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COLLEGE OF ENGINEERING

PUTRAJAYA CAMPUS

TEST 2

SEMESTER 2, 2013 / 2014

PROGRAMME : Bachelor of Engineering (Honours)

SUBJECT CODE : MATB 143

SUBJECT : Differential Equations

DATE : 20thDecember 2013

TIME : 5.15 pm 6.30 pm (1 hours)

INSTRUCTIONS TO CANDIDATES:

1. This paper contains FOUR (4) questions in ONE (1) page.

2. Answer all questions.

3. Write all answers in the answer booklet provided.

4. Write answer to each question on a new page.

5. A formula sheet on basic derivatives and integrals is provided on Page 4.

THIS QUESTION PAPER CONSISTS OF 3 PRINTED PAGES INCLUDING THIS

COVER PAGE.

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Question 1 [15 marks]

(a) Use the reduction of order formula to find a second solution, 2 ( )y x , for the

homogeneous differential equation below.

2 212 2 0; ( )x y xy y y x x + = =

Then write the general solutionfor the problem.

[5 marks]

(b) Show that the Wronskian value for the functions below equals zero. Are these two

functions linearly independent?

( ) cos , ( ) sin ;2

f x x g x x x

= + = < <

[5 marks]

(c) Suppose 2, 2, 5 and 3 4i are roots of an auxiliary equation. Write down the

general solution ( )y x , of the corresponding homogeneous linear differentialequation if it is an equation with constant coefficients. [5 marks]

Question 2 [10 marks]

Use the superposition approachto find the general solution of the differential equation:

2 5 xy y xe + + =

Question 3 [15marks]

(a) Two roots of a cubic auxiliary equation with real coefficients are 1 2m = and

2 1m i= + . What is the corresponding homogeneous linear differential equation?

[5 marks]

(b)

Solve the Cauchy-Euler equation: 21

4 6x y xy yx

+ = . [10 marks]

Question 4 [10 marks]

A spring with spring constant 8 N/m is attached to the ceiling. A 2 kg mass is attached to the

spring. The mass is released from a point 1m above the equilibrium position with a

downward velocity of 0.04 m/s. Answer the following:

(a)

Write the initial conditions and differential equation that models the system. [3 marks]

(b)

Determine the equation of motion, ( )x t . [7 marks]

---END OF PAPER---

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Formulas for Derivatives and Integrals ( ( )u f x= , ( )v g x= , fand gare differentiable functions)

Derivatives

1. Power rule:1

( )n nd du

u n udx dx

= , n 10. cos sind du

u udx dx

=

2. Product rule: ( )d dv du

uv u vdx dx dx= + 11.2

tan secd du

u udx dx=

3. Quotient rule:2

du dvv u

d u dx dx

dx v v

=

12.2cot csc

d duu u

dx dx=

4. Chain rule:dy dy du

dx du dx= , ( )f u=

13.sec sec tan

d duu u u

dx dx=

5. ( ) 0,d

cdx

= c is any constant 14. csc csc cotd du

u u udx dx

=

6. ( )u ud du

e edx dx

= 15.1

2

1sin

1

d duu

dx dxu

=

7. ( ) lnu ud du

a a adx dx= , 0a> 16.

1

2

1

cos1

d du

udx dxu

=

8.1

lnd du

udx u dx

= 17.1

2

1tan

1

d duu

dx dxu

=+

9. sin cosd du

u udx dx

= 18.1

2

1sec

1

d duu

dx dxu u

=

Integrals

1. Integration by parts: u dv uv v du= 10. sec tan secu u du u c= +

2.1

, 11

nn u

u du c nn

+

= + +

11. csc cot cscu u du u c= +

3.1

lndu u cu

= + 12. tan ln cos ln secu du u c u c= + = +

4.u u

e du e c= + 13. cot ln sinu du u c= +

5.ln

uu aa du c

a= + , 0a> 14. sec ln sec tanu du u u c= + +

6. sin cosu du u c= + 15. csc ln csc cotu du u u c= +

7. cos sinu du u c= + 16.1

2 2

1sin

udu c

aa u

= +

8.2

sec tanu du u c= + 17.1

2 2

1 1tan

udu c

a aa u

= + +

9.2csc cotu du u c= + 18.

1

2 2

1 1sec

udu c

a au u a

= +