JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B. Tech II Year MATHEMATICS-III Previous Question Paper (April/May - 2023 - R15)

B. Tech | II Year | I Semester | R15 | April/May - 2023

Answer PaperPaper Code: 123AHMaximum Marks: 75Time: 3 hours

Note: i) Question paper consists of Part A, Part B. ii) Part A is compulsory, which carries 25 marks. In Part A, Answer all questions. iii) In Part B, Answer any one question from each unit. Each question carries 10 marks and may have a, b as sub questions.

Part A

Find the complementary function of

x^2 (d^2y/dx^2) - x(dy/dx) - 3y = log x

2 marks

Answer

The given equation is a Cauchy-Euler equation of the form

x^2(d^2y/dx^2) + ax(dy/dx) + by = f(x)

. To find the complementary function, substitute

x = e^z

(or

z = log x

), which transforms the equation into a linear differential equation with constant coefficients. Let

D = d/dz

. The equation becomes

[D(D-1) - D - 3]y = 0

, simplifying to

(D^2 - 2D - 3)y = 0

. The auxiliary equation is

m^2 - 2m - 3 = 0

, which factors as

(m-3)(m+1) = 0

, yielding roots

m = 3, -1

. Thus, the complementary function is

y_c = c_1e^{3z} + c_2e^{-z}

. Substituting back

e^z = x

, we obtain

y_c = c_1x^3 + c_2/x

Find the P.I of

(x^2 (d^2y/dx^2) - 3x(dy/dx) + 3y = x^2 + 3x + 1)

3 marks

Answer

The given equation is a Cauchy-Euler equation:

x²(d²y/dx²) - 3x(dy/dx) + 3y = x² + 3x + 1

To solve for the Particular Integral (P.I.), we substitute x = e^z, which implies z = ln(x). Let D = d/dz. The operators transform as x(dy/dx) = Dy and x²(d²y/dx²) = D(D-1)y. The equation becomes:

[D(D-1) - 3D + 3]y = (e^z)² + 3e^z + 1

(D² - 4D + 3)y = e^2z + 3e^z + 1

The P.I. is given by:

P.I. = 1/(D² - 4D + 3) * (e^2z + 3e^z + 1)

Evaluating each term:

For e^2z: 1/(2² - 4(2) + 3) * e^2z = 1/(4 - 8 + 3) * e^2z = -e^2z.
For 3e^z: Since D=1 makes the denominator zero, we use the rule 1/f(D) * e^az = z * 1/f'(D) * e^az. Here, f'(D) = 2D - 4. So, z * 1/(2(1) - 4) * 3e^z = -3/2 * z * e^z.
For 1 (which is e^0z): 1/(0² - 4(0) + 3) * 1 = 1/3.

Substituting z = ln(x) back into the expression:

P.I. = -x² - (3/2)x ln(x) + 1/3

Prove that

P_n'(1) = n(n+1)/2

2 marks

Answer

Prove that

∫x J_0^2(x) dx = x/2 [J_0^2(x) + J_1^2(x)]

3 marks

Answer

Write the Cauchy Riemann equations in polar form.

2 marks

Answer

Find the analytic function whose real part is

3 marks

Answer

Find the zeros of the function

sin(1/z)

2 marks

Answer

Show that the function

e^z

has an essential singularity at

z = ∞

3 marks

Answer

Find the fixed points of

w = (z-3)/(z+1)

2 marks

Answer

Find the region to which

w = 1/z

transforms to region

|z| ≤ 2

3 marks

Answer

Part B

Solve

(1+x)^2 (d^2y/dx^2) + (1+x)(dy/dx) + y = 4cos(log(1+x))

10 marks

Answer

Obtain the series solution of the equation

x^2 (d^2y/dx^2) + x(dy/dx) + (x^2-4)y = 0

10 marks

Answer

Prove that

∫ P_n(x)P_n'(x) dx = (P_n(x)^2)/(2n+1)

10 marks

Answer

m_1, m_2

are roots of

J_n(x) = 0

, then prove that

∫ x J_n(m_1 x) J_n(m_2 x) dx = 0

10 marks

Answer

State and prove Cauchy’s integral formula.

10 marks

Answer

Evaluate

∫_c (z^2 - x + iy) dz

, where C consists of the line segments from

z = 0

z = i

and the other from

z = i

z = 1+i

10 marks

Answer

State and prove Laurent series for the function

f(z)

10 marks

Answer

Evaluate

∫_0^π (dθ)/(a+b cos θ)

;

(a>b>0)

10 marks

Answer

Show that every bilinear transformation maps the circles in the z-plane onto the circles in the w-plane.

10 marks

Answer

Find the graph of the region

1 < x < 2, π < y < 2π

under the mapping

w = e^z

10 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B. Tech II Year MATHEMATICS-III Previous Question Paper (April/May - 2023 - R15)

Part A

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Part B

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