CONFIDENTIAL
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
BFC 14003
PART A Q1
A periodic function is defined by
2, 2 x 0 f ( x) x, 0 x 2 f ( x) f x 4 . (a)
Sketch the graph of the function over 4 x 4 . (2 marks)
(b)
Determine whether the function is even, odd or neither. (1 marks)
(c)
Show that the Fourier series of the function f ( x) is
1 1n 3 2 n x 2 n x c o s s i n 2 n n 1 2 . n2 2 n 1
(17 marks)
Q2
(a)
Consider the series,
1 5
1 25
1 125
...................
(i)
Give the sigma notation of the series above.
(ii)
Determine whether the series converge or diverge. (6 marks)
(b)
Given that 4
lim lim
n
(n 1) 4 / 4 n1 n 4 / 4n
1 1 4 n (n 1) 4 1 n lim . lim lim li m n 1 4 n
Determine whether the series
n 1
4
n4 4n
n
n
4
4
conditionally convergent, absolutely
convergent, or divergent. (3 marks) (c)
Given
n 1
( x 3) n n2
2
.
BFC 14003
By using the ratio test for absolute convergence, find (i)
The interval of converges,
(ii)
The radius of converges.
(11 marks)
PART B Q3
(a)
Solve the differential equation by using the method of separation of variables,
1 4 x 2 dy y 3 xdx . Hence, find the particular solution when y (2)
3 5
. (5 marks)
(b)
By using the substitution of y xv and
x (Hints:
dx a 2 x2
dy dx
dy dx
x
dv dx
v , find the solution of
y x2 y 2 . x
sinh sinh1 ( ) c, c is const onstan ant, t, sinh 1 ( x) ln | x x 2 1 | .) a
(6 marks) (c)
The temperature of a dead body when it was found at 3 o’clock in the morning is 85 F . The surrounding temperature at that time was 68 F . After two
hours, the temperature of the dead body decreased to 74 F . Determine the
time of murdered. (Hints:
dT dt
k T T s , where
dT dt
is the temperature change in a body , k
is the proportionality constant, T is temperature in the body and T s is the temperature constant of the surrounding medium. The normal human body temperature is 98.6 F .)
(9 marks)
3
BFC 14003
Q4
(a)
Use the method of variation of parameters to solve y 2 y y
e x ( x 1) 2
.
(10 marks) (c)
A spring is stretched 0.25 m ( l ) when a 4 kg mass ( M ) is attached. The weight is pushed up 1/2 m and released. The damping constant equals c 2 . If the general equation describing the spring-mass system is Mu cu ku 0, find an equation for the position of the spring at any time t . (Hints: weight, W Mg ,
g 9.8, k
W
l
.) (10 marks)
Q5
(a)
Find
e
(i)
(ii)
2t
cos3t t 3e 2t
(t 1)
2
H (t 2) (10 marks)
(b)
By using Laplace transform, solve
y 2 y 5 y 20,
y(0) 0, y(0) 10. (10 marks)
Q6
(a)
Find
3 s 2 2 2 . s 9 s 4
(i)
1
(ii)
1
1 3 s . 2 ( s 4 ) 5 (10 marks)
(b)
Find the general solution for the second order differential equation
y 3 y 2 y 3e x 10 cos x, by using the undetermined coefficient method. (10 marks) 4
BFC 14003
FINAL EXAMINATION SEMESTER / SESSION : SEM II / 2012/2013 SUBJECT : CIVIL ENGINEERING MATHEMATIC II
COURSE : 1 BFF / 2 BFF SUBJECT CODE : BFC 14003
FORMULA Second-order Second-order Differential Equation The roots of characteristic equation and the general solution for differential equation a y b y cy 0.
Characteristic equation: am2 bm c 0. Case The roots of characteristic equation 1. Real and different roots: m1 and m2
m m1 m2
2.
Real and equal roots:
3.
