Upload
others
View
6
Download
0
Embed Size (px)
Citation preview
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 1
UNIT–1
1. a) What is periodic function, gave an example of periodic function. 2
Solution: A function f (x) is said to be periodic if there exist a least positive integer T such that Suppose
( ) ( ) ( 2 ) ...... ( )f x f x T f x T f x nT , n I
Where T is called period of f (x).
Example: sin x, cos x of period 2 while period of tan x is
b) What are the Dirichlet’s conditions for a Fourier series expansion? 2
Solution: Suppose f (x) is any finite function defined in (–L, L), then the Fourier series of f (x) is exists only if
the following conditions are satisfied:
(i). f (x) is periodic i.e. f (x) = f (x + 2L), where 2L is the period of f (x)
(ii). f (x) and its integral are finite and single valued.
(iii). f (x) has a finite number of discontinuities.
(iv). f (x) has a finite number of maxima and minima.
These conditions are known as Dirichlet’s conditions.
c) Find the Fourier transform of 2a xe , where a > 0 3
Solution: Given the function: 2
( )x
F x e
The Fourier transform of a function F(x) is given by
1
( ) ( )2
ipxf p F x e dx
2
21 1( )
2 2
x ipxx ipxf p e e dx e dx
2 2 2 22
4 4 2 41 1( )
2 2
p p ip px ipx x
f p e dx e dx
[Add and subtract p
2/4]
22
241
( )2
ipp x
f p e e dx
Putting , so that 2
ipx t dx dt
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 2
2
24
1( )
2
p
tf p e e dt
2 2
4 41 1
( )2 2
p p
f p e e
2
since t
e dt
Thus 2
24
1( )
2
p
xF e e f p
By change of scale property, we have
22 a xa xF e F e
1( )
pF F ax f
a a
2 21
4 41 1 1 1
2 2
p p
a ap
f e ea a a a
Thus 2
241
2
p
a x aF e ea
Answer
d) Expand f (x) = x sin x, 0 < x < 2 in a Fourier series. 7
Solution: Here, 2L = 2 – 0 i.e. L = Suppose the Fourier series of f (x) with period 2L is,
0
1 1
( ) cos sin2
n nn n
a n x n xf x a b
L L
0
1 1
( ) cos sin2
n nn n
af x a nx b nx
---- (1) [Since L = ]
Now, 2 2 20 00 0
1 1 1( ) sin ( cos ) 1( sin ) 2a f x dx x x dx x x x
and 2 2
0 0
1 1( ) cos sin cosna f x nx dx x x nx dx
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 3
2 2
0 0
1 12 cos sin sin ( 1) sin ( 1)
2 2x nx x dx x n x n x dx
…. (2)
2
2 2
0
1 cos( 1) cos( 1) sin( 1) sin( 1)1
2 1 1 ( 1) ( 1)
n x n x n x n xx
n n n n
2
2; 1
1na n
n
For n = 1, from equation (2), we get
22
1 00
1 1 cos 2 sin 2 1sin 2 1
2 2 2 4 2
x xa x x dx x
and 2 2
0 0
1 1( ) sin sin sinnb f x nx dx x x nx dx
2 2
0 0
1 12 sin sin cos( 1) cos( 1)
2 2x nx x dx x n x n x dx
…. (3)
2
2 2
0
1 sin( 1) sin( 1) cos( 1) cos( 1)1
2 1 1 ( 1) ( 1)
n x n x n x n xx
n n n n
0; 1nb n
For n = 1, from equation (3), we get
2
22
1 00
1 1 sin 2 cos 21 cos 2 1
2 2 2 2 4
x x xb x x dx x x
1b
From equation (1), we have
01 1
2 2
( ) cos cos sin sin2
n nn n
af x a x a nx b x b nx
2
2
1 2sin 1 cos cos sin
2 1n
x x x nx xn
Answer
Or
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 4
d) Express f (x) = x as a 7
(i). Half range cosine series in 0 < x < 2
(ii). Half range sine series in 0 < x < 2
Solution: (i). Suppose the half range cosine series is,
0
1
( ) cos2 2
nn
