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 , x 0 2 x, 0 x f ( x) f x 2 .
f ( x)
(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 f ( x) is
3 2
1n 1 1 2 cos nx 2 sin nx . 2 n 1 n 1
n
n
(17 marks)
Q2
(a)
Given that cos x 1
x 2 2!
x 4 4!
x 6 6!
.
(i)
Find the first three nonzero terms of a power series for
(ii)
Hence, evaluate
1
0
sin x x
sin x x
.
dx by using the series expansion. (6 marks)
(b)
(i)
Determine whether the series
n2
n 1 e
n
converges or diverges by
using ratio test.
(ii)
Determine whether the series
n 1
(1) n n
converges
absolutely, converges conditionally, or diverges by using a suitable convergence test. (8 marks)
(c)
Find the radius of convergence of
3 ( x 2)
n
.
n0
(6 marks) 2
BFC 14003
PART B Q3
(a)
Solve the differential equation by using the method of separation of variables,
1 x
dy dx
x 1 y .
Hence, find the particular solution when y(1) 0 . (7 marks) (b)
By using the substitution of y xv and
dy dx
x2 y 2 3xy
dy dx
x
dv dx
v , find the solution of
. (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 dt
kN , where N denotes the number of bacteria at time, t hours
and k is the proportionality factor.) (7 marks)
Q4
(a)
Use the method of variation of parameters to solve
y y sec x tan x , which satisfies the initial conditions y(0) 0 and y(0) 2 . (Hints: sec2 x tan 2 x 1 ,
sec x dx tan x.) 2
(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 describi ng the spring-mass system is
3
BFC 14003
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 (i)
3t 2 4 e t 5t . e
(ii)
t cosh 4t t (t 3) .
(iii)
sinh(3t ) e
3
H (t 3) .
3t
(10 marks) (b)
By using Laplace transform, solve
y 2 y y et ,
y (0) 2, y (0 (0) 3. (10 marks)
Q6
(a)
Find (i)
(ii)
1
5 s 6 2 . 2 s 4 (s 2) 4 16 s 2 . 2 ( s 3 ) ( s 1 )
1
(10 marks) (b)
Find the general solution for the second order differential equation
y 3 y 2 y x 2 2 5 sin x, by using the undetermined 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: am 2 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.
i , m2 i Complex roots: m1
General solution
y Aem1 x Bem2 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 An 1 x n 1
x r ( Bn x n Bn1 x n1
A1x A0
B1x B0 )
Ce x C cos x or C sin x
x r ( Pe x )
Pn ( x )e x
x r ( Bn x n Bn1 x n 1
B1x B0 )e x
cos x Pn ( x ) sin x cos x Ce x sin x cos x Pn ( x )e x sin x
x r ( Bn x n Bn1x n1
B1x B0 ) cos x
x r (Cn x n Cn1 x n 1
C1x C0 ) sin x
x r ( P cos x Q sin x)
x r e x ( P cos x Q sin x) x r ( Bn x n Bn1x n 1
B1x B0 )e x cos x
x r (Cn x n Cn1 x n 1
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 )e st dt F (s ) 0
f (t )
F ( s)
f (t )
a
a
H (t a)
s 1
e at
s a a
sin at
2
s a s
cos cos at sinh at cosh at
2
t n , n 1, 2, 3, ... at
e f (t ) t f (t ) , n 1, 2, 3, ... n
s a n!
