# BPUT SAMPLE PAPER FOR MATH-III

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MATH-III

FULL MARKS:-70

Q.1 is compulsory (comprise 2 marks for each bit )

Q.2 – Q.7 (do any five)  comprises 10 marks each

1.

(a) Find the general solution of the partial differential solution 2p + 3q=1

p=ΔZ / ΔX, q=ΔZ / ΔY

Ans:- Pp + Qq = R. dX /P = dY / Q = dZ / R

(dX / 2) = (dY / 3) = (dZ / 1)

=> 3X – 2Y = C1, Y- 3Z = C2

F(3X-2Y, Y-3Z) = 0

(b) solve Δ^2 Z / (ΔX / ΔY) = sin X

Ans:- ΔZ / ΔY = – cosX + C1(Y)

Z= -YcosX + C2(Y) + C3

(c) Convert the following partial differential equation into an ordinary differential equation into an ordinary differential equation by applying Laplace Transform:

(Δu/Δx) + 2x (Δu/Δt) = 2x, u(x,0) = 1,u (0,t)=1

Ans:- L(Δu / Δx) + L (2x. Δ / Δt) = L(2x)

=> L(Δu / Δx) + 2x {sL [u(x,t)] – u(x,0) }  = 2x / s

=> Δ / Δx . L [u(x,t)] + 2xsL[u(x,0)] = 2x + (2x / s)

=> Δ / Δx. [u(x,s)] + 2xsu (x,s) = 2x [1 + (1/s)]

L[u(x,t)] = u(x,s)

(d) Reduce the equation (Δ^2)u / Δ(t^2) = (C^2) { [ ( Δ^2) u / (Δt^2) ] + [ 1/r (Δu/Δx) ] into two ordinary differential equations by the method of separation of variables.

Ans:- LET u(r,x,t) = F(r,x) . G(t)

Δ^G / Δt + K^2 e^2G=0

Therefore,

Frr + 1/r(Fx) + K^2F =0

(e) Find 3√8i

Ans:- 3√8i

= 3√8 { cos [ (π/2 + 2kπ) / 3] + i sin [ (π/2 + 2kπ) / 3 }

or, K= 0,1,2

Thus it will be, 2e^(i.π/6), 2e^(5i. π/6, 2e^(3i. π/2)

(f) Solve the partial differential equation Uyy =Uy

Ans:- Let Uy = P.Py =P

Integrate w.r.t. y, we get

Loan p =y + i(x)

P = (e^y) . h(x)

Uy = (e^y) . h(x).

Integrate w.r.t. y we get

U = (e^y) . h(x) + k(y)

(g) Define an Analytic function. Give example of a function which is differentiable at a point but not analytic there.

Ans:- A function f(z) is sad to be analytic in a domain D if f(z) is defined and differentiable at all points of D.

Eg:- f(z) = z^2 is differentiable at z = 0 but not analytic there.

(h) Prove that:

Mod of (W) = mod of { (2z-1) / (z-2) }

Ans:- W= mod of { (2z-1) / (z-2) } < or =mod of 1

=> √(x^2 + y^2) < or = mod of 1

=>mod of z < or = 1

(i) What type of  singularities the functions sin (1/z) and tan (z) have at z=0?state with reason.

Ans:- sin(1/z), z=0 isolated essential singular point

tan(z),z =0 is not a singular point

(j) When is an equation of the form

AUxx + 2BUxy + CUyy = F(x,y,U,Ux,Uy) said to be elliptic, parabolic and hyperbolic?

Ans:- A Uxx +  2 B Uxy + C Uyy = F(x,y U,Ux,Uy) is said to be elliptic, parabolic and hyperbolic only if [AC – (B^2 ) ] > 0,    [AC – (B^2) ] = 0    and    [AC – (B^2) ]< 0 respectively.

2. Find complete solution of the following:

2xz- px^2 -2qxy +pq =0

Where, p =Δz/Δx, q=Δz/Δy

Ans :- f( x,y,z,p,q)) = 0

f = 2xz – px^2 – 2qxy +pq

fx = 2z – 2qy -2px, fy = -2qx, f=2x

fp = -x^2 + q, fq= -2xy + p

Charpit’s auxillary equation is :-

(dx / fp) = (dy / fq) = [dz / (pfp + qfq) = [dp / -( fx + pfz) ] = [dq / -(fy + qfz)]

dq = 0

=>q = a , p ={ [ 2x (z-ay) / (x^2 – a)}

dz = pdx + qdy

=>z = ay + b(x^2 –a), where a and b are two arbitrary constants.

3. A silver bar is of length 10 cm and is of constant cross section of area 1 cm^2. It is perfectly insulted laterally and its ends are kept at 0 c. Its density, thermal conductivity and specific heat are respectively 10.6 gm/cm^3, 1.04 cal/(cm sec 0C), 0.056 cal/(gm 0C). Its initial temperature in 0C is given by f(x) = x(10-x). Find the temperature u(x,t) in the bar.

