Engineering Mathematics 2 questions & answers

Engineering Mathematics 2 questions & answers

1. Mock Exam. January-April 2013Page 1 of 8 EE203 Mathematical Methods for Engineers II


EE203 Mock Final Examination
\
1(a) Show that the following ordinary differential equation is an exact differential equation.
Then find its general solution:

2
3
x
e
y
dx
dy
x
x

(10 marks)

(b) Show that x
exy
2
1)(

 is one solution of the following homogenous differential equation:

044  yyy .

Determine the second solution by using Reduction of Order method or otherwise.
Find the general solution of the above differential equation.

(8 marks)


2(a) Find the general solution for the following non-homogenous differential equation:
tyyy 2sin35'2'' 

(14 marks)

(b) Solve the following separable variables differential equation:

x
xy
dx
dy )1(2 

(6 marks)


3(a) Consider the following IVP consisting of the differential equations with initial conditions:

3)0(and10)0( ;016  yyyy

Find the solution )(ty by using Laplace and Inverse Laplace method.
(12 marks)

(b) Determine F(s), which is the Laplace transform of f(t) defined by:









t
tt
tf
,0
0,sin
)(
(8 marks)

2. Mock Exam Jan-April 2013 Page 2 of 8 EE203Methematical Methods for Engineers II





4 Obtain the solution to the following differential equation by Laplace transform,

)(23 tryyy  ; 0)0(y and 0)0(y

where r(t) is a square wave defined by: 








2,0
21,2
10,0
)(
t
t
t
tr
(10 marks)


5 Use power series method to obtain the solution for the following differential equation:

0''')1(
2
 yxyyx
(15 marks)


6 A periodic function is defined for one period as:








x
x
xf
0if
0if
)(


(a) Justify whether f(x) is even function, odd function or neither.
(4 marks)

(b) Expand f(x) in a Fourier series.
(13 marks)

3. Mock Exam Jan-April 2013 Page 3 of 8 EE203Methematical Methods for Engineers II





The Following FORMULA SHEET Will be given during the final exam


The Fourier series of a function f(x) of period p=2L is given by:










1
sincos)(
0
n
L
xn
b
L
xn
aaxf
nn

where:



L
L
dxxf
L
a )(
2
1
0 


L
L
n dx
L
xn
xf
L
a

cos)(
1



L
L
n dx
L
xn
xf
L
b

sin)(
1

The Fourier series of an even function of period 2L is a “Fourier cosine series” 









1
cos)(
0
n
L
xn
aaxf
n


with coefficients 

L
dxxf
L
a
0
0 )(
1
and 

L
dx
L
xn
xf
Ln
a
0
cos)(
2 

The Fourier series of an odd function of period 2L is a”Fourier sine series” 









1
sin)(
n
L
xn
n
bxf


with coefficients 

L
dx
L
xn
xf
Ln
b
0
sin)(
2 


Power series:
0
n
n
n
xay




4. Mock Exam Jan-April 2013 Page 4 of 8 EE203Methematical Methods for Engineers II



LAPLACE TRANSFORM TABLE

)(tf




0
)()( dtetfsF
st
)(tfe
at
)(asF )()( atuatf 
)(sFe
as )('tf
)0()(fssF )(''tf
)0(')0()(
2
fsfsFs 



t
u(t)

5. Mock Exam Jan-April 2013 Page 5 of 8 EE203Methematical Methods for Engineers II



The Solution Scheme

1. (a). Solve



1.(b).
Given that x
exy
2
1)(

 is one solution of 044  yyy .
To get the second solution, let x
exuxy
2
2 )()(

 .
Then,
ueeuy
xx 22
2
2

 and xxx
euueeuy
222
44



Substitute these into the differential equation to get
04)2(444
222222

 xxxxxx
ueueeueuueeu

Some cancellations occur because x
e
2 is one solution, leaving
0
2

x
eu
Or 0u
Two integrations yield dcxxu )( . Since we only need one second solution 2y , we only need
one u, so we will choose c=1 and d=0. This gives xxu)( and x
xexy
2
2
)(

 .