Complex roots: m1 i , m2 i
General solution
y Aem1 x Be m2 x y ( A Bx)e mx y e x ( A cos x B sin x)
The method of undetermined undetermined coefficients For non-homogeneous second order differential equation ay by cy f ( x ), the particular
solution is given by y p ( x) :
f ( x)
y p ( x)
Pn ( x) An x n An1 xn 1 A1 x A0
x r ( Bn x n Bn 1 x n 1 B1x B0 )
Ce x C cos x or C si sin x
x r ( Pe x )
x r ( P cos x Q sin x)
Pn ( x )e x
x r ( Bn x n Bn 1 x n 1 B1x B0 )e x
cos x Pn ( x ) sin x cos x x Ce sin x cos x Pn ( x )e x sin x
x r ( Bn x n Bn 1x n 1 B1x B0 ) cos x x r (Cn x n Cn 1 x n1 C1x C0 ) sin x
x r e x ( P cos x Q sin x ) x r ( Bn x n Bn 1x n 1 B1x B0 )e x cos x x r (Cn x n Cn 1 x n1 C1x C0 )e
x
sin x
Note : r is is the least non-negative integer ( r = = 0, 1, or 2) which determine such that there is no terms in particular integral y p ( x) corresponds to the complementary function y c ( x) .
The method of variation of parameters
If the solution of the homogeneous equation a y b y cy 0 is yc Ay1 By2 , then the particular solution for a y b y cy f ( x) is
y uy1 vy2 , where u
y 2 f ( x) aW
dx A, v
y1 f ( x) aW 5
dx B and W
y1
y 2
y1 y 2
y1 y 2 y 2 y1 .
BFC 14003
FINAL EXAMINATION SEMESTER / SESSION : SEM II / 2012/2013 SUBJECT : CIVIL ENGINEERING MATHEMATIC II
COURSE : 1 BFF / 2 BFF SUBJECT CODE : BFC 14003
Laplace Transform L { f (t )}
f (t )
F ( s )
a
a
f (t )e st dt F (s) 0
f (t )
s a a
sin at
2
2
2
2
s a s
cos cos at
s a a
sinh at
t n , n 1, 2, 3, ... at
e f (t ) t f (t ) , n 1, 2, 3, ... n
a) (t
e as
f (t ) (t a )
e as f (a )
t 0
s F ( s a)
(1)
ds n
F ( s) G( s)
f (u ) g (t u ) du
n 1
d n
s
e as F ( s)
s 2 a 2 n!
n
as
f (t a) H (t a)
s 2 a 2 s
cosh at
e
H (t a)
s 1
e at
F ( s )
y (t )
Y ( s)
y(t )
sY ( s ) y(0)
y(t )
s 2Y ( s) sy(0) y (0)
F ( s )
Fourier Series Fourier series expansion of periodic function with period 2 L 1 n x n x bn sin f ( x) a0 an cos 2 L L n 1 n 1 where 1 L a0 f ( x) dx L L 1 L n x an f ( x) cos dx L L L 1 L n x bn f ( x) sin dx L L L
6
Fourier half-range series expansion
f ( x) where 2 a0 L 2 an L 2 bn L
1
a0
2
L
0
L
0
L
0
n x
an cos L bn sin
n 1
n 1
f ( x) dx f ( x) cos
n x
dx L n x f ( x) sin dx L
n x L
BFC 14003
Marking Scheme (M – method, method, A – answer) answer)
Q1 (a)
Q1 (b)
Neither
x T x 4 T 2L 4 2 L L 2
a0
Mark
Total
A2
2
A1
1
A1
1
L L f ( x)dx L 1
02 2dx 02 xdx 2 1
M1
2 x 2 2 x 2 2 2 0 0
A1 17
Q1 (c)
1
4 2
A1
3
A1
2
an
1 L n x f ( x )cos dx L L L
