(SEM II) THEORY EXAMINATION 2024-25 ENGINEERING MATHEMATICS-II

About this previous year paper

B.Tech (Semester II) Theory Examination 2024–25

Maximum Marks: 70 | Time: 3 Hours

All symbols have their usual meaning.

This paper evaluates problem-solving ability in ordinary differential equations (ODEs), Laplace transforms, series convergence, complex analysis, contour integration, Fourier series, and analytic functions. The question paper is divided into three sections: A, B, and C.

SECTION A — Short Answer / Concept Questions (Q1a–Q1g)

Total = 14 marks (2 marks × 7 questions)

This section contains brief questions from basic ODEs, Laplace transforms, harmonic functions, residue theorem, and convergence of sequences.

Q1(a)

Find general solution of differential equation:

d3ydx3+dydx=0\frac{d^3y}{dx^3} + \frac{dy}{dx} = 0dx3d3y+dxdy=0

Q1(b)

Find particular integral of:

y′′−8y′+16y=e4xy'' - 8y' + 16y = e^{4x}y′′−8y′+16y=e4x

Q1(c)

Find Laplace Transform of:

f(t)=sin⁡2tcos⁡3tf(t)= \sin 2t \cos 3tf(t)=sin2tcos3t

Q1(d)

Find inverse Laplace transform of:

F(s)=s−1s2+3s+2F(s)=\frac{s-1}{s^2+3s+2}F(s)=s2+3s+2s−1

Q1(e)

Test convergence of sequence:

an={1,n=2p for some p∈N1n,otherwisea_n = \begin{cases} 1, & n = 2^p \text{ for some } p \in \mathbb{N} \\ \frac{1}{n}, & \text{otherwise} \end{cases}an={1,n1,n=2p for some p∈Notherwise

Q1(f)

Show that the function

h(x,y)=x2+xy−y2h(x,y)=x^2 + xy - y^2h(x,y)=x2+xy−y2

is harmonic.

Q1(g)

Find the residue at the simple pole of:

f(z)=8z3(z−1)(z+1)3f(z)=\frac{8z^3}{(z-1)(z+1)^3}f(z)=(z−1)(z+1)38z3

SECTION B — Descriptive Questions (Attempt any 3)

Total = 21 marks (7 marks × 3)

Q2(a)

Solve the ODE:

y′′−2y′+2y=x+excos⁡xy'' -2y' +2y = x + e^x \cos xy′′−2y′+2y=x+excosx

Q2(b)

Solve using Laplace Transform:

y′′′+2y′′−y′−2y=0y''' + 2y'' - y' - 2y = 0y′′′+2y′′−y′−2y=0

with initial conditions:

y(0)=1, y′(0)=2, y′′(0)=2y(0)=1,\; y'(0)=2,\; y''(0)=2y(0)=1,y′(0)=2,y′′(0)=2

Q2(c)

Test convergence of series:

∑n=1∞1⋅3⋅5⋯(2n−1)2⋅4⋅6⋯(2n)x2n\sum_{n=1}^{\infty} \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} x^{2n}n=1∑∞2⋅4⋅6⋯(2n)1⋅3⋅5⋯(2n−1)x2n

Q2(d)

Given analytic function f(z)=u+ivf(z)=u+ivf(z)=u+iv satisfying:

ucos⁡y−vsin⁡yex=cosh⁡y−cos⁡x\frac{u\cos y - v\sin y}{e^x} = \cosh y - \cos xexucosy−vsiny=coshy−cosx

Find f(z)f(z)f(z) such that:

f(π2i)=3f\left(\frac{\pi}{2i}\right)=3f(2iπ)=3

Q2(e)

Evaluate using contour integration:

∮Cz2−7(z−1)(z−2)(z−3)dz\oint_C \frac{z^2 - 7}{(z-1)(z-2)(z-3)} dz∮C(z−1)(z−2)(z−3)z2−7dz

where CCC is the circle ∣z∣=2|z|=2∣z∣=2.

SECTION C — Long Answer / Advanced Problems (Q3–Q7)

Total = 35 marks (7 marks each)

Q3 – ODEs / Simultaneous Differential Equations

Q3(a)

Solve:

x2y′′−xy′+4y=xsin⁡(log⁡x)x^2 y'' - x y' + 4y = x \sin(\log x)x2y′′−xy′+4y=xsin(logx)

Q3(b)

Solve the system:

dxdt=3x+8y\frac{dx}{dt} = 3x + 8y dtdx=3x+8y dydt=−x−3y\frac{dy}{dt} = -x - 3ydtdy=−x−3y

Q4 – Laplace / Convolution

Q4(a)

Find Laplace transform of:

∫0teτsin⁡ττdτ\int_0^t \frac{e^\tau \sin \tau}{\tau} d\tau∫0tτeτsinτdτ

Q4(b)

Using convolution theorem, evaluate inverse Laplace transform:

L−1[p2(p2+4)(p2+9)]L^{-1} \left[ \frac{p^2}{(p^2+4)(p^2+9)} \right]L−1[(p2+4)(p2+9)p2]

Q5 – Series Convergence / Fourier Series

Q5(a)

Test convergence of series:

1+α+1β+1+(α+1)(2α+1)(β+1)(2β+1)+(α+1)(2α+1)(3α+1)(β+1)(2β+1)(3β+1)+⋯1 + \frac{\alpha+1}{\beta+1} + \frac{(\alpha+1)(2\alpha+1)}{(\beta+1)(2\beta+1)} + \frac{(\alpha+1)(2\alpha+1)(3\alpha+1)}{(\beta+1)(2\beta+1)(3\beta+1)} + \cdots1+β+1α+1+(β+1)(2β+1)(α+1)(2α+1)+(β+1)(2β+1)(3β+1)(α+1)(2α+1)(3α+1)+⋯

Q5(b)

Find Fourier series of f(x)=x2f(x)=x^2f(x)=x2 on −π≤x≤π-\pi \le x \le \pi−π≤x≤π.
Hence prove:

∑n=1∞1n2=π26\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}n=1∑∞n21=6π2

Q6 – Analyticity / Complex Functions

Q6(a)

Show that function

f(z)={x3y5(x+iy)(x6+y10),z≠00,z=0f(z)= \begin{cases} \frac{x^3 y^5}{(x+iy)(x^6+y^{10})}, & z\ne 0 \\ 0, & z=0 \end{cases}f(z)={(x+iy)(x6+y10)x3y5,0,z=0z=0

is not analytic at z=0z=0z=0 even though Cauchy-Riemann equations are satisfied at origin.

Q6(b)

Show that

f(z)=z∣z∣f(z)=z|z|f(z)=z∣z∣

is nowhere analytic.

Q7 – Complex Integration / Cauchy’s Theorem

Q7(a)

State Cauchy’s Integral Theorem.
Verify Cauchy theorem for function

f(z)=eizf(z)=e^{iz}f(z)=eiz

around rectangle with vertices

1+i, 1−i, −1−i, −1+i1+i,\; 1-i,\; -1-i,\; -1+i1+i,1−i,−1−i,−1+i

in counter-clockwise direction.

Q7(b)