# Engineering Mathematics

 Question 1
$A 5 \mathrm{~cm}$ long metal rod $A B$ with initially at a uniform temperature of $T_{0}{ }^{\circ} \mathrm{C}$. Thereafter, temperature at both the ends are maintained at $0^{\circ} \mathrm{C}$. Neglecting the heat transfer from the lateral surface of the rod, the heat transfer in the rod is governed by the one-dimensional diffusion equation $\frac{\partial T}{\partial t}=D \frac{\partial^{2} T}{\partial x^{2}}$, where $D$ is the thermal diffusively of the metal, given as $1.0 \mathrm{~cm}^{2} / \mathrm{s}$.
The temperature distribution in the rod is obtained as
$T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \frac{n \pi x}{5} e^{-\beta n^{2} t}$ where $x$ is in $\mathrm{cm}$ measured from $A$ to $B$ with $x=0$ at $A, t$ is $s, C_{n}$ are constants in ${ }^{\circ} \mathrm{C}, \mathrm{T}$ is in ${ }^{\circ} \mathrm{C}$ and $\beta$ is in $\mathrm{s}^{-1}$.

The value of $\beta$ (in $\mathrm{s}^{-1}$, rounded off to three decimal places) is ___
 A 0.395 B 0.125 C 0.254 D 0.685
GATE CE 2023 SET-2      Partial Differential Equation
Question 1 Explanation:
Given: Diffusion equation:
$\frac{\partial T}{\partial t}=\mathrm{D} \frac{\partial^{2} \mathrm{~T}}{\partial \mathrm{x}^{2}}$
And also,
$D=1$
$\mathrm{T}(0, \mathrm{t})=0$
$\mathrm{T}(5,5)=0$
$T(x, 0)=T_{0}$
We know, solution of equation is given by,
$T(x, t)=\left(C_{1} \cos \alpha x+C_{2} \sin \alpha x\right) C_{3} e^{-\alpha^{2} t} \;\;\; ...(1)$

For $\mathrm{T}(0, t)=0$
$\left(C_{1}+0\right) C_{3} e^{-\alpha^{2} t}=0$
$\Rightarrow \mathrm{C}_{1}=0$

For $\mathrm{T}(\mathrm{s}, \mathrm{t})=0$
$\left(C_{2} \sin \alpha 5\right) C_{3} \mathrm{e}^{-25 t}=0$
$\Rightarrow \sin 5 \alpha=0$
$\Rightarrow \alpha=\frac{\mathrm{n} \pi}{5}$

Now, from equation (1)
$T(x, t)=C_{2} C_{3} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2}}{25} t}$
$=b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}$

Generalize solution
$T(x, t)=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}$
For $\mathrm{T}(\mathrm{x}, 0)=\mathrm{T}_{0}$
$\Rightarrow T_{0}=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right)$

From fourier series,
$b_{n}=0$ for even value of $n$
$\therefore \mathrm{T}(\mathrm{x}, \mathrm{t})=\sum_{\mathrm{n}=13,5 \ldots .}^{\infty} \mathrm{b}_{\mathrm{n}} \sin \left(\frac{\mathrm{n} \pi \mathrm{x}}{5}\right) \mathrm{e}^{-\frac{\mathrm{n}^{2} \pi^{2} \mathrm{t}}{25}}$

Given: $T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\beta n^{2} t}$
On comparison;
$\beta=\frac{\pi^{2}}{25}=0.395 \mathrm{sec}^{-1}$
 Question 2
Cholesky decomposition is carried out on the following square matrix $[\mathrm{A}]$.