a n xf x a
…. (1)
Now,
22
2 2
0 0 00
2( ) 2
2 2
xa f x dx x dx
and 2 2
0 0
2( ) cos cos
2 2 2n
n x n xa f x dx x dx
2
2 2 2 2 2 20
2 4 4 ( 1) 4sin 1 cos 0 0
2 2
nn x n xx
n n n n
2 2
4( 1) 1n
nan
Putting in equation (1) we get
2 2
1
( 1) 14( ) 1 cos
2
n
n
n xf x
n
Answer
(ii). suppose the half range sine series is,
1
( ) sin2
nn
n xf x b
…. (2)
Now, 2 2
0 0
2( ) sin sin
2 2 2n
n x n xb f x dx x dx
2
2 20
2 4 4 ( 1)cos 1 sin 0 0 0
2 2
nn x n xx
n nn
4
nbn
Putting in equation (2), we get
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 5
1
4 ( 1)( ) sin
2
n
n
n xf x
n
Answer
UNIT–2
2. a) 2
22 sin cos sin 2
4L t t L t
p
2
b) Write the Linearity property of Laplace Transform 2
Solution: If a and b are any constants and F (t) and G(t) be any two function of t, then
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )L a F t b G t a L F t b L G t or L a F t b G t a f p b g p
c) Find the Laplace Transform of f (t), 3
2, 0 2
( ) 1, 2 3
7, 3
t
f t t t
t
Solution: By definition of Laplace transform,
0
( ) ( ) ptL f t f t e dt
2 3
0 2 32 ( 1) 0.pt pt pte dt t e dt e dt
32
20 2
2 ( 1) 1 0pt pt pte e e
tp p p
3 2
2 3 2
2 2
2 2 11
p pp p pe e
e e ep p pp p
2 2 3 3
2 2
2 2( )
p p p pe e e e
L f tp p pp p
Answer
d) Evaluate: 2
1
3 2
6 22 16
6 11 6
s sL
s s s
7
Solution: Suppose 2 2
3 2
6 22 16 6 22 16
( 1)( 2)( 3) 1 2 36 11 6
s s s s A B C
s s s s s ss s s
…. (1)
Where A, B and C determine by
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 6
2
1
6 22 16 6 22 160
( 2)( 3) (1)(2)s
s sA
s s
2
2
6 22 16 24 44 164
( 1)( 3) ( 1)(1)s
s sB
s s
and 2
3
6 22 16 54 66 162
( 1)( 2) ( 2)( 1)s
s sC
s s
Putting in equation (1) and taking Inverse Laplace Transform, we get
2
1 1 1
3 2
6 22 16 1 14 2
2 36 11 6
s sL L L
s ss s s
2
1 2 3
3 2
6 22 164 2
6 11 6
t ts sL e e
s s s
Answer
Or
d) Using Convolution theorem, Evaluate 1
2 2 2( )
sL
s a
Solution: Suppose 2 2
( )s
f ss a
and 2 2
1( )g s
s a
1 1
2 2( ) cos ( )
sL f s L at F t
s a
And 1 1
2 2
1 sin( ) ( )
atL g s L G t
as a
By Convolution theorem of Inverse Laplace transform, we have
1
0 ( ) ( ) ( ) ( )
tL f s g s F x G t x dx
1
2 0 02 2
sin[ ( )] 1cos . 2 cos .sin ( )
2
t ts a t xL ax dx ax at ax dx
a as a
0
1sin sin [ ( )]
2
tax at ax ax at ax dx
a
2 cos sin sin( ) sin( )A B A B A B
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 7
0 0 0
1 1 sin sin 2 sin 1 sin 2
2 2
t t tat ax at dx at dx ax at dx
a a
0
cos 21 sin
2 2
tax at
x ata a
1 cos cos
sin 0 2 2 2
at att at
a a a
1 sin
2t at
a
Thus
1
22 2
sin
2
s t atL
as a
UNIT–3
3. a) Explain the ordinary point and singular point of differential equation. 2
Solution: Suppose the second order differential equation is
2
0 1 22
y y( ) ( ) ( ) y 0
d dP x P x P x
dxdx ------ (1)
Where P0(x), P1(x) and P2(x) be polynomial function of independent variable x.
(i). Ordinary Point:
A point x = 0, is said to be an ordinary point of equation (1) if P0(0) (ii). Singular Point:
A point x = 0, is said to be an singular point of equation (1) if P0(0)
b) Give the complete solution of differential equation when the roots of indicial equations are equal.
