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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.
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. 

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