e as
f (t ) (t a)
e as f (a)
t 0
f (u ) g (t u ) du
2
s F ( s a)
(1)
d n ds n
s
(t a)
n 1
n
as
e as F ( s)
s 2 a 2 s
e
f (t a) H (t a)
2
s 2 a 2 a
F ( s )
F ( s) G( 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
Fourier half-range series expansion
f ( x) where 2 a0 L 2 an L 2 bn L
6
1
a0
2
L
0
L
0
L
0
n x
an cos L bn sin
n 1
f ( x) dx n x dx L n x f ( x) sin dx L f ( x) cos
n 1
n x L
BFC 14003
Marking Scheme (M – method, method, A – answer) answer)
Mark
Total
A2
2
Neither
A1
1
x T x 2 T 2L 2 2 L L
A1
Q1 (a)
Q1 (b) Q1 (c)
a0
1
L L f ( x)dx L
1
0 2 dx 0 2 xdx
M1
1 0 2 x x 2 0
A1
1
A1
3
A1
2 2 2
1 L n x f ( x ) co c o s dx L L L 1 0 2 cos(nx)dx 0 2 x cos(nx)dx
an
7
M1
17
BFC 14003
(M – method, method, A – answer) answer)
Mark
u 2 x
dv cos(nx)
2
1
sin(nx) n 1 2 cos(nx) n
0
0 1 2 2 2 x sin(nx) sin(nx) 2 cos(nx) n n n 0
12 2 c o s ( n ) n 2 n 2
M1
A1
A1
1 L n x f ( x ) si s i n dx L L L 1 0 2 sin(nx)dx 0 2 x si s in(nx)dx
bn
M1
u 2 x
dv sin(nx)
2
cos(nx)
1
0
n 1
n
2
sin( nx)
0 1 2 2 2 x cos(nx) cos(nx) 2 sin(nx) n n n 0
f ( x)
M1
A1
2 A1
n
1 2 2 2 2 cos(n ) 2 cos(nx) sin nx 2 n 1 n n n
3
8
A1M1
Total
BFC 14003
(M – method, method, A – answer) answer)
sin x
d dx
Mark
2 2 2 (1)n 1 co cos(nx) sin(nx) 2 n 1 n n
3
3 2
1n 1 1 2 cos nx 2 sin nx 2 n 1 n 1
(a) (i)
A1
n
n
A1
cos x
d x 2 x 4 x 6 1 dx 2! 4! 6!
Q2
Total
M1
2 x 4 x 3 6 x 5 4! 6! 2 ! x
x 3
6
x 5 120
3
A1
So,
x
sin x
x 3
6
120 x
x
1
0
sin x x
dx
Q2
x 5
1
0
x
x
x 5 / 2 6
6
x 9 / 2 120
x 9 / 2 120
A1
dx
M1 1
2 2 2 x11 / 2 x 3 / 2 x 7 / 2 42 1320 3 0
(a) (ii)
2
2
3 42 0.621 .
2 1320
3
A1
n
( n 1) 2 lim
n
(b) (i)
e n 1 n2 e
lim
n
e
n
e n 1
(n 1)
2
1 e
3
n 2
1 ;
M1
n2
n 1 1 1 lim 1 li lim e n n e n n 1
A1
n2
n 1 e
Q2
x 5 / 2
2
the series converges. 9
A1A1
BFC 14003
(M – method, method, A – answer) answer)
(1) n
n
n 1
1
1
1
2
1
3
4
1
Mark
Total
5
Use alternating series test:
1
(a) 1 Q2 (b) (ii)
(b)
1
2
1
3 1
lim lim
n
4
1
M1
5
0
n
5
A1
the series converges.
Use p-series: p
1
M1
0 p 1
2
A1
Conclusion, the series converges conditionally.
A1
the series diverges.
For x 2 , let u n 3( x 2) n .Then lim
u n 1
n
lim n
un
3( x 2)
1
M1
n
lim | x 2 |
M1
| x 2 |
A1
n
Q2 (c)
Q3 (a)
3( x 2) n
Thus, the series converges absolutely if | x 2 | 1 or 1 x 3 . The series diverges if x 1 or x 3
M1 A1
Therefore, the radius of convergence of the series is R 1 .
