MATH 2250—7-3 #36 Solution
From System (22) on page 423 in the text, we have the following equations
describing the system:
10
10
dx1
= − x1 + x3
dt
20
20
dx2
10
10
= x1 − x2
dt
20
50
dx3
10
10
= x2 − x3
dt
50
20
x1 (0) = 18, x2 (0) = 0, x3 (0) = 0
or


−0.5
0
0.5

0
0 
x =  0.5 −0.2
 x,
0
0.2 −0.5


18

x(0) =  0 

0
We first find the eigenvalues:
−0.5 − λ
0
0.5
0.5
−0.2 − λ
0
0
0.2
−0.5 − λ
=0
Using a cofactor expansion across the top row, we get
(−0.5 − λ) −0.2 − λ
0
0.2
−0.5 − λ
+ (0.5) 0.5 −0.2 − λ
0
0.2
(−0.5 − λ)(−0.2 − λ)(−0.5 − λ) + (0.5)(0.2)(0.5) = 0
−λ3 − 1.2λ2 − 0.45λ − 0.05 + 0.05 = 0
λ(λ2 + 1.2λ + 0.45) = 0
The roots of this equation are
λ = 0,
λ = −0.6 ± 0.3i
For λ = 0, the eigenvector equation is





−0.5
0
0.5
a
0





0 
 0.5 −0.2
 b  =  0 
0
0.2 −0.5
c
0
=0
which is solved by the eigenvector (2, 5, 2).
For λ = −0.6 + 0.3i, the eigenvector equation is





0.1 − 0.3i
0
0.5
a
0





0.5
0.4 − 0.3i
0

 b  =  0 
0
0.2
0.1 − 0.3i
c
0
This becomes easier to deal with if we multiply both sides by 10, which leaves
us with





1 − 3i
0
5
a
0





5
4 − 3i
0 

 b  =  0 
0
2
1 − 3i
c
0
By design, the determinant of the matrix is zero and we should have one
free variable. That means we can fix one element of the unknown vector to
a convenient value. For example, we can set a = 1 + 3i, so that the top row
equation is (1 − 3i)(1 + 3i) + 5c = 0, or 10 + 5c = 0, so c = −2. Then both
the second and the third row confirm that b = 1 − 3i. So our eigenvector is
(1 + 3i, 1 − 3i, −2).
To find the real solutions associated with the complex eigenvalues, we
start with


1 + 3i

 −0.6t
(cos(0.3t) + i sin(0.3t))
 1 − 3i  e
−2
which gives us




3 cos(0.3t) + sin(0.3t)
cos(0.3t) − 3 sin(0.3t)



−0.6t 
e
 cos(0.3t) + 3 sin(0.3t)  + i  −3 cos(0.3t) + sin(0.3t) 
−2 sin(0.3t)
−2 cos(0.3t)
The real part and imaginary part are two linearly independent solutions.
Now our general solution is






3 cos(0.3t) + sin(0.3t)
2
cos(0.3t) − 3 sin(0.3t)




−0.6t 
−0.6t 
x = c1  5 +c2 e
 −3 cos(0.3t) + sin(0.3t) 
 cos(0.3t) + 3 sin(0.3t) +c3 e
2
−2 cos(0.3t)
−2 sin(0.3t)
The initial conditions give us








2
1
3
18







x(0) = c1 
 5  + c2  1  + c3  −3  =  0 
2
−2
0
0
Solving this gives c1 = 2, c2 = 2, and c3 = 4.
Putting everything together, our final solution is
x1 (t) = 4 + e−0.6t [14 cos(0.3t) − 2 sin(0.3t)]
x2 (t) = 10 + e−0.6t [−10 cos(0.3t) + 10 sin(0.3t)]
x3 (t) = 4 + e−0.6t [−4 cos(0.3t) − 8 sin(0.3t)]
As t → ∞, the time-dependent functions go to zero (because the coefficient in the exponent is negative), so in the limit there will be 4, 10, and 4
pounds of salt in tanks 1, 2, and 3, respectively.