Solution: When the roots of indicial equation are equal i.e. m1 = m2
The Complete solution is 1
1
1 2y
y ym m
m m
c cm
c) Solve the differential equation 3
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 8
2
2
y y2 tan 5y sec
xd dx e x
dxdx
Solution: Given the differential equation is,
2
2
y y2 tan 5y sec
xd dx e x
dxdx ….. (1)
Here, P = –2 tan x, Q = 5 and R = ex sec x
Now this problem solve by Removable of first derivative method.
Suppose the complete solution is,
y = v y1 ….. (2)
Where v is a function of x only.
Now we can find the value of y1 such as
1 1( 2 tan )
log sec2 21y sec
Pdx x dxxe e e x
and 2 2 2 2 21
1 1 1 15 ( 2 tan ) ( 2 sec ) 5 tan sec 6
4 2 4 2
dPQ Q P x x x x
dx
11
sec
y sec
xxR e x
R ex
The normal form of Removable of first derivative is,
2
1 12
d vQ v R
dx
2
26
xd vv e
dx …. (3)
This is LDR of higher order.
The A.E. is
2 6 0m
6m i
Therefore 1 2. . cos 6 sin 6C F c x c x
Now, 2 2
1 1. .
76 1 6
xx x e
P I e eD
The solution of equation (3) is,
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 9
1 2. . . . cos 6 sin 67
xev C F P I c x c x
Putting in equation (2), which our complete solution
1 2y cos 6 sin 6 sec7
xec x c x x
Answer
d) Solve 2
2
2
y y(1 ) 2 y 0
d dx x
dxdx in series solution. 7
Solution: Given the differential equation is,
2
2
2
y y(1 ) 2 y 0
d dx x
dxdx ….. (1)
Here, P0(x) = 1– x2
Clearly at x = 0, P0(x) = 1– 0 = 1 ≠ 0
Therefore x = 0 is ordinary point.
Suppose the complete solution of equation (1) by Power series method is
2 30 1 2 3y = + ........a a x a x a x ….. (2)
0
y = kk
k
a x
…. .(3)
Differentiating both sides w.r.t. x, we get
1
0
y = k
k
k
da k x
dx
and
2 2
2 0
y ( 1) k
k
k
da k k x
dx
Putting the value of y, yd
dxand
2
2
yd
dx in equation (1), we get
2 2 1
0 0 0
(1 ) ( 1) 2 0k k kk k k
k k k
x a k k x x a k x a x
2
0 0 0 0
( 1) ( 1) 2 0k k k kk k k k
k k k k
a k k x a k k x a k x a x
…. (4)
Equating the coefficient of x0 on both sides in equation (4), we get
2 02 (2 1) 0 0 0a a
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 10
02
2
aa
Equating the coefficient of x1 on both sides in equation (4), we get
3 1 13(3 1) 0 2 0a a a
13
2
aa
Equating the coefficient of x2 on both sides in equation (4), we get
0 024
1
4 4 2 8
a aaa
Equating the coefficient of x3 on both sides in equation (4), we get
5 3 3 35 (5 1) 3(3 1) 6 0a a a a
3 1 15
1
20 20 2 40
a a aa
Putting the values of a2, a3, a4 and a5 in equation (2), we get
2 3 4 50 01 10 1y .....
2 2 8 40
a aa aa a x x x x x
2 4 3 5
0 1y 1 ..... .....2 8 2 40
x x x xa a x
Answer
Or
d) Solve by the method of variation of parameter 7
2 1 yD x
Solution: The given differential equation is,
2
2
yy
dx
dx ….. (1)
Here, P = 0, Q = 1 and R = x
The A.E. is
2 1 0m
m i
Therefore C.F. is
1 2y cos sinc c x c x
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 11
Suppose cosu x and sinv x
sinu x and cosv x
and 2 2cos sincos sin 1 0
sin cos
x xw x x
x x
Suppose the complete solution of equation (1) is
y cos sinAu Bv A x B x ….. (2)
Where A and B determine by the formula,
1 1 1.
sin ( cos ) ( sin )v R
A dx c x x dx c x x x cw
1cos sinA x x x c
and 2 2 2.
cos (sin ) 1.( cos )u R
B dx c x x dx c x x x cw
2sin cosB x x x c
Putting the values of A and B in equation (2), we get
1 2y cos sin cos sin cos sinx x x c x x x x c x
1 2y cos sinc x c x x Answer
UNIT–4
4. a) Find the Partial differential equation by eliminating a and b from the relation 2
2 2 2( ) (y )x a b z c
Solution: Given the equation is,
2 2 2( ) (y )x a b z c …..(1)
Partially differentiating w.r.t., x we get
2 ( ) 0 2 0z
x a zx
x a zp …. (2)
Again equation (1) Partially differentiating w.r.t., y, we get
0 2 (y ) 2 0y
zb z
y b zq …. (3)
Putting in equation (1), we get
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 12
2 2 2 2z p q z c Answer
b). Solve y yz p z x q x 2
Solution: Given differential equation is
y yz p z x q x ….. (1)
This is Lagrange PDE.
