Problem 1. Solve the initial value problem:
x′1 = 4x1 + x2 + 4x3
x′2 = x1 + 7x2 + x3
x′3 = 4x1 + x2 + 4x3
x1(0) = 1
x2(0) = 2
x3(0) = 3
or


 
′
x
4 1 4 x1
 1


 
−
⇀
⇀
′
x = x′2 = 1 7 1 x2 = A−
x
x′3
4 1 4 x3
 

1
 
⇀
−
x (0) = 2
3
To find the eigenvalues of A:
4 − λ
1
4 7−λ
1 1
4
1
4 − λ
7 − λ
1 1 1 7 − λ 1
= (4 − λ) + 4
− 1
4
4 4 − λ
1
1 4 − λ
= −λ(λ − 6)(λ − 9)
So the eigenvalues are λ = 0, 6, 9.
To find the associated eigenvectors:
λ=0

4 1 4

1 7 1
4 1 4


1 0 1
0


0 ↔ 0 1 0
0
0 0 0







−1
−t
0


 

0 →  0  = t  0 
1
t
0
λ=6

−2 1 4
1 1
4 1 −2

 1


0
1 0 −1


0 ↔ 0 1 2
0
0 0 0


0
t
1





0 → −2t = t −2
0
t
1
λ=9

−5 1
4

 1 −2 1
4
1 −5


0
1 0 −1


0 ↔ 0 1 −1
0 0 0
0


 
 
0
1
t
 
 

0 → t = t 1
1
t
0
So we get solutions:


−1


λ = 0 ↔  0  e0t

1
1



λ = 6 ↔ −2 e6t
1
 
1
 
λ = 9 ↔ 1 e9t
1
And general solution:




 
−1
1
1



 6t
  9t
−
⇀
x = c1  0  + c2 −2 e + c3 1 e
1
1
1

e6t
e9t


−1
c1
 

=  0 −2e6t e9t c2
1
e6t
e9t c3
Now, plugging in initial values gives:
 

1
e0

1
1


1
1 0 0


2 ↔ 0 1 0
0 0 1
3
1 1
−2 1
1 1
−1

→ 0

So solution is






 
1
1
c1
 
 

0 → c2 = 0
c3
2
2

9t
2e
−1
1
1
−1


 
 




⇀
−
x = 1  0 +0 −2 e6t +2 1 e9t =  0 +2e9t 
1
1
1
1
2e9t




 

9t
−1 + 2e


=  2e9t 
1 + 2e9t

that is:

c1
−1 1 1 c1




 
−2e0 e0 c2 =  0 −2 1 c2
1
1 1 c3
e0
e0 c3
−1

 
−
⇀
2 = x (0) =  0
3
e0
x1 = −1 + 2e9t
x2 = 2e9t
x3 = 1 + 2e9t



Problem 2. Solve the initial value problem (this
one goes with the three tanks we did in class).
1
1
′
x3
x1 = − x1 + (0)x2 +
10
10
1
1
x1 + − x2 + (0)x3
x′2 =
10
10
1
1
x′3 = (0)x1 +
x2 + − x3
10
10
x1(0) = 10
x2(0) = 5
x3(0) = 10
or
1
− 10
x′1
 1
 ′
⇀
−
′
x =  x2  = 
 10
′
x3
0



0
1
− 10
1
10

 
1
10  x1
⇀
−
0 
 x2 = A x
1
x3
− 10

10
 
⇀
−
x (0) =  5 
10
To find the eigenvalues of A:
1
1
0
− 10 − λ
10
1
1
−
−
λ
0
10
10
1
1
0
−
−
λ
10
10
=−
λ
(3 + 30λ + 100λ2 )
100
Using the quadratic formula, we get eigenvalues
are
√
−3 ± 3i
.
λ = 0,
20
To find the associated eigenvectors:
λ=0
1
− 10
 1