$[A]=\left[\begin{array}{cc} 8 & -5 \\ -5 & a_{22} \end{array}\right]$

Let $\mathrm{I}_{\mathrm{ij}}$ and $\mathrm{a}_{\mathrm{ij}}$ be the $(i, j)^{\text {th }}$ elements of matrices $[L]$ and $[A]$, respectively. If the element $I_{22}$ of the decomposed lower triangular matrix $[\mathrm{L}]$ is 1.968 , what is the value (rounded off to the nearest integer) of the element $a_{22}$ ?
 A 5 B 7 C 9 D 11
GATE CE 2023 SET-2      Linear Algebra
Question 2 Explanation:
We know, cholesky decomposition,
$A=LL^{T}$
Where, $L=$ lower tringular matrix
$\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]$

$\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]$
On comparison on both sides,
$\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}$
$and,\;\; L_{11} L_{21}=-5$
$\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}$
$and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}$
$=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}$
$=6.998 \approx 7$

 Question 3
Two vectors $\left[\begin{array}{llll}2 & 1 & 0 & 3\end{array}\right]^{\top}$ and $\left[\begin{array}{llll}1 & 0 & 1 & 2\end{array}\right]^{\top}$ belong to the null space of a $4 \times 4$ matrix of rank 2 . Which one of the following vectors also belongs to the null space?
 A $\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top}$ B $\left[\begin{array}{llll}2 & 0 & 1 & 2\end{array}\right]^{\top}$ C $\left[\begin{array}{llll}0 & -2 & 1 & -1\end{array}\right]^{\top}$ D $\left[\begin{array}{llll}3 & 1 & 1 & 2\end{array}\right]^{\top}$
GATE CE 2023 SET-2      Calculus
Question 3 Explanation:
Given matrix is $4 \times 4$ and rank of matrix is 2 .
Therefore, rank of matrix $\neq$ No. of variables Thus, there two linearly dependent vectors \& two linearly independent vectors are present.
\begin{aligned} & X_{1}=\left[\begin{array}{llll} 2 & 1 & 0 & 3 \end{array}\right]^{\top} \\ & X_{2}=\left[\begin{array}{llll} 1 & 0 & 1 & 2 \end{array}\right]^{\top} \end{aligned}
$\therefore \quad X=\mathrm{K}_{1}\left[\begin{array}{l}2 \\ 1 \\ 0 \\ 3\end{array}\right]+\mathrm{K}_{2}\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 2\end{array}\right]$
For $\mathrm{K}_{1}=1 \text{ and } \mathrm{~K}_{2}=-1$
$X=\left[\begin{array}{c}1 \\ 1 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{\top}$
 Question 4
The solution of the differential equation.

$\frac{d^{3} y}{d x^{3}}-5.5 \frac{d^{2} y}{d x^{2}}+9.5 \frac{d y}{d x}-5 y=0$

is expressed as $y=C_{1} e^{2.5 x}+C_{2} e^{\alpha . x}+C_{3} e^{\beta x}$, where $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ and $\alpha$ and $\beta$ are constants, with $\alpha$ and $\beta$ being distinct and not equal to 2.5. which of the following options is correct for the values of $\alpha$ and $\beta$ ?
 A 1 and 2 B -1 and -2 C 2 and 3 D -2 and -3
GATE CE 2023 SET-2      Ordinary Differential Equation
Question 4 Explanation:
The differential eqn. can be written as :
$D^{3}-5.5 D^{2}+9.5 D-5=0$
Solving we get,
$D=1,2,2.5$
Hence, the solution is given as
$y=C_{1} e^{2.5 x}+C_{2} e^{1 x}+C_{3} e^{2 x}$
Comparing we get,
\begin{aligned} & \alpha=1 \\ & \beta=2 \end{aligned}
Hence, option (A) is correct.
 Question 5
The steady-state temperature distribution in a square plate $A B C D$ is governed by the 2-dimensional Laplace equation. The side $A B$ is kept at a temperature of $100{ }^{\circ} \mathrm{C}$ and the other three sides are kept at a temperature of $0{ }^{\circ} \mathrm{C}$. Ignoring the effect of discontinuities in the boundary conditions at the corners, the steady-state temperature at the center of the plate is obtained as $T_0 ^{\circ} \mathrm{C}$. Due to symmetry, the steady-state temperature at the center will be same $\left(\mathrm{T_0}{ }^{\circ} \mathrm{C}\right)$, when any one side of the square is kept at a temperature of $100{ }^{\circ} \mathrm{C}$ and the remaining three sides are kept at a temperature of $0{ }^{\circ} \mathrm{C}$. Using the principle of superposition, the value of $T_0$ is ___ (rounded off to two decimal places).
 A 25.35 B 14.25 C 19.92 D 28.25
GATE CE 2023 SET-2      Partial Differential Equation
Question 5 Explanation:
We know, Laplace equation in two dimensional on a unit square with Dirichlet boundary condition :
$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0 \lt (x, y) \lt 1$
This Laplace equation is called harmonic function. Solution of Laplace equation,
$u(x, y)=\left(C_{1} \cos \beta x+C_{2} \sin \beta x\right)\left(C_{3} e^{\text {By }}+C_{4} e^{-\beta y}\right)$
Let square $A B C D$ is unit square.
Now, given
\begin{aligned} & u(x, 0)=0 \\ & u(0, y)=0 \\ & u(1, y)=0 \\ & u(x, 1)=100 \end{aligned}