Here P = y z, Q = z x and R = x y
The Lagrange A.E. is
y
y y
dx d dz
z z x x
Taking first two ratios, we get
yy y
y
dx dx dx d
z z x
Integrate both sides, we get
2 2 2 2
1 1y y
2 2 2
x xc c
Taking Last two ratios, we get
yy y
y
d dzd z dz
z x x
Integrate both sides, we get
2 2 2 2
2 2y y
2 2 2
z zc c
The General solution of equation (1), we get
2 2 2 2y y
, 02 2
x z
Answer
c) Solve 2 2 2
3 2y
2 22
y y
xz z ze
xx
3
Solution: The given Partial differential equation is
2 2 2
3 2y
2 22
y y
xz z ze
xx
….. (1)
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 13
Suppose, ,y
D Dx
From (1), we have 2 2 3 2y2 xD DD D z e
The A.E. is,
2 2 1 0 1, 1m m m
The C.F. is, 1 2. . (y ) (y )C F x x x
3 2y
3 2y 3 2y
2 2 2 2
1 1. .
252 (3) 2(3)(2) (2)
xx x e
P I e eD DD D
The Complete solution is,
3 2y
1 2(y ) (y )25
xe
z x x x
Answer
d) Solve the Charpits method 2 2( ) yp q q z 7
Solution: Given Partial differential equation is,
2 2( ) yp q q z …. (1)
2 2y y 0p q q z
Suppose 2 2y yf p q q z
2 20, , , 2 y, 2 y
y
f f f f fp q q p q z
x z p q
The Charpits A.E.is
y
y
dp dq dz dx d
f f f f f f f fp q p q
x z z p q p q
2 2 2
y
0 (2 y) (2 y ) 2 y 2 y
dp dq dz dx d
p q p p q q z p z qp q q
Taking first two ratios, we get
2
dp dq
p q p
p dp q dq
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
F
Page 14
Integrate both sides, we get
2 2 2p q a …. (2) where a2 = 2c1
Putting in equation (1), we get
2
2 yy
aa q z q
z
Putting in equation (2), we get
22
2 2yap a
z
2 2 4 2
2 2 2 2
2
yy
a z a ap p z a
zz
Since ydz p dx q d
2
2 2 2 yy y
a az a dx d
z z
2
2 2 2
y y
y
z dz a da dx
z a
Integrating both sides, we get
2 2 2yz a a x c Answer
Or
d) Solve 2u u
ux t
by the method of separation of variables, where 3
, 0 6x
u x e 7
Solution: Given differential equation is,
2u u
ux t
….. (1)
With initial condition 3, 0 6
xu x e
Suppose the complete solution is,
( , ) ( ). ( ) .u x t X x T t X T …. (2)
Where X is function of x and T is function of t only.
