Vidyalankar Applied Mathematics - III Prelim Question...

1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 1

Vidyalankar S.E. Sem. III [BIOM]

Applied Mathematics - III Prelim Question Paper Solution

f(t) =

1

2

1L

s 2s 2 =

12

1L

s 1 1 =

t 1

2

1e L

s 1

= et sin t

s1

2

eL

s 2s 2 = e(t) sin (t ) H(t )

ux = ex cos y + 3x2 3y2

uxx = ex cos y + 6x uy = ex sin y 6xy

uyy = ex cos y 6x uxx + uyy = ex cos y + 6x ex cos y 6x

= 0 u satisfies Laplace equation u is Harmonic.

If n is an even integer then since, n m 1 = (m + n)! , we have

Jn(x) =

m 2m n

m 0

( 1) (x / 2)m!(m n)!

Jn (x) =

m 2m n

2m nm 0

( 1) ( x)

m!(m n)!2

If n is even, (2m + n) is even & (x)2m+n = x2m+n

Jn (x) =

m 2m n

m 0

( 1) (x / 2)m!(m n)!

= Jn(x)

Hence, Jn(x) is even. Again, if n is odd then (2m + n) is odd & hence,

(x)2m+n = 2m n(x )

Jn(x) =

m 2m n

m 0

( 1) ( 1) (x / 2)m!(m n)!⋅ ⋅

=

m 2m n

m 0

( 1) (x / 2)m!(m n)!

= Jn(x)

Hence, Jn(x) is odd

1. (a)

1. (b)

1. (c)

Vidyala

nkar

Vidyalankar : S.E. – Maths III


We have,

u = ax4 + bx2 y2 + cy4 + dx2 2y2

ux

= 4ax3 +2bxy2 + 2dx

uy

= 2bx2y + 4cy3 4y

v = 4x3y exy3 + 4xy vx

= 12xy2 ey3 + 4y

vy

= 4x3 3exy2 + 4x

f(z) is analytic. It satisfies Cauchy Riemann equations.

ux

= vy

4ax3 + 2bxy2 + 2dx = 4x3 3exy2 + 4x Comparing both sides, we get

4a = 4 a = 1 2b = 3e …(1) 2d = 4 d = 2

Also uy

=

vx

2bx2y + 4cy3 4y = 12x2y + ey3 4y

Comparing both sides, we get 2b = 12 b = 6 4c = e …(2)

From (1), we get 2b = 3e

12 = 3e e = 4

From (2), we have 4c = e 4c = 4 c = 1

a = 1, b = 6, c = 1, d = 2, e = 4

a0 =

0

2f(x) dx =

0

2x dx =

2

0

2 x2

=

an =

0

2 n xf(x) cos dx =

0

2 n xx cos dx

1. (d)

2. (a)

Vidyala

nkar

Prelim Question Paper Solution


=

2 2

20

n x n xsin cos

2 xn n

=

n

2 22 2

22

1( 1) 02 0n nn

=

n 2 2

2 2 2 2

2 ( 1)

n n =

n2

2 2

2 1 1

n

=

n

2 2

( 1) 12

n

=

2 2

4 , if n is oddn

0, if n is even

The Half Range cosine series is given by

f(x) =

0n

n 1

a n xa cos

2

=

n

2 2n 1

( 1) 1 n x2 cos

2 n

The curve C is the boundary of the triangle in

xoy plane as z = 0.

F = 2 2ˆ ˆ ˆy i x j (x z) k where z = 0

= 2 2ˆ ˆ ˆy i x j x k

F =

2 2

ˆ ˆ ˆi j k

x y z

y x x

= ˆ ˆ ˆ0 i j ( 1) 2(x y) k

= ˆ ˆj 2(x y) k

(F ) . n̂ =

ˆ ˆj 2(x y) k . k̂ as n̂ = k̂ .

= 2(x y) We have by Stoke’s theorem.

