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JNTU B.Tech II MATHEMATICS-III Previous Question Paper (August/September - 2022 - R15)

B.Tech | II | II-I | R15 | August/September - 2022

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

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Solve the differential equation x2(d2y/dx2) + 4x(dy/dx) + 2y = log x.

7 marks

Answer

The given equation is a Cauchy-Euler equation: x2(d2y/dx2) + 4x(dy/dx) + 2y = log x. Substitute x = ez, so z = log x. Let D = d/dz. The equation transforms to: [D(D-1) + 4D + 2]y = z, which simplifies to (D2 + 3D + 2)y = z.

The auxiliary equation is m2 + 3m + 2 = 0, giving m = -1, -2. The complementary function is yc = c1e-z + c2e-2z = c1/x + c2/x2.

For the particular integral, yp = 1/(D2 + 3D + 2) * z = 1/2 * [1 + (D2 + 3D)/2]-1 * z = 1/2 * [1 - (D2 + 3D)/2] * z = 1/2 * (z - 3/2). Substituting back z = log x, yp = 1/2 * log x - 3/4.

The general solution is y = c1/x + c2/x2 + 1/2 * log x - 3/4.

b

Solve in series (1 - x2)(d2y/dx2) - x(dy/dx) + 4y = 0.

8 marks

Answer

The equation (1 - x2)y'' - xy' + 4y = 0 is a linear homogeneous differential equation. Since x=0 is an ordinary point, we assume a solution of the form y = Σ anxn for n=0 to infinity. Then y' = Σ n anxn-1 and y'' = Σ n(n-1) anxn-2.

Substituting into the equation: Σ n(n-1) anxn-2 - Σ n(n-1) anxn - Σ n anxn + 4 Σ anxn = 0. Extracting the coefficient of xn: (n+2)(n+1)an+2 - [n(n-1) + n - 4]an = 0, which simplifies to an+2 = [(n2 - 4) / ((n+2)(n+1))] an = [(n-2)(n+2) / ((n+2)(n+1))] an = [(n-2)/(n+1)] an.

For n=0: a2 = -2/1 * a0 = -2a0. For n=2: a4 = 0. Thus, all subsequent even terms a4, a6, ... are 0. The even series terminates at a2.

For n=1: a3 = -1/2 * a1. For n=3: a5 = 1/4 * a3 = -1/8 * a1. For n=5: a7 = 3/6 * a5 = -1/16 * a1.

The general solution is y = a0(1 - 2x2) + a1(x - 1/2x3 - 1/8x5 - 1/16x7 - ...).

215 marks
a

Solve the differential equation (2x + 3)2(d2y/dx2) - (2x + 3)(dy/dx) - 12y = 6x.

8 marks

Answer

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b

Using Frobenius method, solve in series the equation x(d2y/dx2) + (dy/dx) + xy = 0.

7 marks

Answer

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315 marks
a

Prove that Pn(x) = (1 / n!2n) * (dn/dxn)(x2 - 1)n.

7 marks

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b

Prove that (i) J1/2(x) = √(2/πx) sin x (ii) J-1/2(x) = √(2/πx) cos x.

8 marks

Answer

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415 marks
a

Prove that ∫01 x Jn(αx)Jn(βx)dx = 0 if α ≠ β and (1/2)[Jn+1(α)]2 if α = β, where α, β are roots of Jn(x) = 0.

8 marks

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b

Show that ∫-11 x Pn(x)Pn-1(x)dx = 2n / (4n2 - 1).

7 marks

Answer

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5

Show that the function f(z) = (z̅)2/z for z ≠ 0 and 0 for z = 0 satisfies C-R conditions at origin but does not have a derivative at origin.

15 marks

Answer

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615 marks
a

Show that v(x, y) = - sin x sinh y is harmonic. Find the conjugate harmonic of v and find an analytic function f = u + iv.

7 marks

Answer

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b

Evaluate ∮c [ez/(z(1-z)3)] dz where c is the circle |z - 1| = 1/2.

8 marks

Answer

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715 marks
a

Expand the function f(z) = 1/(z + 1)2 in power series about the point z = -i.

7 marks

Answer

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b

Evaluate the integral ∫-∞∞ [x2/((x2 + a2)(x2 + b2))] dx = π / (a + b).

8 marks

Answer

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815 marks
a

Discuss the transformation w = sin z. Also find its derivative.

7 marks

Answer

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b

Find the bilinear transformation that maps the points 0, 1, i in z-plane onto the points 1 + i, -i, 2 - i in the w-plane.

8 marks

Answer

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This page contains the solved question paper with answers of MATHEMATICS-III from JNTU for B.Tech MME-MMT II II-I, August/September - 2022, R15.

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