text: # Inverse Laplace transform
  
  1. Sep 12, 2014
  
  ### Mutaja
  
  1. The problem statement, all variables and given/known data
  
  Find the inverse Laplace transform of the expression:
  
  F(S) = $\frac{3s+5}{s^2 +9}$
  
  2. Relevant equations
  
  3. The attempt at a solution
  
  From general Laplace transforms, I see a pattern with laplace transforming sin(t) and cos(t) because:
  
  L{sin(t)+cos(t)} = $\frac{s+1}{s^2 +1}$
  
  All I'm missing here is a couple of constants(?).
  
  I know that the laplace transform works like this:
  
  L{Asin(Bt)+Ccos(Dt)} = $\frac{Cs}{s^2 +D}$ + $\frac{A*B}{s^2 +B^2}$
  
  Looking at my original problem, I can see that I need B to equal D, $D^2$ + $B^2$ = 9, C = 3 and A*B = 5.
  
  If I set A = 5, B = 3, C = 9 and D = 3 I get $3^2$ + $3^2$ = 18, 5
  3 = 15 (putting numbers into my equations above).
  
  Therefore I need to divide the whole expression by 3 to get my correct answer:
  
  L{$\frac{1}{3}($5sin(3t)+9cos(3t))} = $\frac{3s+5}{s^2 +9}$
  
  Is there a simpler way to do this? Or do I just have to use the trial and error method until I find the correct factors/constants?
  
  Any input on this will be greatly appreciated. Thanks
  
  2. Sep 12, 2014
  
  ### HallsofIvy
  
  Staff Emeritus
  What? One of the equations above said A*B= 5 which is NOT satisfied by A= 5, B= 3. You have only three equations for the four values A, B, C, and D. And there is no reason to assume that A, B, C, and D must be integers.
  
  Use the fact that the Laplace transform, and so its inverse, is linear.
  $$\frac{3s+ 5}{s^2+ 9}= 3\frac{s}{s^2+ 9}+ 5\frac{1}{s^2+ 9}$$
  and look up their inverse transforms separately.
  
  The laplace transform of $cos(\omega x)$ is $\frac{s}{s^2+ \omega^2}$ so the inverse transform of $3\frac{s}{s^2+ 9}$ is $3 cos(3x)$
  
  The Laplace transform of $sin(\omega x)$ is $\frac{\omega}{s^2+ \omega^2}$ so the inverse transform of $5\frac{1}{s^2+ 9}= (5/3)\frac{3}{s^2+ 9}$ is $(5/3)sin(3x)$.
explanation: The document is a forum post discussing the inverse Laplace transform of a given function. It contains a math question and an attempt at a solution, followed by a response from another user providing a more straightforward method to solve the problem.
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distilabel_metadata.content: # Inverse Laplace transform
  
  1. Sep 12, 2014
  
  ### Mutaja
  
  1. The problem statement, all variables and given/known data
  
  Find the inverse Laplace transform of the expression:
  
  F(S) = $\frac{3s+5}{s^2 +9}$
  
  2. Relevant equations
  
  3. The attempt at a solution
  
  From general Laplace transforms, I see a pattern with laplace transforming sin(t) and cos(t) because:
  
  L{sin(t)+cos(t)} = $\frac{s+1}{s^2 +1}$
  
  All I'm missing here is a couple of constants(?).
  
  I know that the laplace transform works like this:
  
  L{Asin(Bt)+Ccos(Dt)} = $\frac{Cs}{s^2 +D}$ + $\frac{A*B}{s^2 +B^2}$
  
  Looking at my original problem, I can see that I need B to equal D, $D^2$ + $B^2$ = 9, C = 3 and A*B = 5.
  
  If I set A = 5, B = 3, C = 9 and D = 3 I get $3^2$ + $3^2$ = 18, 5
  3 = 15 (putting numbers into my equations above).
  
  Therefore I need to divide the whole expression by 3 to get my correct answer:
  
  L{$\frac{1}{3}($5sin(3t)+9cos(3t))} = $\frac{3s+5}{s^2 +9}$
  
  Is there a simpler way to do this? Or do I just have to use the trial and error method until I find the correct factors/constants?
  
  Any input on this will be greatly appreciated. Thanks
  
  2. Sep 12, 2014
  
  ### HallsofIvy
  
  Staff Emeritus
  What? One of the equations above said A*B= 5 which is NOT satisfied by A= 5, B= 3. You have only three equations for the four values A, B, C, and D. And there is no reason to assume that A, B, C, and D must be integers.
  
  Use the fact that the Laplace transform, and so its inverse, is linear.
  $$\frac{3s+ 5}{s^2+ 9}= 3\frac{s}{s^2+ 9}+ 5\frac{1}{s^2+ 9}$$
  and look up their inverse transforms separately.
  
  The laplace transform of $cos(\omega x)$ is $\frac{s}{s^2+ \omega^2}$ so the inverse transform of $3\frac{s}{s^2+ 9}$ is $3 cos(3x)$
  
  The Laplace transform of $sin(\omega x)$ is $\frac{\omega}{s^2+ \omega^2}$ so the inverse transform of $5\frac{1}{s^2+ 9}= (5/3)\frac{3}{s^2+ 9}$ is $(5/3)sin(3x)$.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a forum post discussing the inverse Laplace transform of a given function. It contains a math question and an attempt at a solution, followed by a response from another user providing a more straightforward method to solve the problem.
  
  Label: question-and-answer