[june 2015] subject: engg. mathematics-ii paper code be-301 · solution of engineering...

Solution of Engineering Mathematics-II

[June 2015]

Subject: Engg. Mathematics-II Paper Code BE-301

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


[June 2015]


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


[June 2015]


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


[June 2015]


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


[June 2015]


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


[June 2015]


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


[June 2015]


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


[June 2015]


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,


[June 2015]


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


[June 2015]


02

2

aa


3 1 13(3 1) 0 2 0a a a

13

2

aa


0 024

1

4 4 2 8

a aaa


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


[June 2015]


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)



[June 2015]


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


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)


[June 2015]


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


[June 2015]


F


2 2 2p q a …. (2) where a2 = 2c1


2

2 yy

aa q z q

z


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


[June 2015]


. 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



[June 2015]


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


[June 2015]


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


[June 2015]


. 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


[June 2015]


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

[june 2015] subject: engg. mathematics-ii paper code be-301 · solution of engineering...

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