Problem 01 | Multiplication by Power of t

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Problem 01
Find the Laplace transform of   $f(t) = t \cos 2t$.
 

Solution 01
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 2^2}$

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 4}$
 

$\mathcal{L} (t \cos 2t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{s}{s^2 + 4} \right]$

$\mathcal{L} (t \cos 2t) = -\left[ \dfrac{(s^2 + 4)(1) - s(2s)}{(s^2 + 4)^2} \right]$

$\mathcal{L} (t \cos 2t) = -\dfrac{s^2 + 4 - 2s^2}{(s^2 + 4)^2}$

Problem 03 | Change of Scale Property of Laplace Transform

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Problem 03
Supposed that the Laplace transform of a certain function   $f(t)$   is   $\dfrac{s^2 - s + 1}{(2s + 1)^2 (s - 1)}$,   find the Laplace transform of   $f(2t)$.
 

Solution 03
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then by change of scale property,   $\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
 

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{s^2 - s + 1}{(2s + 1)^2 (s - 1)}$

Problem 02 | Change of Scale Property of Laplace Transform

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Problem 02
Given that   $\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$,   find   $\mathcal{L} \left( \dfrac{\sin 3t}{t} \right)$.
 

Solution 02
$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
 

$\mathcal{L} \left( \dfrac{\sin 3t}{t} \right) = 3\mathcal{L} \left( \dfrac{\sin 3t}{3t} \right)$,   thus,   $a = 3$

Problem 01 | Change of Scale Property of Laplace Transform

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Problem 01
Find the Laplace transform of   $f(t) = \cos 4t$   using the change of scale property.
 

Solution 01
$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
 

$\mathcal{L} (\cos t) = \dfrac{s}{s^2 + 1}$
 

Thus,
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{\dfrac{s}{4}}{\left( \dfrac{s}{4} \right)^2 + 1} \right]$

$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{\dfrac{s}{4}}{\dfrac{s^2}{16} + 1} \right]$

Problem 02 | Second Shifting Property of Laplace Transform

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Problem 01
Find the Laplace transform of   $g(t) = \begin{cases}{f(t - 2)^3 & t > 2 \\ 0 & t  

Solution 01
$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$
 

$F(s) = \mathcal{L} (t^3)$   and   $a = 2$

$F(s) = \dfrac{3!}{s^{3 + 1}}$

$F(s) = \dfrac{6}{s^4}$
 

Thus,
$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-2s} \left( \dfrac{6}{s^4} \right)$

Problem 01 | Second Shifting Property of Laplace Transform

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Problem 01
Find the Laplace transform of   $g(t) = \begin{cases}{f(t - 1)^2 & t > 1 \\ 0 & t  

Solution 01

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