1 n x n x 02 2 cos( )dx 02 x cos( )dx 2 2 2
7
M1
BFC 14003
(M – method, method, A – answer) answer)
u x
dv n x cos( ) 2 2 n x sin( ) n 2 4 n x 2 2 cos( ) 2 n
1 0
1 4
Mark
M1
n x n x 4 n x 2x sin( ) sin( ) 2 2 cos( ) n 2 n 2 2 n 2 2 0 0
2
A1
1 4 4 2 2 cos(n ) 2 2 2 n n
bn
A1
1 L n x L f ( x)sin dx L L
1 n x n x 02 2 sin( )dx 02 x sin( )dx 2 2 2
u x
dv n x sin( ) 2 2 n x cos( ) n 2 4 n x 2 2 sin( ) n 2
1
0
1
M1
M1
n x n x 4 n x 2x cos( ) cos( ) 2 2 sin( ) 2 n 2 2 n 2 n 2 0 4
0
2
1 4 4 4 cos( n ) cos(n ) 2 n n n
8
A1
A1
Total
BFC 14003
(M – method, method, A – answer) answer)
1 4 4 n x cos(n ) 2 2 cos( ) 2 2 n 2 3 2 n f ( x ) 2 n1 1 4 4 4 n x cos(n ) cos( n ) sin( n 2 2 n n
A1
3 1 4 4 n x 1 4 n x 2 2 (1) n 2 2 cos( ) s i n ( ) 2 n 1 2 n n 2 2 n 2
M1 A1
1 1n n x 2 n x c o s s i n 2 n n 1 2 n2 2 n 1
A1
3
(i)
Q2 (a)
Mark
n 1
2
Total
1 A1A1
5 n
(1st A1 for limit, 2nd A1 for term)
(ii) By using ratio test, 1 n 1 1 5n 5 lim n1 lim n n 5 1 1
M1M1
6
5n
1 5
A1
1
so the series converge. Or the series represent geometric series where r < < 1, so converge.
Q2
By ratio test for absolute convergence,
(b)
the series converge absolutely as ¼ < 1.
A1
M1M1 3
A1
(i) Apply the ratio test for absolute convergence.
lim n
Q2 (c)
un 1 un
lim
n
( x 3)n 1 (n 1)
2
n2 ( x 3)
n
M1
2 n lim x 3 n n 1
11
2
1 x 3 lim 1 x 3 n n 1 Thus, the series converges absolutely if x 3 1, or 1 x 3 1, or 2 x 4 .
9
A1
M1A1
BFC 14003
(M – method, method, A – answer) answer)
Mark
Total
The series diverges if x < 2 or x > 4. When x 2, (using p-series)
(1)
n
M1
1
1
2
1
1
1
1
3
4
2
A1
n2 2 3 42 n 1 Or the series of absolute values
(1)n n2
n 1
1
1
22
2
M1
2
A1
converge absolutely since p 2 1.
When x 4, (using p-series)
1n
n n 1
2
1
1 22
1 32
1 42
also converge since p 2 1.
A1
So, the interval of convergence is [2, 4].
A1
radius of convergence convergence is R = 1 (ii) The radius
A1
1 4 x 2 dy xy3dx dy x dx 3 2 y 1 4x
1 y 3
x
dy
1 4x
2
dx
A1
assume a 1 4 x da 8 x dx 2
Q3 (a)
M1 5
1 y 3
dy
1 2 y
1 8
1
a
da
1
2
(2 a ) k
8 1
4
Given y (2)
3.45 k
A1
1 4 x 2 k 3 5
, so A1
10
BFC 14003
(M – method, method, A – answer) answer)
x
1
2 y 2
dy
1 4 x 2
Mark
3.45
4
Total
A1
y x2 y 2
dx
dv v xv x 2 x 2v 2 dx
x x
xv x 1 v 2 x v 1 v2
x x Q3 (b)
dv dx dv
v v 1 v2
dx 1
1 v2 dv
1 v2
A1
1 1 v
2
1 x
dv
M1 6
dx
1 x
A1
dx
By using the hints,
sinh 1 (v) ln | x | k ,
ln | v v 2 1 | ln | x | k
ln |
y x
dT
M1
y
( )2 1 | ln | x | k
A1
x
3 am, at t T 98.6 F Q3 (c)
A1
k is a constant.