C

F. dr =

S

ˆ( F) . n ds

= S

2(x y) ds = R

2(x y) dx dy

= 1 x

0 0

(x y) dy dx =

x1 2

0 0

y2 xy dx

2

2. (b)

x

y

z

A B

C

O (0, 0, 0)

(1, 0, 0)

(1, 1, 0)

O x

y

B

A

(1, 0) (0, 0)

C y=x

(1, 1)

Vidyala

nkar



=

1 2

2

0

x2 x dx

2 =

1 2

0

x2 dx

2

=

13

0

x3

= 13

(i)

3L H t

2 =

3

s21

es

as1

1 eH(t a)s

L sin t H t2

=

s2e L sin t

2 L f(t)H(t a) = ase L f(t a)

=

s2e L cos t

=

s2

2

se

s 1

3

L sinH t H t2 2

=

3

L sin t H t L H t2 2

=

s 3s

2 22

s 1e e

ss 1

(ii) Since, n

nd

x J xdx

= xn Jn + 1 (x)

nn

dx J x dx

dx =

nn 1x J x dx + C

xn nJn (x) = n

n 1x J x dx + C … (1)

Put n = 2 in equation (1), 2

3x J x dx = x2J2 (x) + C … (2)

Putting n = 1 in equation (1), 1

2x J x dx = x 1J1 (x) + C … (3)

Now, 3J x dx = x2 (x2) J2 (x) + 122 x J x dx + C

3J x dx = J2 (x)

12J x

x + C

Find the Fourier series for f(x) = | x | in (, ).

Hence deduce that 2

8 =

2 2 2 2

1 1 1 1...

1 3 5 7

a0 =

1

f(x) dx =

0

0

1x dx x dx

3. (a)

2. (C)

Vidyala

nkar



=

02 2

0

1 x x2 2

=

2 210

2 2

= 2

1 =

an =

1

f(x) cos nx dx =

1

| x | cos nx dx =

0

2x cos nx dx

=

20

2 sinnx cos nxx (1)

n n =

2 2

2 cos n 1

n n

=

n

2 2

2 ( 1) 1

n n =

n

2

2 ( 1) 1

n

f(x) is even function bn = 0.

f(x) =

0

n 1

aan cos nx

2

| x | =

n

2n 1

2 ( 1) 1cos nx

2 n =

2 2 2

2 2 2 2cos x cos 3x cos 5x ...

2 1 3 5

= 2 2 2

4 cos x cos 3x cos 5x...

2 1 3 5

Putting x = 0, we get

0 = 2 2 2

4 1 1 1...

2 1 3 5

2

8 =

2 2 2

1 1 1...

1 3 5

t tr e cos t i e sin t j i.e. x = et cos t ; y = et sin t

dx = et (cos t – sin t) dt, dy = et (sin t + cos t) dt t tdr [e (cos t sin t) i e sin t cos t) j ] dt x2 + y2 = e2 t (cos2 t + sin2 t) = e2 t

(x2 + y2)3/2 = (e2 t)3/2 = e3 t

F .dr =

t

3 t

e (cos t i sin t j)

e. et [ (cos t sin t) i + (sin t + cos t) j ] dt

= 2t

3t

e{cos t (cos t sin t) sin t (sin t cos t)} dt

e

= et dt

0t 0-t t

11

e 1F .dr e dt e 1

1 e

1 e

F .dre

3. (b)

Vidyala

nkar



f (x)

0

n nn 1

aa cosnx b sinnx

2 … (1)

an =

1

f(x)dx =

0 π

π 0

1 π dx x dxπ

=

20

0

1 xπxπ 2

=

221 π

π 2

=

21π 2

= 2

an = π

-

1f(x) cosnx dx

π =

0

0

1 cos nx dx x cos nx dx

π

=

0

20

1 sinnx sinnx cosnxx 1

n n n

= 2 2

1 cosn 10 0 (0) 0

n n

=

n

2 2

11 1

n n =

n

2

11 1

n

bn = π

-

1f(x) sin nx dx

π =

0

0

1 sin nx dx x sin nx dx

π

=

0

20

1 cosnx cosnx sinnxx 1

n n n

=

1 1 cosn cosnx0

n n

=

11 cosn cosn

n n =

11 2cosn

n

= n1

1 2 1n

Putting these values in (1), we get

f (x) =

n n

2n 1 n 1

1 1 1 2 11cosnx sinnx

4 nn

= 2 2 2

2 cos x cos3x cos5x.....

4 1 3 5

sinx 13 sin2x sin3x .....

