−3 1
and .
1 1
Form P using these eigenvectors as columns:

−3 1
P = .
1 1
Then

20
−1
P AP = D =
06
with the eigenvalues down the main diagonal. Compute

−1/4 1/4
−1
P = .
1/4 3/4
Now make the change of variables X = PZ in the differential equation:

X = (PZ) = PZ = A(PZ) + G.

Then

PZ = (AP)Z + G.
Multiply this equation on the left by P −1 to get
−1
−1
Z = (P AP)Z + P G

or
−1

Z = DZ + P G.

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October 14, 2010 20:32 THM/NEIL Page-314 27410_10_ch10_p295-342