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MATB 143 Test2 Sem2 1314
Transcript of MATB 143 Test2 Sem2 1314
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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
= +