1 2

… (2)

3. (c)

Vidyala

nkar



Now f(x) is discontinuous at x=0. At a point of discontinuity x = c

f (x) =

x c x c

1lim f(x) lim f(x)

2

f (0) = 1

02

= 2

Hence substituting x=0 in equation (2), we get

2

= 2 2 2

2 1 1 1.........

4 1 3 5

2

8 =

2 2 2

1 1 1........

1 3 5

=

2n 1

1

2n 1

L{sin 3u} = 2

3

s 9

L{u sin 3u} = 1

2

d 31

ds s 9 nL t f(t) = n d

1 L f(t)ds

=

22

3(2s)

s 9

=

22

6s

s 9

t

0

L u sin u du =

22

1 6ss s 9

=

22

6

s 9

t

0

(s)L f(u)du

s

t

4t

0

L e u sin 3u du =

22

6

s 4 9 =

22

6

s 8s 16 9

=

22

6

s 8s 25

(ii) L{1cos t} = L{1} L{cos t}

= 2

1 ss s 1

1 cos tL

t =

2s

1 sds

s s 1

s

f(t)L (s) ds

t

=

2

s

1log s log s 12

= 2

s

12 logs log s 12

4. (a)

Vidyala

nkar



=

2 2s

1log s log s 1

2 =

2

2s

1 slog

2 s 1

=

2

2

1 slog

2 s 1 =

2

2

1 slog

2 s 1

Let u = 3x2y + 2x2 y3 2y2

ux

= 6xy + 4x

uy

= 3x2 3y2 4y

We know that

dv =

v v

dx dyx y

=

u u

dx dyy x

= 2 23x 3y 4y dx 6xy 4x dy

= 2 23x 3y 4y dx 6xy 4x dy Integrating, we get

v = x3 + 3xy2 + 4xy

orthogonal trajectory of family of curves is given by x3 + 3xy2 + 4xy = C1 x3 3xy2 4xy = C when C = C1

(i) We know that x +iy = z …(1)

x iy z …(2) From (1) and (2) we get

1

x z z2

, i

y (z z)2

x 1z 2

y iz 2

x 1

2z

y i

2z

We know that

z

=

x yx z y z

=

1 ix 2 y 2

=

1i

2 x y … (3)

z

=

x yx yz z

=

1 ix 2 y 2

=

1i

2 x y … (4)

4. (b)

4. (c)

Vidyala

nkar



From (3) and (4) we get

2

z z =

z z

=

1i i

4 x y x y

=

2 2 2 2

2 2

1i i

4 x y x yx y =

2 2

2 2

14 x y

2

4z z

=

2 2

2 2x y

(ii) Note that, log f ' (z) = 21log f ' (z)

2 =

1log [f '(z). f '(z)

2

= 1 1

log f '(z) log [f '(z)]2 2

L.H.S. =

2 2

2 2log| f '(z)|

x y

= 4

2 1 1log f ' (z) log f '(z)

2 2z z

= 2

log f '(z) 2 logf '(z)z z zz

= 2

(0) 2 (0) 0

z z (z, z are independent)

By Green’s Thm.

C

pdx Q dy =

R

Q Pdx dy

x y

Now C

f d r = 2 2

C

x y i xy j dx idy

= 2 2

C

x y dx xy dy

By comparison P = x2y, Q = xy2

Qx

= y2, Py

= x2

C

f d r = 2 2

R

y x dx dy

put x = r cos, y = r sin, dx dy = r dr d

=

r a(1 cos )

2

0 r 0

2 r r dr d

5. (a)

y

x

0

r = a(1+cos)

Vidyala

nkar



=

a (1 cos )4

00

r2 d4

=

4 4

0

1a (1 cos ) d

2

=

442

0

a2 cos d

2 2 =

4 8

0

8a cos d2

Put 2

= t, d = 2dt & using reduction formula,

C

f d r =

/2

4 8

0

8a 2cos t dt

=

4 7 5 3 116a

8 6 4 2 2

= 435 a

16

Since w =

5 4z4z 3

We have 4wz 3w = 5 4z z(4w + 4) = 5 + 3w

z =

5 3w4(w 1)

when |z| = 1,

5 3w4(w 1)

= 1

5 3(u iv)(u iv 1)

