My Math Forum How to solve this inverse transform Laplace?

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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: 19,062 Thanks: 1619 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: 11 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!

 Tags inverse, laplace, solve, transform

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post szz Differential Equations 1 November 2nd, 2014 03:18 AM szz Calculus 1 November 1st, 2014 03:14 PM Rydog21 Calculus 2 October 15th, 2013 09:35 AM Deiota Calculus 1 April 28th, 2013 10:28 AM defunktlemon Real Analysis 1 April 14th, 2012 02:19 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top