Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech II Year MATHEMATICS-III Previous Question Paper (July/August - 2021 - R15)

B.Tech | II Year | II-II | R15 | July/August - 2021

Answer PaperPaper Code: 124DEMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

1

Solve the equation

xy'' + y = 0
in series.

15 marks

Answer

To solve the differential equation

xy'' + y = 0
, we observe that x = 0 is a regular singular point. We seek a power series solution of the form
y = ∑n=0∞ anxn+r
, where
a0 ≠ 0
.

Differentiating term by term:

y' = ∑n=0∞ an(n+r)xn+r-1
y'' = ∑n=0∞ an(n+r)(n+r-1)xn+r-2

Substituting into the original equation

xy'' + y = 0
:

∑n=0∞ an(n+r)(n+r-1)xn+r-1 + ∑n=0∞ anxn+r = 0

Adjusting indices to align powers of

x
:

a0r(r-1)xr-1 + ∑n=1∞ [an(n+r)(n+r-1) + an-1]xn+r-1 = 0

The indicial equation is obtained by setting the coefficient of the lowest power

xr-1
to zero:

r(r-1) = 0 ⇒ r1 = 1, r2 = 0

The recurrence relation for

n ≥ 1
is:

an = -an-1 / [(n+r)(n+r-1)]

For

r = 1
:

an = -an-1 / [(n+1)n]

This yields

an = (-1)n a0 / [(n+1)!n!]
. The first solution is
y1(x) = x ∑n=0∞ [(-1)n xn / (n!(n+1)!)]
.

For

r = 0
, the recurrence relation leads to a second linearly independent solution involving a logarithmic term due to the integer difference between roots
r1 - r2 = 1
. The general solution is
y(x) = c1y1(x) + c2y2(x)
, where
y2(x) = y1(x)ln(x) + ∑n=0∞ bnxn
.

2

Find the series solution of

xy'' + y' + 4y = 0
.

15 marks

Answer

To find the series solution of the differential equation

xy'' + y' + 4y = 0
, we assume a power series solution of the form
y = Σ anxn
for n=0 to ∞. Differentiating term by term, we get
y' = Σ n anxn-1
and
y'' = Σ n(n-1) anxn-2
.

Substituting these into the differential equation:

x Σ n(n-1) anxn-2 + Σ n anxn-1 + 4 Σ anxn = 0
Σ n(n-1) anxn-1 + Σ n anxn-1 + 4 Σ anxn = 0

Combining the first two sums:

Σ n2 anxn-1 + 4 Σ anxn = 0

Shifting the index of the first sum by letting k = n-1, so n = k+1:

Σ (k+1)2 ak+1xk + 4 Σ akxk = 0

Equating the coefficients of xk to zero gives the recurrence relation:

ak+1 = -4ak / (k+1)2

Calculating the coefficients in terms of a0:

  • For k=0: a1 = -4a0 / 12 = -4a0
  • For k=1: a2 = -4a1 / 22 = -4(-4a0) / 4 = 4a0
  • For k=2: a3 = -4a2 / 32 = -4(4a0) / 9 = -16a0 / 9

The general solution is

y = a0(1 - 4x + 4x2 - (16/9)x3 + ...)
, which can be expressed using the Bessel function of the first kind of order zero,
y = a0J0(4√x)
.

3

Show that

P0(x) = 1, P1(x) = x, P2(x) = (3x2 - 1)/2
and hence express
2x2 - 4x + 2
as a Legendre polynomial.

15 marks

Answer

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4

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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515 marks
a

Evaluate

∫c (sec z)/(z - 1) dz
where c is the ellipse
4x2 + 9y2 = 9
.

8 marks

Answer

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b

Evaluate

∫c (ez)/(z2 + 1) dz
where c is the circle
|z - i| = 3
.

7 marks

Answer

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6

Evaluate

∫-∞∞ (x2)/(x4 + 10x2 + 9) dx
.

15 marks

Answer

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7

Find the bilinear transformation that maps the points

(∞, i, 0)
into the points
(0, i, ∞)
.

15 marks

Answer

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8

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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About this paper

This page contains the solved question paper with answers of MATHEMATICS-III from Jawaharlal Nehru Technological University Hyderabad for B.Tech MME-MMT II Year II-II, July/August - 2021, R15.

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