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June 16th, 2017, 10:49 AM   #1
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How to solve this inverse transform Laplace?


Knowing the Laplace transform of L(sen t-cos t) may help you.
Macoleco is offline  
June 16th, 2017, 11:28 AM   #2
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Knowing the Laplace transform of t*cos(t) is particularly helpful.
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June 18th, 2017, 07:58 AM   #3
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$\displaystyle \mathcal{L}^{-1} \left \{ \frac{1}{ \left ( s^{2}+1 \right )^{2} } \right \}$

$\displaystyle \mathcal{L}^{-1} \left \{ \frac{2a^{3}}{ \left ( s^{2}+a^{2} \right )^{2} } \right \}=\sin \left ( at \right )-at \cos \left ( at \right )$

We can use this definition by making a couple convenient substitutions and algebraic manipulations to make the problem look like our definition.

$\displaystyle a=1$

Then our problem becomes:
$\displaystyle \mathcal{L}^{-1} \left \{ \frac{a^{3}}{ \left ( s^{2}+a^{2} \right )^{2} } \right \}$

We now need a 2 in the numerator to complete the definition, so multiply top and bottom by 2:

$\displaystyle \mathcal{L}^{-1} \left \{ \frac{2a^{3}}{ 2 \left ( s^{2}+a^{2} \right )^{2} } \right \}$

Factor out a $\displaystyle \frac{1}{2}$:

$\displaystyle \frac{1}{2} \mathcal{L}^{-1} \left \{ \frac{2a^{3}}{ \left ( s^{2}+a^{2} \right )^{2} } \right \}$

We now have the definition, so substitute:
$\displaystyle =\frac{1}{2} \left [ \sin \left ( at \right )-at \cos \left ( at \right ) \right ]$

Finally, substitute $\displaystyle a=1$ back into the solution to get:

$\displaystyle =\frac{1}{2} \left [ \sin \left ( t \right )-t \cos \left ( t \right ) \right ]$

Hope that helps!
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