GATE ME 2013 | Question: 1

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

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piyag476 asked Feb 19, 2017

GATE ME 2013 | Question: 27
The function $f(t)$ satisfies the differential equation $\dfrac{d^2f}{dt^2}+f=0$ and the auxiliary conditions, $f(0)=0$, $\dfrac{d(f)}{d(t)}(0)=4$. The Laplace transform of $f(t)$is given by $\dfrac{2}{s+1} \\$ $\dfrac{4}{s+1} \\$ $\dfrac{4}{s^2+1} \\$ $\dfrac{2}{s^4+1}$

The function $f(t)$ satisfies the differential equation $\dfrac{d^2f}{dt^2}+f=0$ and the auxiliary conditions, $f(0)=0$, $\dfrac{d(f)}{d(t)}(0)=4$. The Laplace transform ...

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Arjun asked Feb 26, 2017

GATE2017 ME-1: 3
Consider the following partial differential equation for $u(x, y)$, with the constant $c > 1$: $\dfrac{\partial u}{\partial y}+c\dfrac{\partial u}{\partial x}=0$ Solution of this equation is $u(x, y) = f (x+cy)$ $u(x, y) = f (x-cy)$ $u(x, y) = f (cx+y)$ $u(x, y) = f (cx-y)$

Consider the following partial differential equation for $u(x, y)$, with the constant $c 1$:$\dfrac{\partial u}{\partial y}+c\dfrac{\partial u}{\partial x}=0$Solution of...

Arjun

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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).

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Arjun asked Feb 17, 2018

GATE2018-2-3
Consider a function $u$ which depends on position $x$ and time $t$. The partial differential equation $\frac{\partial u}{\partial t} = \frac{\partial^2 u }{\partial x^2}$ is known as the Wave equation Heat equation Laplace's equation Elasticity equation

Consider a function $u$ which depends on position $x$ and time $t$. The partial differential equation $$\frac{\partial u}{\partial t} = \frac{\partial^2 u }{\partial x^2}...

Arjun

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Arjun asked Feb 17, 2018

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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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