MAT351T_WR2_2014A

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QUESTION 1 [25]

1.1 Derive the Laplace transform of the function 2( ) tf t e from first principles. (3)

2

0

( ) st tF s e e dt

2

0

s te dt

2

0

1

2

s te

s

1 1

02 2s s

1.2 Determine the inverse Laplace transform of 2( )

4 1

ss eF s

s s

(5)

2

2 2( )

2 5

ss

F s es

2 1 2 12

5( ) cosh 5 1 sinh 5 1 1

t tf t e t e t H t

1.3 Consider the following second order linear differential equation

'' 4 8y y t with initial conditions 0 2y and ' 0 1.y

1.3.1 Use Laplace transforms to solve the differential equation. (8)

2 0s Y s s y

2

' 0y

1

2

1 84Y s

s s

3 2

2

2 2

8 1 2 8 14 2 1

s s s sY s s s

s s

3 2

2 2

2 8 1

4

s s sY s

s s

or

2 2 2

8 1 2 1

4 4

s s

s s s

31

4 4

2 2

02

4

s

s s s

(Partial fractions)

1 14 4

2 2 2

22 2 1

4 4

s s

s s s s

314 8

2 sin 2y t t t

MAT351T_WR2_2014A

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1.3.2 Use the method of undetermined coefficients to solve the equation given in 1.3:

'' 4 8y y t with initial conditions 0 2y and ' 0 1.y (9)

Complimentary function: 2 4 0m 2m j cos2 sin 2cy A t B t

Particular integral: y Ct D 0 4 4 8Ct D t

'y C 1

4C and 2D

'' 0y

14

2py t

General solution: 14

cos2 sin 2 2y A t B t t

' 14

2 sin 2 2 cos2y A t B t

0 2 :y 2 0 0 2 0A A

' 0 1y : 314 8

1 0 2B B

Particular solution: 3 18 4sin 2 2y t t

QUESTION 2 Linear Algebra [15]

2.1 Determine the eigenvalues and corresponding eigenvectors of matrix 3 5

1 1A

(5)

Eigenvalues 3 5

01 1

0 3 1 5

20 4 8

2 2j

Eigenvectors for 1 2 2j

1

2

3 2 2 5 0

1 1 2 2 0

j k

j k

OR

1

2

1 2 5 0

1 1 2 0

kj

kj

1 21 2k j k and 1

1 2

1

jK

or 1 21 2 5j k k and 1

5

1 2K

j

for 2 2 2j and 2

1 2

1

jK

or for 2 2 2j and 1

5

1 2K

j

MAT351T_WR2_2014A

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2.2 Consider the following system of differential equations

2

x x y z

y y

z y z

with solution vectors 1

1

0

0

tX e

, 2

1

0

2

tX e

and 2

3

4

3

1

tX e

2.2.1 Use the information given to determine the eigenvalues of the system (no calculations). (1)

1 1 , 2 1 and 3 2

2.2.2 Prove that the solution vectors are linearly independent. (3)

2

2

2 2

2

2

40 3

0 0 3 0 6 6 02

0 2

t t t

t

t t t t t

t t

t t

e e ee

W e e e e ee e

e e

Therefore the solution vectors are linearly independent.

2.2.3 Write the general solution of the system as three separate equations. (2)

2

1 2 3

1 1 4

0 0 3

0 2 1

t t tX c e c e c e

therefore

2

1 2 3

2

3

2

2 3

4

3

2

t t t

t

t t

x t c e c e c e

y t c e

z t c e c e

2.2.4 Determine the particular solution of the system if

6

0 6

0

X

(4)

2

1 2 34t t tx t c e c e c e 0 6 :x 1 16 1 4 2 3c c

2

33 ty t c e 0 6 :y 3 36 3 2c c

2

2 32 t tz t c e c e 0 0 :z 2 20 2 2 1c c

Particular solution: 2

1 1 4

3 0 1 0 2 3

0 2 1

t t tX e e e

Total [40]