JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II MATHEMATICS-III Previous Question Paper (March - 2021 - R15)

B.Tech | II | I | R15 | March - 2021

Answer PaperPaper Code: 123AH/113AHMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

1

Obtain the series solution of the equation

y'' + xy = 0
.

15 marks

Answer

To find the series solution for the differential equation y'' + xy = 0, we assume a power series solution of the form:

y = ∑_{n=0}^{∞} a_n x^n

Differentiating term by term:

y' = ∑_{n=1}^{∞} n a_n x^{n-1}
y'' = ∑_{n=2}^{∞} n(n-1) a_n x^{n-2}

Substituting these into the original equation:

∑_{n=2}^{∞} n(n-1) a_n x^{n-2} + x ∑_{n=0}^{∞} a_n x^n = 0
∑_{n=2}^{∞} n(n-1) a_n x^{n-2} + ∑_{n=0}^{∞} a_n x^{n+1} = 0

Adjusting indices to equate powers of x to x^k:

  1. For the first sum, let k = n - 2, so n = k + 2. When n=2, k=0.
  2. For the second sum, let k = n + 1, so n = k - 1. When n=0, k=1.

The equation becomes:

2(1)a_2 + ∑_{k=1}^{∞} [(k+2)(k+1)a_{k+2} + a_{k-1}] x^k = 0

Equating coefficients to zero:

  1. a_2 = 0
  2. a_{k+2} = -a_{k-1} / [(k+2)(k+1)] for k ≥ 1.

Using the recurrence relation:

  1. If k=1: a_3 = -a_0 / (3·2)
  2. If k=2: a_4 = -a_1 / (4·3)
  3. If k=3: a_5 = -a_2 / (5·4) = 0 (since a_2 = 0)
  4. If k=4: a_6 = -a_3 / (6·5) = a_0 / (6·5·3·2)

The general solution is y = a_0 y_1(x) + a_1 y_2(x), where:

y_1(x) = 1 - x^3/6 + x^6/180 - ...
y_2(x) = x - x^4/12 + x^7/504 - ...
2

If m1, m2 are roots of

Jn(x) = 0
, then prove that
∫01 x Jn(m1x)Jn(m2x) dx = 0
.

15 marks

Answer

The Bessel function Jn(x) satisfies the Bessel differential equation:

x2y'' + xy' + (x2 - n2)y = 0

Let u = Jn(m1x) and v = Jn(m2x). These functions satisfy the following differential equations:

x2u'' + xu' + (m12x2 - n2)u = 0
x2v'' + xv' + (m22x2 - n2)v = 0

Dividing by x, we obtain:

(xu')' + (m12x - n2/x)u = 0
(xv')' + (m22x - n2/x)v = 0

Multiplying the first equation by v and the second by u, then subtracting the two equations, we get:

v(xu')' - u(xv')' + (m12 - m22)xuv = 0

Integrating both sides from 0 to 1:

(m22 - m12) ∫01 x Jn(m1x)Jn(m2x) dx = [x(u'v - uv')]01

Evaluating the right-hand side at the limits:

  1. At x = 0, the expression vanishes because x = 0 and Jn(0) is finite.
  2. At x = 1, the expression becomes 1 * (m1Jn'(m1)Jn(m2) - m2Jn'(m2)Jn(m1)).

Since m1 and m2 are roots of Jn(x) = 0, we have Jn(m1) = 0 and Jn(m2) = 0. Thus, the right-hand side is zero. Since m1 ≠ m2, the integral must be zero:

∫01 x Jn(m1x)Jn(m2x) dx = 0
3

Verify Cauchy’s theorem for the integral of

z3
taken over the boundary of the rectangle with vertices -1, 1, 1+ i, -1+ i.

15 marks

Answer

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4

Evaluate

∫c (y - x - 3x2i) dz
, where c consists of the line segments from z = 0 to z = i and the other from z = i to z = 1+i.

15 marks

Answer

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5

Find Laurent expansion of

1 / (z2 - 4z + 3)
for
1 < |z| < 3
.

15 marks

Answer

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6

Evaluate

∫02π (sin2θ / (a + b cos θ)) dθ
; (a > b > 0).

15 marks

Answer

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7

Determine the region of the w-plane into which the region bounded by

1/2 ≤ x ≤ 1
and
1/2 ≤ y ≤ 1
is mapped under the transformation
w = z2
.

15 marks

Answer

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8

Find the bilinear transform which maps the points z = 0, -i, -1 into the points w = i, 1, 0. Find the image of the line

y = mx
under this transformation.

15 marks

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About this paper

This page contains the question paper of MATHEMATICS-III from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech EEE II I, March - 2021, R15.

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