# Engineering Mathematics

 Question 1
The initial value problem

$\frac{dy}{dt} +2y=0,\;\;y(0)=1$

is solved numerically using the forward Euler's method with a constant and positive time step of $\Delta t$.
Let $y_n$ represent the numerical solution obtained after $n$ steps. The condition $|y_{n+1}| \leq |y_n|$ is satisfied if and only if $\Delta t$ does not exceed _____. (Answer in integer)
 A 0 B 1 C 2 D 3
GATE ME 2023      Differential Equations
Question 1 Explanation:
Solution of differential equation given by forward Euler's method is
$y_{n+1}=y_{n}+h f\left(x_{n}, y_{n}\right)$
Given: $\frac{d y}{d t}+2 y=0$
$\Rightarrow \quad \frac{d y}{d t}=-2 y$
and step size, $\mathrm{h}=\Delta \mathrm{t}$
$\therefore \quad y_{n+1}=y_{n}+\Delta t\left(-2 y_{n}\right)$
$\mathrm{y}_{\mathrm{n}+1}=\mathrm{y}_{\mathrm{n}}(1-2 \Delta \mathrm{t})$
$\frac{y_{n+1}}{y_{n}}=1-2 \Delta t \;\;\;...(i)$

Given:
$\left|y_{n+1}\right| \leq\left|y_{n}\right|$
$\Rightarrow \quad\left|\frac{y_{n+1}}{y_{n}}\right| \leq 1$

From equation (i)
$\therefore \quad|1-2 \Delta t| \leq 1$

Case (i) when $(1-2 \Delta t) \gt 0$ :
Then, $\quad 1-2 \Delta t \leq 1$
$\Delta t \geq 0$

Case (ii) when $(1-2 \Delta t) \lt 0$ :
then $-(1-2 \Delta t) \leq 1$
$1-2 \Delta \mathrm{t} \geq-1$
$\Delta \mathrm{t} \leq 1$
$\therefore \quad$ Value of $\Delta t: 0 \leq \Delta t \leq 1$
 Question 2
Consider the second-order linear ordinary differential equation

$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=0,\;\; x \geq 1$

with the initial conditions

$y(x=1)=6,\;\;\left.\begin{matrix} \frac{dy}{dx} \end{matrix}\right|_{x=1}=2$

The value of $y$ at $x=2$ equals _________. (Answer in integer)
 A 6 B 9 C 13 D 17
GATE ME 2023      Differential Equations
Question 2 Explanation:
$\frac{x d^{2} y}{d x^{2}}+\frac{x d y}{d x}-y=0, x \geq 1$
This is an Euler-cauchy equation and has solutions of the form
\begin{aligned} & y=x^{n} \\ & y^{\prime}=n x^{n-1} \\ & y^{\prime \prime}=n(n-1) x^{n-2} \\ & x^{2}(n)(n-1) x^{n-2}+x\left(n x^{n-1}\right)-x^{n}=0 \\ & x^{n}[x(n-1)+n-1)=0 \\ & n^{2}-1=0 \\ & n=\pm 1 \\ & y_{1}=x \\ & y_{2}=\frac{1}{x} \\ & y=C_{1} x+\frac{C_{2}}{x} \end{aligned}
Given $y(x=1)=6$
\begin{aligned} 6 & =C_{1}(1)+\frac{C_{2}}{(1)} \\ C_{1}+C_{2} & =6 \;\;\;...(i)\\ \frac{d y}{d x} & =C_{1}-\frac{C_{2}}{x^{2}}=2 \\ x & =1 \end{aligned}
$C_{1}=C_{2}=2$
$C_{1}=4, C_{2}=2$
$y(x=2)=4(2)+\frac{2}{(2)}=9$

 Question 3
The smallest perimeter that a rectangle with area of 4 square units can have is ______ units. (Answer in integer)
 A 4 B 6 C 8 D 10
GATE ME 2023      Calculus
Question 3 Explanation:

Given $\quad B \times L=4$
Perimeter $(P)=2 B+2 L$
perimeter to be smallest possible
$\frac{d P}{d B}=0=\frac{d}{d B}\left(2 B+2 \times \frac{4}{B}\right)=0$
$2-\frac{8}{B^{2}}=0$
$B=\pm 2 \Rightarrow B=2 \quad L=2$
Smallest perimeter $=2(B+L)=2(2+2)=8$
 Question 4
Which one of the options given is the inverse Laplace transform of $\frac{1}{s^3-s}$
$u(t)$denotes the unit-step function.
 A $\left ( -1+\frac{1}{2}e^{-t}+\frac{1}{2}e^{t} \right )u(t)$ B $\left (\frac{1}{3}e^{-t}-e^{t} \right )u(t)$ C $\left ( -1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)} \right )u(t-1)$ D $\left ( -1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)} \right )u(t-1)$
GATE ME 2023      Differential Equations
Question 4 Explanation:
Inverse Laplace of $\frac{1}{s^{3}-s}$
$\qquad L^{-1}\left[\frac{1}{s\left(s^{2}-1\right)}\right](u(t))$
\begin{aligned} & =\mathrm{L}^{-1}\left[\frac{1}{s} \times \frac{1}{(s-1)} \times \frac{1}{(s+1)}\right] u(t) \\ & =L^{-1}\left[\frac{-1}{s}+\frac{1}{2} \times \frac{1}{(s-1)}+\frac{1}{2(s+2)}\right] u(t) \\ & =\left[\frac{e^{t}}{2}+\frac{e^{-t}}{2}-1\right] u(t) \end{aligned}
 Question 5
The value of $k$ that makes the complex-valued function

$f(z)=e^{-kx}(\cos 2y -i \sin 2y)$

analytic, where $z=x+iy$, is _________. (Answer in integer)
 A 1 B 2 C 3 D 4
GATE ME 2023      Complex Variables
Question 5 Explanation:
\begin{aligned} & f(z)=e^{-k x} \cos 2 y-i e^{-k x} \sin 2 y \\ & \text { Suppose }=e^{-k x} \cos 2 y=4(x, y) \\ & =-i e^{-k x} \sin 2 y=v(x, y) \end{aligned}
If function is analytical then it satisfy the equation
\begin{aligned} & \frac{\partial u}{\partial x}=-k e^{-k x} \cos 2 y &...(i)\\ & \frac{\partial u}{\partial y}=-2 e^{-k x} \sin 2 y &...(ii)\\ & \frac{\partial v}{\partial x}=-k e^{-k x} \sin 2 y &...(iii)\\ & \frac{\partial v}{\partial y}=-2 e^{-k x} \cos 2 y &...(iv) \end{aligned}
Cauchy-Riemann equation $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
by putting equation (i), (iv) and solve then $k=2$

There are 5 questions to complete.