= 4

5 3u i3vu 1 iv

= 4

2 25 3u 9v = 2 216 u 1 v

25 + 30u + 9u2 + 9v2 = 16(u2 + 2u + 1 + v2) 16u2 9u2 + 32u 30u + 16 25 + 16v2 9v2 = 0 7u2 + 2u 9 + 7v2 = 0 7u2 + 2u + 7v2 = + 9

2 22u u v

7 =

97

2 22 1u u v

7 49 =

9 17 49

221

u v7

= 6449

centre =

1,0

7, radius =

87

5. (b)

Vidyala

nkar



i is root of J0(x) = 0 Let the Bessel fourier series be,

x2 = f(x) = 0Ci J ( ix)

Multiplying by xJ0 (nx) on both sides and integrating from 0 to 1.

1

20

0

x x J ( nx) = 1

0 n 0 ii 1 0

Ci x J ( x) J ( x) dx

=

12

n 0 n0

C x J ( x)

= 21 n

CnJ ( )

2

Cn = 1

30 n2

1 n 0

2x J ( x) dx

J ( )

nn

dx J (x)

dx = xn Jn1 (x)

xnJn (x) = nn 1x J (x) dx

1

30 n

0

x J ( x) dx =

131 n

n 0

x J ( x)

= 1 n

n

J ( )

Cn = 21 n

2

J ( ) 1 n

n

J ( )

= n 1 n

2J ( )

x2 = 2i

0 i3i 1 i 1 i

42 J ( x)

J ( )

Consider f (t) = cos 4t cos 3t

L

f(t)t

= 2 2 2 2s

s sds

s 4 s 3

=

2 2 2 2

s

1log (s 4 ) log (s 3 )

2 =

2 2

2 2s

1 s 3log

2 s 4

=

2 2

2 2

1 s 3log

2 s 4 …(1)

By definition of Laplace Transform The equation (1) means

st

0

cos4t cos3te dt

t =

2 2

2 2

1 s 3log

2 s 4

Putting s = 0

0

cos4t cos3tdt

t =

2

2

1 3log

2 4 =

3log

4

6. (a)

5. (c)

Vidyala

nkar



= 3z + 4 + 2i Let = u + iv and z = x + iy u + iv = 3 (x + iy) + 4 + 2i

= 3x + 3iy + 4 + 2i u + iv = (3x + 4) + i (3y + 2) u = 3x + 4 and v = 3y + 2

Now | z | = k i.e. 2 2x y k

x2 + y2 = k2 This is a circle with centre (0, 0) and radius k.

x = 2 2k y

x = k if y = 0 (k, 0) x = 0 if y = k (0, k) u = 3k + 4 and v = 3k + 2 and u = 4 and v = 2 u = 3k + 4 and v = 2 and u = 4 and v = 3k + 2 are the images of (k, 0) and (0, k) for | z | = k.

a0 =

2

0

1f(x) dx =

22

0

1 xdx

2 =

23

0

1 ( x)4x 3

=

3 31

12 =

3212

= 2

6

an =

2

0

1f(x) cos nx dx =

22

0

1 xcos nx dx

4

=

22

2 30

1 sinnx cos nx sinnxx 2 2( 1)( 1)x 1

4 n n n

= 2 2

1 cos 2 n 20 2 0 0 0

4 n n

= 2 2

1 2 24 n n

= 2

1

n [ cos 2n = 1]

bn =

2

0

1f(x) sinnx dx =

2 2

0

1 ( x)sinnx dx

4

=

22

2 30

1 cos nx sinnx cos nxx 2 x 1 2( 1)( 1)

4 n n n

=

2 2

3 3

1 cos 2 n 2 cos 2n 20 0

4 n nn n

6. (b)

6. (c)

Vidyala

nkar



bn =

2 2

3 3

1 2 24 n nn n

= 0

Now, f(x) =

0

n 1

aan cos nx bn sinnx

z

2x

2 =

2

2n 1

cos nx12 n

=

2

2 2 2 2

cos x cos 2x cos 3x...

1 1 2 3

Put x = 0, we get

2

4 =

2

2 2 2 2 2

1 1 1 1...

1 1 2 3 4

2

6=

2 2 2 2

1 1 1 1...

1 2 3 4

Vidyala

nkar

Vidyalankar Applied Mathematics - III Prelim Question...

Documents

Transcript of Vidyalankar Applied Mathematics - III Prelim Question...