8/12/2019 Final Exam Set A
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA
FINAL EXAMINATION
SEMESTER II
SESSION 2012/2013
COURSE NAME : CIVIL ENGINEERING MATHEMATICS II
COURSE CODE : BFC 14003
PROGRAMME : 1 BFF, 2 BFF
EXAMINATION DATE : JUNE 2013
DURATION : 3 HOURS
INSTRUCTION : ANSWER ALL QUESTIONS IN PART A
AND THREE (3)QUESTIONS IN PART B
THIS QUESTION PAPER CONSISTS OF SIX (6) PAGES
CONFIDENTIAL
CONFIDENTIAL
8/12/2019 Final Exam Set A
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BFC 14003
2
PART A
Q1 A periodic function is defined by
2 , 0
( ) 2 , 0
xf x x x
2)( xfxf .
(a) Sketch the graph of the function over x .(2 marks)
(b) Determine whether the function is even, odd or neither.(1 marks)
(c) Show that the Fourier series of the function )(xf is
21 1
1 13 12 cos 2 sin
2
n
n nnx nx
n n
.
(17 marks)
Q2 (a) Given that !6!4!2
1cos642 xxx
x .
(i) Find the first three nonzero terms of a power series for x
xsin
.
(ii) Hence, evaluate dxx
x
1
0
sinby using the series expansion.
(6 marks)
(b) (i) Determine whether the series
1
2
nne
nconverges or diverges by
using ratio test.
(ii) Determine whether the series
1
)1(
n
n
nconverges
absolutely, converges conditionally, or diverges by using a
suitable convergence test.
(8 marks)
(c) Find the radius of convergence of0
3( 2)n
n
x
.
(6 marks)
8/12/2019 Final Exam Set A
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BFC 14003
3
PART B
Q3 (a) Solve the differential equation by using the method of separation of variables,
1 1dy
x x y
dx
.
Hence, find the particular solution when (1) 0y .
(7 marks)
(b) By using the substitution of and ,dy dv
y xv x vdx dx
find the solution of
2 2
.3
dy x y
dx xy
(6 marks)
(c) In a certain culture of bacteria, the rate of increase is proportional to the
number of present. If it is found that the number doubles in 4 hours, how many
may expected at the end of 12 hours?
(Hints:dN
kNdt
, where N denotes the number of bacteria at time, thours
and kis the proportionality factor.)
(7 marks)
Q4 (a) Use the method of variation of parameters to solve
sec tany y x x ,
which satisfies the initial conditions (0) 0y and (0) 2y .
(Hints: 1tansec22
xx ,2
sec tan .)x dx x
(10 marks)
(b) A spring is stretched 0.1 m ( )l when a 4 kg mass ( )M is attached. The
weight is then pulled down an additional 0.2 m and released with an upward
velocity of 4 m/s. Neglect damping, c. If the general equation describing the
spring-mass system is
8/12/2019 Final Exam Set A
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BFC 14003
4
0,Mu cu ku
find an equation for the position of the spring at any time t.
(Hints: weight, , 9.8, .W
W Mg g k l
)
(10 marks)
Q5 (a) Find
(i) 3 25
4.t
te t
e
(ii) 3cosh 4 ( 3) .t t t t
(iii) 3sinh(3 ) ( 3) .tt e H t
(10 marks)
(b) By using Laplace transform, solve
2 , (0) 2, (0) 3.ty y y e y y (10 marks)
Q6 (a) Find
(i)1
2 2
5 6.
4 ( 2) 4
s
s s
(ii)2
1
2
16.
( 3)( 1)
s
s s
(10 marks)
(b) Find the general solution for the second order differential equation23 2 2 5sin ,y y y x x
by using the undetermined coefficient method.
(10 marks)
8/12/2019 Final Exam Set A
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BFC 14003
5
FINAL EXAMINATION
SEMESTER / SESSION : SEM II / 2012/2013 COURSE : 1 BFF / 2 BFF
SUBJECT : CIVIL ENGINEERING MATHEMATIC II SUBJECT CODE : BFC 14003
FORMULA
Second-order Differential EquationThe roots of characteristic equation and the general solution for differential equation
.0 cyybya
Characteristic equation: .02 cbmam Case The roots of characteristic equation General solution
1. Real and different roots: 1m and 2m xmxm BeAey 21
2. Real and equal roots: 21 mmm mxeBxAy )(
3. Complex roots: im 1 , im 2 )sincos( xBxAey x
The method of undetermined coefficients
For non-homogeneous second order differential equation ( ),ay by cy f x the particular
solution is given by ( )py x :
( )f x ( )py x
1
1 1 0( ) n n
n n nP x A x A x A x A
1
1 1 0( )r n n
n nx B x B x B x B
xCe ( )r xx Pe
cos or sinC x C x ( cos sin )rx P x Q x
( ) xn
P x e 11 1 0
( )r n n xn n
x B x B x B x B e
cos( )
sinn
xP x
x
1
1 1 0
1
1 1 0
( )cos
( )sin
r n n
n n
r n n
n n
x B x B x B x B x
x C x C x C x C x
cos
sin
x x
Cex
( cos sin )r xx e P x Q x
cos( )
sin
x
n
xP x e
x
1
1 1 0
1
1 1 0
( ) cos
( ) sin
r n n x
n n
r n n x
n n
x B x B x B x B e x
x C x C x C x C e x
Note : ris the least non-negative integer (r= 0, 1, or 2) which determine such that there
is no terms in particular integral )(xyp corresponds to the complementary function
)(xyc .
