GATE ME 2013 | Question: 27

piyag476 asked Feb 19, 2017 • recategorized Mar 3, 2021 by Lakshman Bhaiya

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piyag476 asked Feb 19, 2017

GATE ME 2013 | Question: 1
The partial differential equation $\dfrac{\partial u }{\partial t}+u\dfrac{\partial u}{\partial x}=\dfrac{\partial^2 u}{\partial x^2}$ is a linear equation of order $2$ non-linear equation of order $1$ linear equation of order $1$ non-linear equation of order $2$

The partial differential equation $\dfrac{\partial u }{\partial t}+u\dfrac{\partial u}{\partial x}=\dfrac{\partial^2 u}{\partial x^2}$ is alinear equation of order $2$no...

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Arjun asked Feb 17, 2018

GATE2018-1-38
$F(s)$ is the Laplace transform of the function $f(t) =2t^2 e^{-t}$. $F(1)$ is _______ (correct to two decimal places).

$F(s)$ is the Laplace transform of the function $f(t) =2t^2 e^{-t}$. $F(1)$ is _______ (correct to two decimal places).

Arjun

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Arjun asked Feb 9, 2019

GATE2019 ME-1: 27
A harmonic function is analytic if it satisfies the Laplace equation. If $u(x,y)=2x^2-2y^2+4xy$ is a harmonic function, then its conjugate harmonic function $v(x,y)$ is $4xy-2x^2+2y^2+ \text{constant}$ $4y^2-4xy + \text{constant}$ $2x^2-2y^2+ xy + \text{constant}$ $-4xy+2y^2-2x^2+ \text{constant}$

A harmonic function is analytic if it satisfies the Laplace equation. If $u(x,y)=2x^2-2y^2+4xy$ is a harmonic function, then its conjugate harmonic function $v(x,y)$ is$4...

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Arjun asked Feb 9, 2019

GATE2019 ME-2: 27
A diffferential equation is given as $x^2 \frac{d^2y}{dx^2} – 2x \frac{dy}{dx} +2y =4$ The solution of the differential equation in terms of arbitrary constants $C_1$ and $C_2$ is $y=C_1x^2 +C_2 x+2 \\$ $y=\dfrac{C_1}{x^2} +C_2x+2 \\$ $y=C_1x^2+C_2x+4 \\$ $y=\dfrac{C_1}{x^2}+C_2x+4$

A diffferential equation is given as $$x^2 \frac{d^2y}{dx^2} – 2x \frac{dy}{dx} +2y =4$$ The solution of the differential equation in terms of arbitrary constants $C_1$...

Arjun

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Arjun asked Feb 9, 2019

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piyag476 asked Feb 19, 2017

GATE ME 2013 | Question: 46
The solution to the differential equation $\dfrac{d^2u}{dx^2}-k\dfrac{du}{dx}=0$ where $k$ is a constant, subjected to the boundary conditions $u(0)$ = $0$ and $u(L)$ = $U$, is $u=U\dfrac{x}{L} $ $u=U\left(\dfrac{1-e^{kx}}{1-e^{kL}}\right) $ $u=U\left(\dfrac{1-e^{-kx}}{1-e^{-kL}}\right)$ $u=U\left(\dfrac{1+e^{kx}}{1+e^{kL}}\right)$

The solution to the differential equation $\dfrac{d^2u}{dx^2}-k\dfrac{du}{dx}=0$ where $k$ is a constant, subjected to theboundary conditions $u(0)$ = $0$ and $u(L)$ = $...

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