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 June 16th, 2017, 10:49 AM #1 Newbie   Joined: Jan 2017 From: Costa Rica Posts: 3 Thanks: 0 How to solve this inverse transform Laplace? 1/(s^2+1)^2 Knowing the Laplace transform of L(sen t-cos t) may help you.
 June 16th, 2017, 11:28 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 Knowing the Laplace transform of t*cos(t) is particularly helpful.
 June 18th, 2017, 07:58 AM #3 Newbie   Joined: Dec 2016 From: Austin Posts: 16 Thanks: 1 Solution Problem: $\displaystyle \mathcal{L}^{-1} \left \{ \frac{1}{ \left ( s^{2}+1 \right )^{2} } \right \}$ Recall: $\displaystyle \mathcal{L}^{-1} \left \{ \frac{2a^{3}}{ \left ( s^{2}+a^{2} \right )^{2} } \right \}=\sin \left ( at \right )-at \cos \left ( at \right )$ We can use this definition by making a couple convenient substitutions and algebraic manipulations to make the problem look like our definition. Let: $\displaystyle a=1$ 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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