The method of variation of parameters
If the solution of the homogeneous equation 0 cyybya is ,21 ByAyyc then the
particular solution for )(xfcyybya is
,21 vyuyy
where ,
)(2AdxaW
xfy
u BdxaWxfy
v )(1
and 122121
21
yyyyyy
yy
W .
8/12/2019 Final Exam Set A
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BFC 14003
6
FINAL EXAMINATION
SEMESTER / SESSION : SEM II / 2012/2013 COURSE : 1 BFF / 2 BFF
SUBJECT : CIVIL ENGINEERING MATHEMATIC II SUBJECT CODE : BFC 14003
Laplace Transform
L
0
)()()}({ sFdtetftf st
)(tf )(sF )(tf )(sF
a
s
a )( atH
s
e as
ate as
1 )()( atHatf )(sFe as
atsin 22 as
a
)( at as
e
atcos 22 as
s
( ) ( )f t t a ( )ase f a
atsinh 22 as
a
0( ) ( )
t
f u g t u du )()( sGsF
atcosh 22 as
s
( )y t )(sY
nt , ...,3,2,1n 1!ns
n ( )y t )0()( yssY
)(tfeat )( asF ( )y t )0()0()(2 ysysYs
)(tftn , ...,3,2,1n )()1( sFds
dn
nn
Fourier Series
Fourier series expansion of periodic
function with period L2
0
1 1
1( ) cos sin
2 n n
n n
n x n xf x a a b
L L
where
L
Ldxxf
La )(
10
L
Ln dx
L
xnxf
La
cos)(
1
L
Ln dx
L
xnxf
Lb
sin)(
1
Fourier half-range series expansion
110
sincos2
1)(
n nn n L
xnb
L
xnaaxf
where
L
dxxfL
a00
)(2
L
n dxL
xnxf
La
0cos)(
2
L
n dxL
xnxf
Lb
0sin)(
2
8/12/2019 Final Exam Set A
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BFC 14003
7
Marking Scheme
(Mmethod, Aanswer) Mark Total
Q1
(a)
A2 2
Q1
(b)Neither A1 1
Q1
(c) 2x T x 2T L
2 2L L
0
1( )LLa f x dx
L
0
0
12 2dx xdx
0 2
0
12 x x
2 21 2
3
1( ) cosLn L
n xa f x dx
L L
0
0
12 cos( ) 2 cos( )nx dx x nx dx
A1
M1
A1
A1
A1
M1
17
8/12/2019 Final Exam Set A
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BFC 14003
8
(Mmethod, Aanswer) Mark Total
u dv
2x cos( )nx
2 1sin( )nx
n
02
1cos( )nx
n
0
2
0
1 2 2 2sin( ) sin( ) cos( )
xnx nx nx
n n n
2 2
1 2 2cos( )nn n
1( ) sinLn L
n xb f x dx
L L
0
0
12 sin( ) 2 sin( )nx dx x nx dx
u dv
2x sin( )nx
2 1cos( )nx
n
02
1sin( )nx
n
0
2
0
1 2 2 2cos( ) cos( ) sin( )xnx nx nxn n n
2
n
2 213 1 2 2 2
( ) cos( ) cos( ) sin2 n
f x n nx nxn n n
M1
A1
A1
M1
M1
A1
A1
A1M1
8/12/2019 Final Exam Set A
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BFC 14003
9
(Mmethod, Aanswer) Mark Total
213 2 2
( 1) 1 cos( ) sin( )2
n
nnx nx
n n
21 1
1 13 12 cos 2 sin
2
n
n nnx nx
n n
A1
A1
Q2
(a) (i)
1206
!6
6
!4
4
!2
2
!6!4!21
cossin
53
53
642
xxx
xxx
xxx
dx
d
xdx
dx
So,
1206
1206sin2/92/5
53
xxx
x
xxx
x
x
M1
A1
A1
3
Q2
(a)
(ii)
.621.0
1320
2
42
2
3
2
1320
2
42
2
3
2
1206
sin
1
0
2/112/72/3
1
0
2/92/51
0
xxx
dxxx
xdxx
x
M1
A1
A1
3
Q2
(b) (i)
1
2
nne
n
2
21
2 1 2
2 2
( 1)
( 1)lim lim
1 1 1 1 lim lim 1
1 1 ; the series converges.
