Topic: Chapter 8: System of Linear 1st-Order_Matrix Approach

Example 1: Solve

4 2dx x ydt

= +

5 22

dyx y

dt= +

Solution

Let 4 2

,

5 2 2x

y

= =

X A

Hence, the system become

4 25 2 2

x xdy ydt

=

( )( )4 2 54 2 2 05 2 2 2

= = =

A I

2 2 3 0 + = 1 21 , 3 = =

For 1 1 = 1 *

15 1 22

25

4 1 2 5 2 1 2 5 1 2 55 2 2 1 5 2 1 1 2 5 0 0

R R RR

+

1 22 05

k k =

1 1

2 2;

5 5te

= =

K X

For 2 3 = ( )

( )*

1 1 22

25

4 3 2 1 2 1 2 1 25 2 2 3 5 2 5 1 2 0 0

R R RR

+

1 22 0k k =

32 2

2 2;

1 1te

= =

K X

Therefore, 31 22 25 1

t tc e c e

= +

X

Example 2: Solve

3dx x ydt

=

9 3dy x ydt

=

Solution

Let 3 1 3 1

,

9 3 9 3x x xdy y ydt

= = =

X A

( )( ) ( )3 1 3 3 9 1 09 3

= = =

A I

1 2 0 = =

For 0 : = 3 1 3 19 3 0 0

01 1

1 1 1;

3 3 3te

= = =

K X

Since there exist one eigenvector, let 12

pp

=

P

( ) =A I P K ( )0 =A I P K 3 1 1 3 1 19 3 3 0 0 0

Infinitely many solution

Hence, 1 23 1p p =

1 21 13 3

p p= + 0 021 1 1 1 1

,

2 3 3 3 3t tte e t

= = + = +

P X

Choose 2 2p = 1 1p =

Therefore,

1 2

1 1 12 3 3

c c t

= + +

X

Example 3: Solve

6dx x ydt

=

5 2dy x ydt

= +

Solution

Let 6 1 6 1

,

5 2 5 2x x xdy y ydt

= = =

X A

( )( ) ( )6 1 6 2 5 1 05 2

= = =

A I

2 8 17 0 + = 4 i =

For 4 i = + ( )

( )( ) *1 1 226 4 1 2 1 5 2 5 2

5 2 4 5 2 5 2 0 0i R R Ri i i i

i i i+

+ + =

+

( )1 25 2 0k i k+ = Choose 2 5k =

( )1 2

25

ik k

+ = 1 2k i= +

( ) ( )1 1 1 2 12 2 1Re ; Im5 5 0i+

= = = = =

K B K B K

4 41 2

2 1 1 2cos sin ; cos sin

5 0 0 5t tt t e t t e

= = +

X X

Therefore,

41 1 2 2 1 2

2cos sin cos 2sin5cos 5sin

tt t t tc c c c et t

+ = + = +

X X X

Example 4: Solve the IVP ( ) ( )16 0 , 0 2 , 0 2y y y y + = = =

Solution Let 1x y= ; 1 2x y x = = ; ( )1 0 2x = 2x y= ; 2 116x y x = = ; ( )2 0 2x =

1 1 1

2 2 2

0 1 0 1;

16 0 16 0x x xdx x xdt

= = =

X A

21 16 016

= = + =

A I

4 i =

For 4 i =

( ) 1 1 244 1 16 4 16 416 4 16 4 0 0

i R R Ri i ii i

( )1 216 4 0k i k+ = 1 2

14

k k=

( ) ( )1 1 1 2 10 1Re Im4 1 0i

= = = = =

K B K B K

( )1 1 2 0 1cos sin cos 4 sin4 0tt t e t t = =

X B B

( )2 2 1 1 0cos sin cos 4 sin 40 4tt t e t t = + = +

X B B

1 1 2 2 1 2sin 4 cos 4

4cos 4 4sin 4t t

c c c ct t

= + = +

X X X

1 2 1 2

2 0 1 1, 2

2 4 0 2c c c c

= + = =

sin 4 cos 41 2

4cos 4 4sin 42t t

t t

=

X

Since 11

sin 4 2cos 42

x y t t= = +

Example 5: Consider two tanks A and B, as shown in Figure 1. Suppose Tank A and B contain 50 gallons of water. Initially, 25 pounds of salt is dissolved in tank A. Liquid is pumped into and out of the tanks as indicated in the figure; the mixture exchanged between the two tanks and the liquid pumped out of tank B are assumed to be well-stirred. (a) Use the information given in the figure to construct a mathematical model for the

number of pounds of salt ( )1x t and ( )2x t at time t in tanks A and B, respectively. (b) Solve using systematic elimination method.

