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

Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech II MATHEMATICS-III Previous Question Paper (November/December - 2020 - R13)

B.Tech | II | II | R13 | November/December - 2020

Answer PaperPaper Code: 114DEMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks.

Main

Solve

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

15 marks

Answer

The given differential equation is a second-order linear differential equation with variable coefficients:

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

To solve this, we first transform it into a linear differential equation with constant coefficients by substituting

x = e^z

, which implies

z = log x

. Using the operator

D = d/dz

, we have the following substitutions:

x(dy/dx) = Dy

and

x²(d²y/dx²) = D(D-1)y

. Multiplying the original equation by

x²

, we get:

x²(d²y/dx²) - x(dy/dx) + xy = log x

. Substituting the operators:

[D(D-1) - D + x]y = z

. Since

x = e^z

, the equation becomes

(D² - 2D + e^z)y = z

. However, recognizing the standard form for Cauchy-Euler equations, we observe the homogeneous part

x²y'' - xy' + y = 0

has the auxiliary equation

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

, which simplifies to

(m-1)² = 0

. Thus, the complementary function is

y_c = (C₁ + C₂ log x)x

. To find the particular integral, we use the method of variation of parameters where

y_p = u₁y₁ + u₂y₂

with

y₁ = x

and

y₂ = x log x

. The Wronskian is

W = x(1 + log x) - log x(x) = x

. The particular integral is calculated as

y_p = -x ∫ (x log x * (log x / x²) / x) dx + x log x ∫ (x * (log x / x²) / x) dx

. Evaluating these integrals yields

y_p = -x ∫ (log x)²/x² dx + x log x ∫ (log x)/x² dx

. Integrating by parts, we obtain

y_p = (log x + 1) + (log x + 1) = 2 log x + 2

. The general solution is

y = (C₁ + C₂ log x)x + log x + 2

Solve in series the equation

y'' + xy' + y = 0

15 marks

Answer

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

y = ∑_{n=0}^{∞} a_n x^n

Differentiating term by term, we obtain:

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

Simplifying the terms:

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

Extracting the constant term (n=0):

2 a_2 + a_0 + ∑_{n=1}^{∞} [(n+2)(n+1) a_{n+2} + (n+1) a_n] x^n = 0

Setting coefficients to zero, we find the recurrence relation:

a_{n+2} = -a_n / (n+2)

Calculating coefficients:

For n=0: a_2 = -a_0 / 2
For n=1: a_3 = -a_1 / 3
For n=2: a_4 = -a_2 / 4 = a_0 / (4 * 2) = a_0 / 8
For n=3: a_5 = -a_3 / 5 = a_1 / (5 * 3) = a_1 / 15

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

y_1(x) = 1 - x^2/2 + x^4/8 - ...

y_2(x) = x - x^3/3 + x^5/15 - ...

If m1, m2 are roots of

J_n(x) = 0

, then prove that

∫₀¹ x J_n(m₁x)J_n(m₂x) dx = 0

15 marks

Answer

415 marks

Find an analytic function whose imaginary part is

e^x(x sin y + y cos y)

8 marks

Answer

Show that

xy²

cannot be real part of an analytic function.

7 marks

Answer

Find Laurent expansion of

1 / (z² - 4z + 3)

for:

15 marks

|z| < 1

7 marks

Answer

|z| > 3

8 marks

Answer

Evaluate

∫₀^2π (sin²θ / (a + b cos θ)) dθ

; (a > b > 0)

15 marks

Answer

Find the image of

|z| > 1

under the transformation

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

15 marks

Answer

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

Jawaharlal Nehru Technological University Hyderabad B.Tech II MATHEMATICS-III Previous Question Paper (November/December - 2020 - R13)

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