For $u(x, 0)=0$
$\quad 0=\left(C_{1} \cos \beta x+0\right)(C_3+C_4)$
$\Rightarrow \quad \mathrm{C}_{3}=-\mathrm{C}_{4}$

For $u(0, y)=0$;
$0=\left(C_{1}+0\right)\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right)$
$\Rightarrow \quad \mathrm{C}_{1}=0$

Now,
\begin{aligned} u(x, y) & =0+C_{2} \sin \beta x\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right) \\ & =a_{x} \sin \beta x\left(e^{\beta y}-e^{-\beta y}\right) \quad ...(1) \end{aligned}
$\left[\because \mathrm{C}_{3}=-\mathrm{C}_{4}\right.$ and Let $\left.\mathrm{C}_{2} \mathrm{C}_{3}=\mathrm{a}_{\mathrm{x}}\right]$

For $u(1, y)=0$
$0=a_{x} \sin \beta x\left(e^{\beta y}-e^{\beta y}\right)$
$\Rightarrow \quad \beta=\mathrm{n} \pi$

For $u(x, 1)=100$
$a_{x} \sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)=100$
$a_{x}=\frac{100}{\sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)}$

From eqn. (1), we get
$\begin{gathered} u(x, y)=\sum_{n=1}^{\infty }\frac{100}{\sin n \pi x\left(e^{n\pi}-e^{-n\pi}\right)} \times \sin n\pi x\left(e^{n \pi y}-e^{- n\pi y}\right) \\ =\sum_{n=1}^{\infty }\frac{100}{\left(e^{n\pi}-e^{-n \pi}\right)}\left[e^{n\pi y}-e^{-n\pi y}\right] \end{gathered}$
Now, at mid point $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$
$u\left(\frac{1}{2}, \frac{1}{2}\right)=T_{0} \text { [given] }$
\begin{aligned} T_0&=\sum_{n=1}^{\infty }\frac{100}{\left(\mathrm{e}^{n\pi}-\mathrm{e}^{-n\pi}\right)}\left[\mathrm{e}^{n\pi / 2}-\mathrm{e}^{-n\pi / 2}\right] \\ &=100\left [ \left ( \frac{\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2}}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}} \right ) +\left ( \frac{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}{\mathrm{e}^{2\pi}-\mathrm{e}^{-2\pi}} \right )+...\right ]\\ &=100\left [ \frac{1}{(\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2})}+\frac{1}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}+\frac{1}{\mathrm{e}^{3\pi / 2}-\mathrm{e}^{-3\pi / 2}} +...(neglect) \right ]\\ &=100[0.199+0.043+0.008]\\ &\approx 25^{\circ}C \end{aligned}

There are 5 questions to complete.

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1. Sir only 10 ques are given but other questions ?