nn
nn n
n
n n
n
e ne
n e n
e
n
e n e n
e
M1
A1A1
3
8/12/2019 Final Exam Set A
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BFC 14003
10
(Mmethod, Aanswer) Mark Total
Q2
(b)
(ii)
5
1
4
1
3
1
2
11
)1(
1n
n
n
Use alternating series test:
(a) 5
1
4
1
3
1
2
11
(b) 01
lim nn
the series converges.
Usep-series:
2
1p 10 p
the series diverges.
Conclusion, the series converges conditionally.
M1
A1
M1
A1
A1
5
Q2
(c)
For 2x , let nn xu )2(3 .Then
|2|
|2|lim)2(3
)2(3limlim
11
x
xx
x
u
u
n
n
n
nn
n
n
Thus, the series converges absolutely if
| 2 | 1x or 1 3x .
The series diverges if 1x or 3x
Therefore, the radius of convergence of the series is 1R .
M1
M1
A1
M1
A1
A1
6
Q3
(a) 1 1
dyx x y
dx
11
dy xdx
xy
1
11
xdy dx
xy
2 1 ln(x 1) k y x , k is constant
Given (1) 0y , so
A1
M1
M2
A1
7
8/12/2019 Final Exam Set A
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BFC 14003
11
(Mmethod, Aanswer) Mark Total
ln(2) 3k
2 1 ln(x 1) ln(2) 3y x
A1
A1
Q3(b) 2 2
3
dy x y
dx xy
2 2 2
3 ( )
dv x x vx v
dx x xv
21
3
v
v
2
13
dv vx vdx v
21 2
3
v
v
2
3 1
1 2
vdv dx
v x
2
3 1
1 2
vdv dx
v x
assume21 2a v
4da
vdv
1
4da vdv
3 1 1
4da dx
a x
23 ln 1 2 ln4
v x k
23
ln 1 2 ln4
yx k
x
A1
M1
A1
M1
A1
A1
6
Q3
(c) at 0,t 0N N ; 4t , 02N N ; 12t , ?N
dN kNdt
7
8/12/2019 Final Exam Set A
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BFC 14003
12
(Mmethod, Aanswer) Mark Total
dNkdt
N
1
dN kdt N
ln | |N kt c , c is constant
ktN Ae , cA e
at 0,t 0A N
So 0ktN N e
at 4t , 40 02 kN N e 42 ke
4 ln | 2 |k
1l n | 2 |
4k
So
1ln|2|
4
0
t
N N e
at 12t , 3ln|2|0N N e
08N
At the end of 12 hours, the number of bacteria is 8 times of the
original number.
M1
A1
A1
A1
A1
M1
A1
Q4
(a)
Step 1:
1, ( ) sec tana f x x x
Step 2:2
1 2
1 2
1 0
Thus,
cos sin
cos sin
sin cos
h
m
m i
y A x B x
y x y x
y x y x
M1
A1
10
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(Mmethod, Aanswer) Mark Total
Step 3:
2 2cos sin
cos sin 1.sin cos
x xW x x
x x
Step 4:
2
2
sin sec tan
tan
1 sec
( tan )
u x x x dx
x dx
x dx
x x C
cos sec tan
tan
sin
cos
sin
sin
1
ln | cos |
v x x x dx
x dx
xdx
x
x du
u x
duu
x D
Step 5:1 2
(tan )cos ( ln | cos |)sin
cos sin sin cos sin ln | cos |
sin cos cos sin cos sin tan
cos ln | cos |
y uy vy
x x C x D x x
C x D x x x x x x
y C x D x x x x x x x
x x
When (0) 0,y
0.C When (0) 2,y
2.D
Thus,
2sin sin cos sin ln | cos | .y x x x x x x
M1A1
M1
A1
M1
A1
A1
A1
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(Mmethod, Aanswer) Mark Total
Q4
(b)
4(9.8) 39.2
39.2392
0.1
0
W mg
Wk
l
c
Thus, from 0,mu cu ku
We get 4 392 0u u or 98 0.u u
The characteristic equation is 2 98 0.m
So, 7 2 .m i
Therefore, ( ) cos(7 2 ) sin(7 2 ).u t A t B t
or ( ) cos(9.9 ) sin(9.9 ).u t A t B t
( ) 7 2 sin(7 2 ) 7 2 cos(7 2 ).u t A t B t
The spring is released after pulling it down 0.2 m.
