# Differential Equations

 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   Engineering Mathematics
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
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   Engineering Mathematics
Question 2 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 3
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   Engineering Mathematics
Question 3 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}
 Question 4
A quadratic function of two variables is given as

$f\left(x_{1}, x_{2}\right)=x_{1}^{2}+2 x_{2}^{2}+3 x_{1}+3 x_{2}+x_{1} x_{2}+1$

The magnitude of the maximum rate of change of the function at the point $(1,1)$ is ____(Round off to the nearest integer).
 A 10 B 12 C 8 D 16
GATE EE 2023   Engineering Mathematics
Question 4 Explanation:
Given :
$\mathrm{F}\left(\mathrm{x}, \mathrm{x}_{2}\right)=\mathrm{x}_{1}^{2}+2 \mathrm{x}_{2}^{2}+\mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}+1$
$\nabla \cdot F=\left(2 x_{1}+3+x_{2}\right) \hat{a}_{x_{1}}+\left(4 x_{2}+3+x_{1}\right) \hat{a}_{x_{z}}$

At $(1,1)$
\begin{aligned} \nabla \cdot F & =6 \hat{a}_{x_{1}}+7 \hat{a}_{x_{2}} \\ |\nabla \cdot F| & =\sqrt{6^{2}+8^{2}}=10 \end{aligned}
 Question 5
The differential equation,

$\frac{\mathrm{du}}{\mathrm{dt}}+2 \mathrm{tu}^{2}=1$

is solved by employing a backward differnce scheme within the finite difference framework. The value of $u$ at the $(n-1)^{\text {th }}$ time-step, for some $n$, is 1.75. The corresponding time $(\mathrm{t})$ is $3.14 \mathrm{~s}$. Each time step is $0.01 \mathrm{~s}$ long. Then, the value of $(u_{n}-u_{n-1})$ is ____ (round off to three decimal places).
 A -0.125 B 0.125 C -0.182 D 0.182
GATE CE 2023 SET-1   Engineering Mathematics
Question 5 Explanation:
\begin{aligned} \frac{d u}{d t} & =1-2 t u^{2} \\ \frac{d u}{d t} & =f(t, u) \\ \Rightarrow \quad f(t, u) & =1-2 \mathrm{tu}^{2} \end{aligned}
Eulers backward difference scheme
\begin{aligned} u_{n}-u_{n-1} & =h f\left(t_{n}, u_{n}\right)=h\left(1-2 t_{n} u n^{2}\right) \\ & =0.01\left(1-2 \times 3.14 \times 1.75^{2}\right) \\ & =-0.1822 \end{aligned}

There are 5 questions to complete.

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