Solution

(a) Tank A ( ) ( ) ( )2 210 lb/gal 3 gal/min lb/gal 1 gal/min50 50in

xR x = + =

( ) ( )1

22lb/gal 4 gal/min

50 4 4 25outxR x

t

= = +

( )1 2 2 11 2 , 0 2550 25in outdx R R x x xdt

= = =

Tank B

( )1 12lb/gal 4 gal/min50 25inxR x = =

( ) ( )1

22lb/gal 4gal/min

50 4 4 25outxR x

t

= = +

( )2 1 2 22 2 , 0 025 25in outdx R R x x xdt

= = =

(b) Let 1 1 12 2 2

2 25 1 50 2 25 1 50;

2 25 2 25 2 25 2 25x x xdx x xdt

= = =

X A

22 25 1 50 2 1 02 25 2 25 25 625

= = =

A I

1 21 3

,

25 25 = =

For 11

:25

=

1 1 2

2522

251 25 1 50 1 1 2 1 1 22 25 1 25 1 1 2 0 0

R R RR

+

125

1 1

1 1,

2 2t

e

= =

K X

For 23

:25

=

1 1 2

2522

251 25 1 50 1 1 2 1 1 22 25 1 25 1 1 2 0 0

R R RR

+

325

2 2

1 1,

2 2t

e

= =

K X

Hence, the general solution is

3125 25

1 2

1 12 2

t tc e c e

= +

X

At 0,t =

1 2 1 2

25 1 1 25 25,

0 2 2 2 2c c c c

= + = =

( ) ( )3 31 125 25 25 251 225 25 , 25 252 2t t t t

x t e e x t e e = + =

Example 6 Solve

2 3 7dx x ydt

= +

2 5dy x ydt

= +

Solution

Let ( )2 3 7 2 3 7, ,1 2 5 1 2 5

x x xdt

y y ydt

= = = = +

X A F

2

1 2

2 31 0 1, 1

1 2

= = = = =

A I

For 1 1: =

*

1 21 1

1 3 1 3 3 31 3 0 0 1 1

R R te+

= =

K X

For 2 1: =

113 1 2

2 2

3 3 1 1 1 1 1 11 1 1 1 0 0 1 1

R R R te

= =

K X

Thus,

1 2

3 11 1

t tc c e c e

= +

X

To find particular solution, we can use either method of undetermined coefficient or variation of parameter.

Undetermined Coefficient

Since ( )tF are constant, assuming p AB

=

X

2 3 7 2 3 71, 3

1 2 5 1 2 5A A Ad A BB B Bdt

= + = = =

Hence,

13p

=

X

Therefore,

1 2

3 1 11 1 3

t tc p c e c e

= + = + +

X X X

Variation of Parameter

( ) 3t t

t t

e et

e e

=

1 13 1 3

t t

t t

e e

e e

=

+

1 72 252 3 2 4

t t t

t t t

e e e

e e e

= =

F

( ) ( ) ( )1 34

t t t

p t t t

e e et t t dt dt

e e e

= =

X F

34

13

t t t

t t t

e e e

e e e

=

=

Therefore,

1 2

3 1 11 1 3

t tc p c e c e

= + = + +

X X X

Example 7 Solve

4 1 3 39 6 10

te

= +

X X

Solution

( )( ) ( )( )24 1 3 14 6 9 10 21 3 7 09 6 3

= = = + = =

A I

1 23 ; 7 = =

When 1 3: =

1 29 31 1

1 1 3 1 1 3 1 19 3 0 0 3 3

R R te +

= =

K X

When 2 7 : =

113 1 29 7

2 2

3 1 3 1 1 9 1 1 9 1 19 1 9 1 0 0 9 9

R R R te

+ = =

K X

Hence, the complimentary solution:

3 71 1 2 2 1 2

1 13 9

t tc c c c e c e

= + = +

X X X

Undetermined Coefficient

Assume ;t t

p pt t

Ae AeBe Be

= =

X X

Substitute pX and pX into the system, we obtain

4 1 3 39 6 10

t t t

t t t

Ae Ae eBe Be e

= +

( )

14 33

9 6 10

t t

t t

e A B Ae

e A B Be

+ =

+ + =

5536

194

A

B

=

=

Therefore,

3 71 2

1 1 55 363 9 19 4

t t tc p c e c e e

= + = + +

X X X

Variation of Parameter

( ) ( )3 7

3 7

3,

3 9 10

t t t

t t t

e e et t

e e e

= =

F

3 37 7 3 11 4 2

10 10 7 73 3 1 14 12

919 3 3

t tt t

t t t tt t

e ee e

e e e ee e

= =

( ) ( ) ( )3 33 7 3 1

1 4 127 73 7 1 1

4 12

33 9 10

t tt t t

p t tt t t

e ee e et t t dt dt

e ee e e

= =

X F

3 7 3712

3 7 112

23 7 37 5524 36

63 7 19172 4

3 9

3 9

tt t

tt t

t tt t

t tt t

ee e dtee e

e ee e

e ee e

=

= =

Therefore,

3 71 2

1 1 55 363 9 19 4

t t tc p c e c e e

= + = + +

X X X