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The Laplace transform of a function $f(t)$ is $L( f )=\dfrac{1}{(s^{2}+\omega ^{2})}.$ Then, $f(t)$ is$f\left ( t \right )=\dfrac{1}{\omega ^{2}}\left ( 1-\cos\:\omega t ...
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Arjun asked Feb 26, 2017
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The Laplace transform of $te^{t}$ is$\dfrac{s}{(s+1)^{2}} \\$$\dfrac{1}{(s-1)^{2}} \\$$\dfrac{1}{(s+1)^{2}} \\$$\dfrac{s}{(s-1)}$

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The Laplace transform of a function $f(t)$ is $L( f )=\dfrac{1}{(s^{2}+\omega ^{2})}.$ Then, $f(t)$ is$f\left ( t \right )=\dfrac{1}{\omega ^{2}}\left ( 1-\cos\:\omega t ...
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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} $
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If the Laplace transform of a function $f(t)$ is given by $\frac{s+3}{\left ( s+1 \right )\left ( s+2 \right )}$, then $f(0)$ is$0$$\frac{1}{2}$$1$$\frac{3}{2}$
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Arjun asked Feb 26, 2017
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The Laplace transform of $te^{t}$ is$\dfrac{s}{(s+1)^{2}} \\$$\dfrac{1}{(s-1)^{2}} \\$$\dfrac{1}{(s+1)^{2}} \\$$\dfrac{s}{(s-1)}$
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