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GATE2015-2-4
Arjun
asked
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Differential Equations
Feb 24, 2017
recategorized
Mar 4, 2021
by
Lakshman Patel RJIT
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The Laplace transform of $e^{i5t}$ where $i=\sqrt{-1}$, is
$\dfrac{s-5i}{s^2-25} \\$
$\dfrac{s+5i}{s^2+25} \\$
$\dfrac{s+5i}{s^2-25} \\$
$\dfrac{s-5i}{s^2+25} $
gateme-2015-set2
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Arjun
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Differential Equations
Feb 24, 2017
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Mar 4, 2021
by
Lakshman Patel RJIT
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The solution of $\dfrac{d^2y}{dt^2}-y=1,$ which additionally satisfies $y \bigg \vert_{t=0} = \dfrac{dy}{dt} \bigg \vert_{t=0}=0$ in the Laplace $s$-domain is $\dfrac{1}{s(s+1)(s-1)} \\$ $\dfrac{1}{s(s+1)} \\$ $\dfrac{1}{s(s-1)} \\$ $\dfrac{1}{s-1} \\$
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Laplace transform of $\cos(\omega t)$ is $\dfrac{s}{s^2+\omega ^2}$. The Laplace transform of $e^{-2t} \cos(4t)$ is $\dfrac{s-2}{(s-2)^2+16} \\$ $\dfrac{s+2}{(s-2)^2+16} \\$ $\dfrac{s-2}{(s+2)^2+16} \\$ $\dfrac{s+2}{(s+2)^2+16}$
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Laplace transform of the function $f(t)$ is given by $F(s)=L\begin{bmatrix} f(t) \end{bmatrix}=\int_{0}^{\infty }f(t)e^{-st}dt$ . Laplace transform of the function shown below is given by $\displaystyle{\frac{1-e^{-2s}}{s}} \\$ $\displaystyle{\frac{1-e^{-s}}{2s}} \\$ $\displaystyle{\frac{2-2e^{-s}}{s}} \\$ $\displaystyle{\frac{1-2e^{-s}}{s}}$
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Consider the following differential equation: $\dfrac{dy}{dt}=-5y$; initial condition: $y=2$ at $t=0$. The value of $y$ at $t=3$ is $-5e^{-10}$ $2e^{-10}$ $2e^{-15}$ $-15e^{2}$
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