A1
1 x
dy
dy
1 y
dx
1 1 y
6
x 1 y x
1 x
dy
dx
x 1 x
A1
M1
dx
2 1 y x ln(x 1) 1) k , k is constant Given y(1) 0 , so 10
M2 A1
7
BFC 14003
(M – method, method, A – answer) answer)
Q3 (b)
Mark
k ln(2) 3
A1
2 1 y x ln(x 1) ln(2) 3
A1
dy dx
x
x2 y 2
dv dx
3xy x2 x2v2
v
3 x( xv)
x
dv dx
3v 1 2v 2
A1
3v
v
3v 1 2v 2
M1
3v
dv
3v 1 2v
1 v2
1 v2
Total
2
1 x
dv
dx 1 x
A1
dx
2 assume a 1 2v da 4v dv 1 da vdv 4
3 1
1
4 a
x
6
M1
da dx 3
A1
ln 1 2v2 ln x k 4
2
y ln 1 2 ln x k 4 x 3
Q3 (c)
A1
4 , N 2 N 0 ; t 0, N N 0 ; t at t 12 , N ? dN dt
7
kN 11
BFC 14003
(M – method, method, A – answer) answer)
dN
1 N
Total
kdt
N
Mark
dN kdt M1
ln | N | kt c , c is constant N Aekt , A ec
A1
0, A N 0 at t So N N 0e kt
A1
at t 4 , 2 N0 N 0e 4k
2 e4k 4k ln | 2 | 1 k l n | 2 | 4 1
So N N 0e 4
A1
t ln|2|
A1
at t 12 , N N 0e3ln|2|
8 N 0
M1 A1
At the end of 12 hours, the number of bacteria is 8 times of the original number. Step 1: a 1, f ( x) sec x tan x Step 2:
M1
2 m 1 0
mi Q4 (a)
Thus,
10
yh A cos x B sin x y1 cos x
y2 sin x
y1 sin x
y2 cos x
A1
12
BFC 14003
(M – method, method, A – answer) answer) Step 3: cos x W sin x
sin x cos x
cos2 x sin 2 x 1.
Mark
M1A1
Step 4:
tan x dx 1 sec x dx
u sin x sec x tan x dx
M1
2
2
( x tan x ) C
A1
tan x dx dx
v cos x sec x tan tan x dx
sin x cos x sin x
u 1
M1
dx
du sin x
du u ln | cos x | D
A1
Step 5: y uy1 vy2
(tan x x C ) cos x (D ln | cos x |) sin x C cos x D sin x sin x x cos x sin x ln | cos x |
A1
y C sin x D cos x cos x x sin x cos x sin x tan x
co cos x ln | co cos x |
When y(0) 0, 0. C When y(0) 2, D 2. Thus, y 2 sin x sin x x cos x sin x ln | cos x | .
13
A1
Total
BFC 14003
(M – method, method, A – answer) answer)
W mg 4(9. 4(9.8) 8) 39.2 39.2
k
W
l
39.2 0.1
Mark
Total
A1
392
A1
c0 Thus, from mu cu ku 0, We get 4u 392u 0 or u 98u 0.
A1
The characteristic equation is m2 98 0.
m 7 2 i.
So, Q4
Therefore, u (t ) A cos(7 2 t ) B sin(7 2 t ).
(b)
or
u (t ) A cos(9.9 t ) B sin(9.9 t ). u(t ) 7 2 A sin(7 2 t ) 7 2 B cos(7 2 t ).
The spring is released after pulling it down 0.2 m. So, we have u (0) 0.2 . Since it’s set in motion with an upward velocity of 4 m/s, we have u(0) 4 . (upward is negative)
M1 A1 A1 A1
10
A1 A1
These initial conditions give us the equation for the positi on of the spring at any time t as as
u (t ) 0.2 cos(7 2 t ) 0.4041sin(7 2 t ).
A1
or
u (t )
Q5(a) (i)
1 5
cos(7 2 t)
2 2 7
sin(7 2 t ). ).