 10

0
0
1
− 10
1
10
1
10

0
1 0 −1


↔
0
0 1 −1

0 0 0
0
0
1
− 10
 


0

0
0
 
t
1
 
 
→ t = t 1
t
1
√
−3 − 3i
λ=
20
√
−3−
1
3i
−
−
20
 10

1

10


0

1 0

↔
0 1
0 0
√
1− 3i
2√
1+ 3i
2
0
0

1
10
√
1 − −3− 3i
− 10
20
0
1
10
√
1 − −3− 3i
− 10
20
0
0


0
√ 
√


1−
1−
3i
3i
−
−
0
t
2√
2√ 



 →  1+ 3i  = t  1+ 3i 
− 2 
− 2 t 
0

0

t
1
So the solutions are:
 
1
 
λ = 0 ↔ 1 e0t
1
√ 
1−
3i
√
√
−
2√  −3+ 3i

−3 − 3i
 e 20 t
1+
3i
↔
λ=
− 2 
20

1
√ 
3i
1−
−
2√  −3


e 20
= − 1+ 3i 

2

1
√ 
1−
3i
−
2√  −3


e 20
= − 1+ 3i 

2

√ !
√ !!
3
3
cos −
t + i sin −
t
20
20
cos
1
√
√ !!
3
3
t − i sin
t
20
20
!
√ √ 
√ 3 t − i sin
3t
1− 3i cos
−


2
20
20
√ √  −3 t

√  e 20
=
3 t − i sin
3t

− 1+ 3i cos


2
20
20

1
√ 
1
3
3
3
− 2 cos 20 t + 2 sin 20 t 

√ √ 
√
 1
 −3 t
3
3
3

=  − 2 cos 20 t − 2 sin 20 t 
 e 20


√ 

3
cos 20 t

 √
√
√
√ 
√ 3 cos − 3 t + 1 sin
3t
 2

20
2
20
 √
√ √ 

 −3 t
1
3
3
3

+i − 2 cos − 20 t + 2 sin 20 t 
 e 20


√ 

− sin 203 t
The real and imaginary part of these functions are
linearly independent.
So,the general solution is the arbitrary sum of the
linearly independent pieces:
 
1
 
⇀
−
x = c1 1
1

√
√
√

√

1
3
3
3
− 2 cos 20 t + 2 sin 20 t 

√ √ 
√
−3
 1
3 t − 3 sin
3t 
 e 20 t
+c2 
−
cos
 2

20
20


√ 2


3
cos 20 t
 √
√
3
3
3
1
 2 cos − 20 t + 2 sin 20 t 
 √
√ √ 
 −3 t

3
3
3
1

+c3 − 2 cos − 20 t + 2 sin 20 t 
 e 20


√ 

− sin 203 t
Now plugging in initial values

 
 
√
1 −1/2
10
√3/2  c1

 
 5  = 1 −1/2 − 3/2 c2
c3
10
1
1
0
√
1 −1/2
√3/2

↔ 1 −1/2 − 3/2
1
1
0


1 0 0
10


5↔
0 1 0
0 0 1
10


25
3 
5 
3 
√5
3
5 , and C = √5 , that is, the
So C1 = 25
,
C
=
2
3
3
3
3
solution is:
 
1
25
 
⇀
−
x =
1
3 1

√
√
√

1
3
3
3
− 2 cos 20 t + 2 sin 20 t 

√ √ 
√
 −3 t
5
1
3
3
3

+  − 2 cos 20 t − 2 sin 20 t 
 e 20
3

√ 

cos 203 t
 √
√
√

3
3
3
1
 2 cos 20 t + 2 sin 20 t 
 √
√ √ 
 −3 t
5 
3
3
3
1

+ √ − 2 cos 20 t + 2 sin 20 t 
 e 20
3

√ 

3
− sin 20 t
or, much simplified:
25
3
25
x2 =
3
25
x3 =
3
x1 =
√ !
√ !!
√
5 3
3
3
5
+
cos
t +
sin
t
3
20
3
20
√ !
10 − 20 t
3
3
cos
−
e
t
3
20
√ !
√ !!
√
3
3
5
5 3
− 20
+e 3t
cos
t −
sin
t
3
20
3
20
− 20
e 3t
I probably wouldn’t have guessed that. Of course,
we see in the limit that each tank has 25/3 pounds
in it. Graphs of the three concentrations with respect to time are:
Sorry the graph is a little blurry: my image editing
software has run amock!