Page 325 - Advanced engineering mathematics

10.2 Solution of X = AX for Constant A 305

Pure water: Mixture:
3 liters/min 3 liters/min

Tank 1 Tank 2

Mixture: Mixture: Mixture:
2 liters/min 4 liters/min 1 liter/min
FIGURE 10.1 Exchange of mixtures in tanks in Example 10.8.

rate of change of x j (t) = x (t) = rate in minus rate out

j

liter
gram liter x 2
gram
= 3 · 0 + 3 ·
min liter min 10 liter

liter x 1
gram liter x 1
gram
− 2 · − 4 ·
min 20 liter min 20 liter
6 3
=− x 1 + x 2 .
20 10
Similarly, with the dimensions excluded,
4 4
x 1 x 2 x 2

x (t) = 4 − 3 − = x 1 − x 2 .
2
20 10 10 20 10

The system we must solve is X = AX with

−3/10 3/10
A = .
1/5 −2/5
The initial conditions are
x 1 (0) = 150, x 2 (0) = 50
or

150
X(0) = .
50
The eigenvalues of A are −1/10,−1/5 with corresponding eigenvalues, respectively,

3/2 −1
and .
1 1
A fundamental matrix is
−t/10 −3t/5
(3/2)e −e
(t) = −t/10 −3t/5 .
e e
The general solution is X(t) = (t)C. To solve the initial value problem, we need C so that

150 3/2 −1
X(0) = = C = C.
50 1 1

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October 14, 2010 20:32 THM/NEIL Page-305 27410_10_ch10_p295-342