T 85 F
;
5 am, t
T 74 F ;
? , t 9
k (T T s )
dt dT
(T T s )
kdt
11
BFC 14003
(M – method, method, A – answer) answer)
1
M1
dT kdt
T T s
ln | T T s | kt c , c is constant
T T s Ae
kt
Mark
, A e
A1
c
3am, at t 85 68 Ae3k Ae3k 17 17 A 3k e 17
So T 68
e3k
e
A1
kt
at t 5am ,
74 68
6 17e e
2 k
17 e
e5k
3k
2 k
6
17 1 6 ln( ) k 2 17 A 17e
3
6
2
17
ln(
T 68 17e
A1
A1
)
3
6
2
17 17
ln(
98.6 68 17e 1.8 e 3
6
2
17 17
ln(
)
3
6
2
17 17
ln( 1
e
2
t ln(
1
)
6 17
)
e
3
6
2
17 17
ln(
t ln(
6 17
M1
1 6 t ln( ) 17
2
)
1
e
2
t ln(
6 17
)
A1
)
e
2
)
ln |1.8 |
M1
t 1.87 A1
1:52 am Therefore, the time of murdered was 1:52am.
12
Total
BFC 14003
(M – method, method, A – answer) answer)
Mark
Total
Step 1:
e x
a 1, f ( x)
( x 1)2
Step 2:
m 2 2m 1 0 m 1
M1
Thus, yh Ae x Bxe x y1 e x
y 2 xe x
y1 e x
y2 xe x e x
A1
Step 3:
W
e x
xe x xe e
x
x
e
x
xe2 x e2 x xe 2 x e 2x . M1A1
Step 4:
e x
xe x
u
e
Q4 (a)
( x 1) 2 2 x
x ( x 1) 2 u 1 u 1
2
M1
dx
10
dx M1
u x 1
du;
u 2 du u
ln | x 1 |
e
( x 1) 2 e 2 x 1
( x 1)2 1 u
x 1
C
A1
e x
x
v
1
2
dx
dx
d u;
M1
u x 1
u 2 du
1 x 1
D
A1
Step 5:
13
BFC 14003
(M – method, method, A – answer) answer)
Mark
Total
y uy1 vy 2
( ln | x 1 |
1 x 1
C ) e x (
Ce Dxe e ln | x 1 | x
x
x
W mg 4(9. 4(9.8 8) 39.2 39.2
k
W
39.2
l 0.25
1 x 1
e x x 1
D ) xe x
(1 x ).
A1 A1
156.8
A1
c2 Thus, from mu cu ku 0, We get 4u 2u 156.8u 0 or 2u u 78.4u 0.
Q4
The characteristic equation is 2m2 m 78.4 0. So, m 0.25 0.25 6.25 6.25i.
(b)
Therefore, u (t ) e 0.25t A cos(6.25 t ) B sin(6.25 t ) .
u(t ) e 0.25t 6.25 A si sin(6.25t ) 6.25B co cos(6.25t )
A1
M1 A1 A1 A1
10
0.25e0.25t A cos(6.25t ) B sin(6.25t ) . The spring is pushed up 1/2 m and released. So, we have u (0) 1 / 2 . (pushed up is negative)
A1
Since the weight is simply released, its initial velocity is zero, u(0) 0 .
A1
These initial conditions give us the equation for the position of the spring at any time t as as
u (t ) e0.25t 0.5co 5cos(6.25 t ) 0.02si 2sin(6.25 t ) .
