My Math Forum Inverse Laplace of 1.5s/(s+100). s in the numerator makes me crazy!

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 April 20th, 2017, 05:35 PM #1 Newbie   Joined: Apr 2017 From: Lithuania Posts: 2 Thanks: 0 Inverse Laplace of 1.5s/(s+100). s in the numerator makes me crazy! Hello, Inverse Laplace of 1.5s/(s+100). i would like to know the steps of solving this problem. s in numerator makes me confused.. which theorem i should use? any information that could help, is highly appreciated thanks in advance!
 April 20th, 2017, 05:37 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,734 Thanks: 605 Math Focus: Yet to find out. Should be something in the tables for $\dfrac{s}{s + a}$. The 1.5 just multiples the whole expression. Thanks from Rkls Last edited by Joppy; April 20th, 2017 at 05:46 PM.
 April 20th, 2017, 05:48 PM #3 Newbie   Joined: Apr 2017 From: Lithuania Posts: 2 Thanks: 0 there is nothing in the tables with first order equations and the s in the numerator... Could you write integral that i need to solve?
 April 24th, 2017, 08:05 PM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,734 Thanks: 605 Math Focus: Yet to find out. Forgot about this sorry. Note that, $\dfrac{1.5s}{s + 100} = 1.5 \left[ 1 - \dfrac{100}{s + 100} \right]$ Since the Laplace transform is a linear operator, you can inverse term by term, $1.5 \left[ \mathcal{L}^{-1} \left[ 1 \right] - \mathcal{L}^{-1} \left[ \dfrac{100}{s + 100} \right] \right] = 1.5(\delta(t) - 100e^{-100t}).$ It is very often the case with inverse Laplace exercises that you need to first split up the terms through partial fractions or other methods.

 Tags 100, 15s or s, crazy, inverse, laplace, makes, numerator

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