So, we have (0) 0.2u .
Since its set in motion with an upward velocity of 4 m/s,
we have (0) 4u . (upward is negative)
These initial conditions give us the equation for the position of the
spring at any time tas
( ) 0.2cos(7 2 ) 0.4041sin(7 2 ).u t t t
or
1 2 2( ) cos(7 2 ) sin(7 2 ).
5 7u t t t
A1
A1
A1
M1
A1
A1
A1
A1
A1
A1
10
Q5(a)
(i)
3 2
5
3 2 5
3
4
4
2 4
( 3) 5
t
t
t t
e te
e t e
s s
M1
A1A1
3
Q5(a)
(ii)
3
3
23
2 2
cosh 4 ( 3)
cosh 4 ( 3)
1627
( 16)
s
t t t t
t t t t
se
s
M1
A1A1
3
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15
(Mmethod, Aanswer) Mark Total
Q5(a)
(iii)
3
3
3[( 3) 3]
9 3( 3)
9 3
2
sinh(3 ) ( 3)
sinh(3 ) ( 3)
sinh(3 ) ( 3)
sinh(3 ) ( 3)
3
9 3
t
t
t
t
s
t e H t
t e H t
t e H t
t e e H t
e
s s
M1
A1
A1A1
4
Q5(b)
2
2
2
2
3 2 2
3
2 , (0) 2, (0) 3
2
1
( ) (0) (0) 2 ( ) (0) ( ) 1
1( ) 2 3 2 ( ) 4 ( )
1
12 1 ( ) 2 1
1
1( 1) ( ) 2 1
1
1 2 1( )
( 1) ( 1) ( 1)
1 2( )
( 1)
t
t
y y y e y y
y y y e
s Y s sy y sY s y Y s s
s Y s s sY s Y ss
s s Y s ss
s Y s ss
sY s
s s s
Y ss
2
2 2
1
1
3 2
2
1
( 1)
Byusing partialfraction,
2 1
( 1) 1 ( 1)
2, 1
( ) ( )
1 2 1
( 1) 1 ( 1)
12
2
t t t
s
s
s A B
s s s
A B
y t Y s
s s s
e t e e t
M1
A1
A1
M1
A1
M1
A1
A1A1A1
10
Q6(a)
(i)
1
2 2
1 1
2 2
2
5 6
4 ( 2) 4
65
4 ( 2) 4
5 cosh 2 3 sin 2
t
s
s s
s
s s
t e t
M1
A1A1
3
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16
(Mmethod, Aanswer) Mark Total
Q6(a)
(ii)
21
2
2
2 2
2 2
2 2
21
2
1 1
16
( 3)( 1)
Bypartialfraction,
16
( 3)( 1) 3 1 ( 1)
16 ( 1) ( 1)( 3) ( 3)
16 ( ) (2 2 ) ( 3 3 )
9, 7, 4
16
( 3)( 1)
9 7
3 1
s
s s
s A B C
s s s s s
s A s B s s C s
s A B s A B C s A B C
A B C
s
s s
s s
1
2
3
4
( 1)
9 7 4t t ts
e e e t
M1
M1
A1
A1
A1A1A1
7
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BFC 14003
17
(Mmethod, Aanswer) Mark Total
Q6(b)
2
2
1 2
2
2
1
2
1
2
1
1
1
2
1 1 1
3 2 2 5sin , (0) 0, (0) 2
3 2 0
2, 1
( ) 2
0 .......(i)
Compare with , no termalike,accept (i)
2
2
3 2 22 3(2
x xh
r
p
p
h
p
p
p p p
y y y x x y y
m m
m m
y Ae Be
f x x
y x Cx Dx E
r y Cx Dx E
y
y Cx D
y C
y y y xC C
2 2
2 2
2
1
2
2
2
2
) 2( ) 2
2 ( 6 2 ) (2 3 2 3) 2
1 3 11, ,
2 2 4
1 3 11
2 2 4
( ) 5sin
cos sin
0 cos sin ............(ii)
Compare with , no termalike,accept (ii)
sin cos
p
r
p
p
h
p
x D Cx Dx E x
Cx C D x C D E x
C D E
y x x
f x x
y x K x L x
r y K x L x
y
y K x L
2
2 2 2
2
2 2
cos sin
3 2 5sin
cos sin 3( sin cos ) 2( cos sin ) 5sin
( 3 ) cos (3 )sin 5sin
3 1,2 2
3 1cos sin
2 2
1 3 11 3 1cos sin
2 2 4 2 2
p
p p p
p
x x
x
y K x L x
y y y x
K x L x K x L x K x L x x
K L x K L x x
K L
y x x
y Ae Be x x x x
M1
A1
M1
A1
A12
A1
M1
A1
A12
A1
A1
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