A1
e cos3t t e e cos3t t e
M1
2t
3
2t
2t
Q5(a) (i)
s 2 ( s 2) 2 9
3
2t
3
6 (s 2) 4
A1A1
14
BFC 14003
(M – method, method, A – answer) answer)
Mark
(t 1) H (t 2) (t 2t 1) 1) H (t 2)
Total
2
2
A1
Let t (t 2) 2 t 2 2t 1 [( [(t 2) 2]2 2[(t 2) 2] 1
M1
(t 2) 2 4(t 2) 4 2( 2(t 2) 4 1
Q5(a) (ii)
7
(t 2) 2 2(t 2) 1
A1
(t 2) 2) H (t 2) 2) 2 (t 2) 2) H (t 2) 2) H (t 2) 2) 2
e2 s t 2 2e2s t H (t 2)
e2 s
2e 2s
e 2s
A1A1A1
s 3 s2 s y 2 y 5 y 20,
M1
y (0) 0, y(0) 10
y 2 y 5 y 20
s 2Y (s ) sy (0) y (0) 2 sY (s ) y (0) 5Y (s ) s 2Y (s ) 10 2sY (s ) 5Y (s ) ( s 2 2s 5)Y (s ) Y (s)
20 s
M1
20 s 20 s
10 20
s ( s 2 2 s 5)
10 s 2 2s 5
A1
By using using partialfract partialfraction, ion, 20 s ( s 2 2 s 5) 5) Q5(b)
A s
Bs C s 2 2s 5
M1
20 ( A B )s 2 (2 A C )s 5A
10
A 4, B 4, C 8
y (t ) 1 Y (s ) 10 4 s 8 4 2 2 s s 2s 5 s 2s 5 4 4 s 2 1 2 s ( s 1) 4 1
4 4[( s 1) 1] 2 1 ( s 1) 2 4 s 4( s 1) 6 4 1 1 1 2 2 s ( s 1) 4 (s 1) 4 4 4e t cos 2t 3e t sin 2t
15
A1
M1
M1
A1 A1A1A1
BFC 14003
(M – method, method, A – answer) answer)
e3 2 3s 2 2 s s 9 s 4 3 2 1 e 1 3s 1 2 2 s 9 s 4 s
Mark
Total
M1
4
1
Q6(a) (i)
2
e3 sin3t 3co 3cosh 2t 3
A1A1A1
1 3 s 2 ( s 4) 5 1 3 s F (s ) ( s 4) 2 5 1 3[ 3[( s 4) 4] ( s 4) 2 5 3( s 4) 13 13 ( s 4) 2 5 Q6(a) 3( s 4) 13 (ii) ( s 4) 2 5 ( s 4) 2 5
1
M1 A1
A1
3( s 4) 13 1 1 2 2 ( s 4) 5 ( s 4) 5 s 13 3e4t 1 2 e4t 1 2 s 5 s 5 3e4t cos 5t
13 5
e
4t
sin 5t
M1
A1A1
16
6
BFC 14003
(M – method, method, A – answer) answer)
y 3 y 2 y 3e
x
Total
0, y (0) 2 10 cos x , y (0) 0,
m2 3m 2 0
M1
m1 2, m2 1 yh Ae 2 x Be x f1 ( x) 3e
Mark
A1
x
y p1 x r Pe x r 0 y p1 Pe x .......(i) Compar Comparee with with yh , no termalik termalike.Ac e.Acce cept pt (i). (i).
M1
y p 1 Pe x y p1 Pe x y p1 3 y p1 2 y p1 3e
A1
x
Pe x 3( Pe x ) 2 Pe x 3e x 6 Pe x 3e x P Q6(b)
1 2
y p1
1 2
e x
A1
f 2 ( x ) 10 co cos x y p 2 x r K cos x L sin x r 0 y p 2 K cos x L sin x.......(ii)
M1
Compar Comparee with with yh , no termalik termalike.Ac e.Acce cept pt (ii). (ii). y p 2 K sin x L cos x y p 2 K cos x L sin x
A1
y p 2 3 yp 2 2 y p 2 10 cos x
K cos x L sin x 3( K sin x L cos x ) 2(K cos x L sin x ) 10 co ( 7 K 9 L ) cos x (9K 7L ) sin x 10 cos x K
7
9
,L 13 13 7 9 y p 2 cos x sin x 13 13 1 7 9 y Ae 2 x Be x e x cos x sin x 2 13 13
17
A1
A1
A1
10