Ans:- L = 10 cm, cross section area = 1 cm^2

ρ = 10.6 gm/cm^3, σ = 0.056 cal/ gm degree Celsius

K = 1.04 cal/ (cm sec degree celcius)

u(0,t) = 0, u(L-t) = 0 for all t.

u(x,0) = f(x) = x(10-x) initial temperature

C^2 = K / ρσ = 1.75 cm^2 / sec.

The temperature u(x,t) in the bar is

u(x,t) = summation of {Bn. Sin(nπx / L). C^(-λn^2x) , λn = (Cnπ / L)

where, Bn = (2/L) integration of{ f(x) sin [ nπx / L] } w.r.t. dx from 0 to 1.]

n= 1,2,3………

Bn = 0, ‘n’ even

= (800 / n^3. Π^3) , for n odd λn^2 = 0.01752 (nπ)^2

u(x,t) ={ [ 800 / (π^3) ] [sin 0.1 πx e(0.01752(π^2t) )]  } + { [1/9] . [ sin 0.3πxe^(0.01752 (3π)^2t] } +…………

4. Determine a and b such that the function u=ax^3 + by^3 is harmonic and find its harmonic conjugate.

Ans:- It is given that function u= ax^3 + by^3 is harmonic.

So, Uxx + Uyy = 0

Uxx = 6ax

Uyy = 6by

So, Uxx + Uyy =0

or,  6ax +6ay =0

or,  ax + by =0

or,  a=0, b=0

Hence u=0,

Therefore, Ux = Vx = 0

And Uy = -Vy = 0

Integrate with respect to y we get

Y = h(x)

or,  Vx = h’(x) =0

or,  h(x) = d

So, V = d (Harmonic conjugate)

So, the corresponding analytic function is

f(z) = u + iv =id

5. Find the poles of f(z) = { 1 / [z^6 +1] }

Ans:- The poles of f(z) = { 1/ [  (z^6) + 1] } are the roots of equation (z^6) + 1 =0

i.e., (z^6) + 1 =0

or,  z = [-1] ^(1/6)

=[cos π + i sin π] ^(1/6)

={ [ cos (2nπ + π) ] + i [ sin (2nπ + π]} ^(1/6)

= cos { [2n+1]π / 6 + i sin { [2n +1]π / 6}

Where, n = o, 1, 2, 3, 4, 5

→ When, n=0, z= cos (π/6) + i sin(π/6

= (√3/2) + i (1/2)

→ When n=1, z= cos(π/2) + i sin (π/2)

= (0 + i1)

→ When n=2, z= cos(5π/6) + i sin (5π/6)

= (-√3 / 2) + i (1/2)

→When n=3, z= cos (7π/6) + i sin (7π/6)

=-(√3/2) – i (1/2)

→ When n=4, z= cos (9 π/6) +i sin (9 π/6)

= 0-i1

→ When n=5, z = cos (11π/6) +i sin (11π/6)

=(√3 /2) – i(1/2)

6. (a) Find the isolated singular point corresponds to the function f(z) = {z-3} / z{z^2 +4}

Ans:- If f(z) = {z – 3} / z{ (z^2) +4}, then z=0, z = – 2i and z = 2i are three isolated singularities of f(z)

(b) Find the location and order of zeros of the function f(z) = cos h^2 Δz

Ans:- Here f(z) = cos (h^2)(2z)

Now f(z) =0

or,  cos (h^2)( 2z) =0

or,  cos h 2z =0

or,  cos 2 iz =0

or,  2iz = (2n+1) (π/2), where n= 0,+-1 ,+-2,……

or,  Z=(2n+1) (π / 4i)

Again f’(z) = 2 cos h 2z sin h 2z.2

= 2 sinh 4z

f’’(z) = 8 cosh 4z

= 8(2cosh^2 . 2z -1)

Therefore, f’(z) = 0, f’’(z) ‡ 0 when z=(2n +1) (π /4i)

and, n= 0, +-1, +-2,……

so, f(z) = cosh^2 2z has a zero of 2nd order at

z = (2n+1) (z/4i), n=0,+-1, +-2,……….

7. Determine α such that the function given bellow is harmonic. Also find its conjugate harmonic U=cosα x cosh 2y.

Ans:- U= cos αx cosh 2y

U is harmonic

or, Uxx + Uyy =0

or, Uxx = -α^2 cos αx cos hxy

or,  Uyy = 4 cos αx cos hzy

or, = +-2

Ux = Vy

and Uy = -Vx (C – R eqn)

or,  V = -sin 2x sin hzy + C1(x), C1(x)

= K (arbitrary constant)

or,  V = -sin 2x sin hzy + K (harmonic conjugate of u).

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