Now equation (2), partially differentiate w.r.t. x and t respectively
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 15
. u dX
T X Tx dx
and . = .u dT
X X Tt dt
Putting the values in equation (1), we get
2 . .X T X T X T
(2. ) X T T T X
2. 2
1X T T X T
kX T X T
(Let)
X
kX
….. (3) and
21
Tk
T
…. (4)
Now from (3) and integrate w.r.t. x, we get
1
1
log log logX
X kx c kxc
1
( ) kx
X x c e
And From equation (4), we have
2 1
1 2
T T kk
T T
Integrate w.r.t. “t” on both sides, we get
2
2
1 1log log log
2 2
k T kT t c t
c
2
1
2( )
kt
T t c e
Putting the values of T(t) and X(x) in equation (2), we get
1 2
1
2( , )
k
k xt
u x t c c e e
….. (5)
By initial condition, 36
xu e
at t = 0
1 2
36
x k xe c c e
Equating both sides, we get
1 2
6c c and 3k
Putting in equation (5), we get
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 16
23
( , ) 6 x t
u x t e Answer
UNIT–5
5. a) If 2 (2 1)u t i t j t k and (2 3)v t i j t k
, find . ?d
u vdt
at t = 1 2
Solution:
Now, 2 3 2 2 3 2. (2 1) . (2 3) 2 3 2 2 5 2u v t i t j t k t i j t k t t t t t t t t
2. 6 10 2 6 10 2 6d
u v t tdt
at t = 1 Answer
b) Find the directional derivative of y yx z z x in the direction of the vector 2 2i j k at the
point (1, 2, 0). 2
Solution: Given y yx z z x
Now ( y y )y
grad i j k x z z xx z
(y ) ( ) (y )z i x z j x k
2 3grad i j k at (1, 2, 0)
Suppose 2 2a i j k
2 2 2 2
31 4 4
a i j k i j ka
a
Directional Derivative = .grad a
2 2 1 10
. . 2 3 (2 2 6)3 3 3
i j kD D i j k
Answer
c) If yr xi j zk , then show that 2n ngrad r n r r
3
Solution: Given yr xi j zk
2 2 2 2yr x z
Partially differentiate w.r.t., x, y and z respectively, we get
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 17
2 2r
r xx
, 2 2y
y
rr
and 2 2r
r zz
r x
x r
,
y
y
r
r
and
r z
z r
Now, y
n ngrad r i j k r
x z
1 1 1n n nr r rn r i n r j n r k
x y z
1 2yn n nx zgrad r n r i j k n r r
r r r
Proved
d) Verify Stoke’s theorem for 2 2( ) 2 yF x y i x j taken round the rectangle bounded by
x = a, y = 0, y = b 7
Solution: Given the function is,
2 2( ) 2 yF x y i x j
By Stoke’s theorem we have,
. . c s
F d r curl F n ds
-----(1)
To evaluate Line integral:
The curve C consists of four lines AB, BC, CD and DA.
. . . . .c AB BC CD DA
F d r F d r F d r F d r F d r
------ (2)
Now , 2 2 2 2 ( y ) 2 y + y ( y ) 2 y y F d r x i x j dx i d j x dx x d
….. (3)
Along line AB,
Here = , = 0x a dx and 0 y b
2 2
0 0. 2 y y y
bb
ABF d r a d a ab
Along the line BC,
Here, y = , y = 0 and b d a x a
3 3
2 2 2 22. ( ) 2
3 3
aa
BC aa
x aF d r x b dx b x ab
Along the line CD,
= 0, = 0 and y 0x dx b
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 18
. 0CD
F d r
since 0F d r
Along the line OA,
y = 0, y = 0 and d a x a
3 3
2
2.
3 3
aa
DA aa
x aF d r x dx
Putting these values in equation (2), we get
32 2
. 3c
aF d r a b 3
2 2 2 2
3
aab a b 2
4 a b --- (3)
To Evaluate surface integral:
Now,
2 2
i j k
y z
+ y 2 y 0
Curl F Fx
x x
= ( 2y 2y) 4y Curl F k k
Since the surface on the xy-plane, then n k (Projection on XY-Plane)
and . 4y . 4yCurl F n k k
0
. 4y y b a
S acurl F n dS dx d
2
2
0
y 4 4
2
b
a
ax ab
…… (4)
From (4) and (3), we get
2. . 4c s
F d r curl F n ds ab
Hence Stoke’s theorem is verified.
Or
d) Prove that 2. ( 1)m m mdiv grad r r m m r 7
Solution: yr xi j zk
2 2 2 2yr x z
Solution of Engineering Mathematics-II
[June 2015]
Subject: Engg. Mathematics-II Paper Code BE-301
Page 19
Partially differentiate w.r.t., x, y and z respectively, we get
2 2r
r xx
, 2 2y
y
rr
and 2 2r
r zz
r x
x r
,
y
y
r
r
and
r z
z r
Now, y
m mgrad r i j k r
x z
1 1 1
y
m m mr r rm r i m r j m r k
x z
1 2ym m mx zgrad r m r i j k m r r
r r r
L.H.S = 2 2m m mdiv grad r div m r r m div r r
2 2 2 4(3) ( 2)m m m mm r div r r grad r m r r m r r
.div a div a a grad
2 2 23 ( 2) ( 1)m m mm r m r m m r = R.H.S. Proved