Finding the inverse Laplace transform

[Rubik]

Homework Statement

Find the inverse Laplace transform of the following function
1/[s(s² + 4)²]

Homework Equations

1/ (s² + [itex]\omega[/itex]²)² = (1/ 2[itex]\omega[/itex]³) (sin[itex]\omega[/itex]t -[itex]\omega[/itex]t cos[itex]\omega[/itex]t)

The Attempt at a Solution

L^-1 = (1/s) (1/ s² +2²)²
= (1/16) (sin 2t - 2t cos 2t) as the Laplace transform of (1/s) = 1

Is this what I am meant to do?

 

[Screwdriver]

The inverse Laplace transform of a product isn't the product of the inverse Laplace transforms in general. You can go for the convolution theorem (might be a mess), or use partial fractions to decompose that, which requires you to use a few other inverse identities once you get the decomposition.

 

[Rubik]

Right.. how silly of me, so doing it by partial fractions I got 0=B=C and A= 1/16
giving me F(s) = (1/16)(1/s) does that seem a little more on track?

 

[Screwdriver]

Well using partial fractions is definitely on track. But remember that when you have a quadratic term in the denominator, you have to have a linear term in the numerator (plus, it's squared, so there are going to be two.) The form for the expansion should be:

[tex]\frac{1}{(s)(s^2+4)^2} = {\frac{A}{s}} + \frac{Bs + C}{s^2 +4} + \frac{Ds + E}{(s^2+4)^2}[/tex]

Give that a shot.

 

[Rubik]

Okay after that I am now getting A = 1/16, B = -1/16, C = 0, D = -1/4, E = 0.

(1/16)(1/s) - (1/16)(s/[s² + 4]) - (1/4)(s/(s² + 4)²

 

[Screwdriver]

That looks good! Now that you have those, you can invert them term-by-term since the inverse Laplace transform is linear, so [itex]\mathcal{L}^{-1}(af(t)+bg(t)) = a\mathcal{L}^{-1}(f(t)) + b\mathcal{L}^{-1}(g(t))[/itex] for constants [itex]a[/itex] and [itex]b[/itex].

 

[Rubik]

Thank you so much! :D

 

[Screwdriver]

You're welcome!

 

[Ray Vickson]

There is a standard result: if f(t) <---> g(s), then integral_{x=0..t} f(x) dx < ---> g(s)/s.

RGV

 

[Rubik]

Now I am trying to solve an IVP using Laplace transform if you feel like pointing me in the right direction feel free to check it out.. :)

https://www.physicsforums.com/showthread.php?p=3464473#post3464473