Now
0
22
)(
4
222
22






x
xxx
xx
e
xeee
xee
xW Consider the previous example, the differential equation to solve is:

2
3
x
e
y
dx
dy
x
x


Solution
Rewrite the equation: 0)3(
32
 dyxdxeyx
x


Thus:
x
eyxM 
2
3 and
3
xN


2
3x
dx
dM
 and
2
3x
dx
dN
 
x
dN
dy
dM

 (the equation is exact)

Then f(x,y) will be a function such that:


x
eyxM
x
df


2
3 and
3
xN
y
df




)()(),(
3
ygeyxygMdxyxf
x

and
33
)(' xNygx
dy
df

0)('
33
 xxyg

Therefore, Cyg)( 
x
eyxyxf 
3
),(

We get the solution: Ceyxyxf
x

3
),(

)(
33
CexyCeyx
xx



6. Mock Exam Jan-April 2013 Page 6 of 8 EE203Methematical Methods for Engineers II




W (x) is NOT zero for all x. Therefore, 1y and 2y form a fundamental set of solutions for all x, and the
general solution of 044  yyy is
xx
xececxy
2
2
2
1
)(




2 (a) tyyy 2sin35'2'' 

The corresponding homogenous equation is: 05'2''  yyy .
Hence the auxiliary equation is:
r
2
+2r+5=0, which has TWO complex roots as follows.

 r1,2 = -1  2 i

The general solution is : pc
Yyy 
The complimentary solution is: )2cos2sin(
212211
tCtCeycycy
t
c



Since ttr 2sin3)( , then we guess that tDtCY
p
2sin2cos
tDtCY
p
2cos22sin2 
tDtCY
P 2sin42cos4 

Substitute into original DE:
tTDttDtCtDtC 2sin3)2sin2(cos5)2cos22sin2(22sin42cos4 
tCDtDC 2sin)34(2sin)4(0 

The above equation is True for all t: Choose two values of t as follows:
0344/
040


CDt
dCt
 




17/3
17/12
D
C
ttY
p 2sin
17
3
2cos
17
12



The general solution: pc
Yyy 
tttCtCey
t
2sin
17
3
2cos
17
12
)2cos2sin(
21 

#

2(b).
Separation of variable: 






x
x
y
dx
dy
2
1
 x
dxx
y
dy )1(2 
 dx
xy
dy







1
12


By integration:

7. Mock Exam Jan-April 2013 Page 7 of 8 EE203Methematical Methods for Engineers II


kxxy lnln2
Kxey
x
lnlnlnln
2


x
exKy ..
2
 #


3(a).
3)0(and10)0( :where016  yyyy 


016310
016)0()0(
2
2


YsYs
YysyYs  222222
4
4
4
3
4
10
4
103
)(







ss
s
s
s
sY

#4sin
4
3
4cos10)( ttty 


3(b)








t
tt
tf
,0
0,sin
)(
f(t) can be rewritten in term of step functions as:   )(.sin)(.sin)()(sin)(   tuttuttututtf
)().sin()(.sin)(  tuttuttf
 )().sin(sin)(  tutttf 
1
1
.
1
1
1
1
)(
222









s
e
e
ss
sF
s
s


##

4. )(23 tryyy 
; 0)0(y and 0)0(y
From the figure, it can be written: )2(2)1(2)(  tututr
ss
e
s
e
s
ssY
22 22
)12(


ss
e
sss
e
sss
Y
2
22
2)
)23(
1
(2)
)23(
1
(







Partial fractions: 2
2/1
1
12/1
)2)(1(
1
)23(
1
2







 sssssssss

Therefore: ss
e
sss
e
sss
Y
2
2
2
2/1
1
12/1
2
2
2/1
1
12/1























    )2(21)1(21)(
)2(2)2()1(2)1(


tueetueety
tttt #

8. Mock Exam Jan-April 2013 Page 8 of 8 EE203Methematical Methods for Engineers II



5. 0''')1(
2
 yxyyx
; Power series method:

0
n
n
n
xay



.'
1
1




n
n
n
xany 2
2
)1(''




n
n
nxanny
2
2
2
)1()1(




n
n
nxannx
+ .
1
1




n
n
n
xanx −
0
n
n
n
xa

 =0
n
nor
n
nxann




0
2
)1)((
+2
2
)1)((



 
n
n
n
xann + .
0
1
n
nor
n
nxan


 −
0
n
n
n
xa

 =0 n
nn
n
n
n
n
n xanaannxann ])1)([()1)((
0
2
2
 





=0
n-2=k
0]1)1)([()1)(2(
00
2  





k
k
k
k
k
k xakkkxakk

0])1)(2()1)(2[(
0
2 



k
k
k
k xakkakk
for k=0,1,2,….
)2(
)1(
2




k
ak
a
k
k

2
1
2
0
2 
a
a
0
2
4
8
1
4
a
a
a  0
3a
0
5
2
3
5

a
a …….
Thus:
xaxxay
1
42
0 .....)
8
1
2
1
1(  #



6. Fourier series: 







x
x
xf
0if
0if
)(

(a). Odd function since f(-x)=-f(x)
an=0 and a0=0

(b). Period: L= )1)1((
2

n
n
n
b
b1=-4; b3=-4/3; b5=-4/5 and so on


 






1
4
sin)1)1((
2
)(
n
nx
n
xf
#