3t 2 4 e t 5t e e3t t 2 4e 5t
2 ( s 3) 3)3
M1 3
4
A1A1
s5
t cosh 4t t (t 3) t cosh 4t t (t 3 ) 3
3
Q5(a) (ii)
s 2 16 ( s 2 16) 2
27e
M1 3
3 s
A1A1
14
BFC 14003
(M – method, method, A – answer) answer)
sinh(3t ) e
Q5(a) (iii)
H (t 3)
sinh(3t ) e3t H (t 3)
sinh(3t ) e3[(t 3)3]H (t 3 )
sinh(3t ) e9e3(t 3) H (t 3) 3 s 9 2
e
Total
3t
Mark
M1 4
A1
9 3 s
s 3
A1A1
y 2 y y et ,
y (0) 2, y (0) 3
y 2 y y et s 2Y ( s ) sy (0) y (0) 2 sY (s ) y (0) Y ( s ) s 2Y ( s ) 2 s 3 2sY ( s ) 4 Y ( s )
s
2
2s 1 Y ( s ) ( s 1) 2 Y ( s ) Y ( s)
Q5(b)
1 M1
s 1
1 s 1 1 s 1 1
s 1 1
A1
2s 1
A1
2s 1
( s 1)3
2 s ( s 1) 2
2 s 1 Y ( s) ( s 1)3 ( s 1) 2
1
M1
( s 1) 2 10
1
A1
By using using partialfract partialfraction, ion, A B 2 s 1 ( s 1) 2 s 1 ( s 1) 2 A 2, B 1 1
y (t ) 1
M1
Y ( s )
1 2 1 3 1) s 1 ( s 1) 1) 2 ( s 1)
A1
1
A1A1A1
et t 2 2et et t 2
5 s 6 2 2 s 4 (s 2) 4 6 s 1 5 1 2 2 s 4 (s 2) 4 1
Q6(a) (i)
M1
5 co cosh 2t 3e 2t sin 2t
A1A1
15
3
BFC 14003
(M – method, method, A – answer) answer) 1
Mark
Total
16 s 2 2 ( s 3)( s 1)
By partialfraction, 16 s
2
( s 3) 3)( s 1)
2
A s 3
B s 1
M1
C ( s 1)
2
16 s 2 A( s 1) 1) 2 B( s 1) 1)( s 3) 3) C ( s 3) 3) Q6(a) (ii)
2 2 16 s ( A B ) s (2 A 2 B C ) s ( A 3 B 3C )
M1 A1 7
A 9, B 7, C 4
1
1
16 s 2 2 ( s 3)( s 1)
9 s 3
1
7 s 1
1
4 2 ( s 1 )
9e3t 7et 4e t t
16
A1 A1A1A1
BFC 14003
(M – method, method, A – answer) answer)
Mark
Total
y 3 y 2 y x 2 2 5 sin x , y (0 (0) 0, y (0) 2 m 2 3m 2 0
M1
m1 2, m2 1
A1
yh Ae 2 x Be x f1 ( x) x 2 2 r 2 y p1 x Cx Dx E
M1
r 0 y p1 Cx Dx E . ......(i) 2
Comp Compar aree with with yh , no termali termalike ke,acc ,accep eptt (i) (i) y p 1 2Cx D y p1 2C
A1
y p1 3 y p1 2 y p1 x 2 2 2C 3( 2C x D ) 2(Cx 2 Dx E ) x 2 2 2Cx (6C 2 D ) x (2 ( 2C 3D 2 E 3) x 2 2
C Q6(b)
1 2
2
,D
y p1
1 2
3 2
, E
x2
3
2 f 2 ( x ) 5 sin x
11
x
4 11 11
A 12
A1
4
y p 2 x r K cos x L sin x r 0 y p 2 K cos x L sin x............(ii)
M1
Comp Compar aree with with yh , no termali termalike ke,acc ,accep eptt (ii) (ii) y p 2 K sin x L cos x y p2 K cos x L sin x
A1
y p2 3 y p 2 2 y p 2 5 sin x
K cos x L sin x 3( K sin x L cos x ) 2(K cos x L sin x ) 5 sin x ( K 3L ) cos x (3K L ) sin x 5 sin x 3 1 K , L 2 2 3 1 y p 2 cos x sin x 2 2 1 3 11 3 1 y Ae 2 x Be x x 2 x cos x sin x 2 2 4 2 2
17
A 12
A1
A1
10
