Inverse LaPlace Transform help

[lpau001]

Homework Statement

Use the First Shifting Theorem (translation on the s axis) to find f(t).

L^-1[(2s-1)/(s²(s+1)³]

Homework Equations

The Attempt at a Solution

This is going to take forever.. Can we assume that I am putting the L^-1 all over the place?

[(2s-1)/(s^2(s+1)^3)] = [(2(s+1)-3)/(s^2(s+1)^3)] */ on this step I don't know how to change the s^2 in the denominator to an (s+1)^2 form.. do I need too? /*

[(2[STRIKE](s+1)[/STRIKE])/(s^2(s+1)^{[STRIKE]3[/STRIKE]2})] - [3/(s^2(s+1)^3)]

Then using partial fractions */ This is where I get confused /*
(As+B)/(s^2) + C/(s+1) + (Ds+E)/(s+1)^2 = (2s-1)/(s^2(s+1)^3)

solving coefficients (after a lot of math)
A=5 B=-1 C=-3 D=-2 E=-6

Putting coefficients back into problem..
(5s-1)/s^2 + (-3)/(s+1) + (-2s-6)/(s+1)^2

Attempted solution..
[(5s)/(s^2) - (1)/(s^2)] + [-3/s|s+1] + [-2s/s^2|s+1] + [-4/s^2|s+1]

I get...
5 - t - 3e^-t - 2e^-t - 4te^-t

Which is wrong. I'm close, kinda, I believe I erred in my partial fractions set up, but any help would be so nice.. Thanks PF.

 

[vela]

You want to expand it as

$$\frac{2s-1}{s^2(s+1)^3} = \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+1}+\frac{D}{(s+1)^2}+\frac{E}{(s+1)^3}$$

Hopefully, you can see the pattern. You should find A=5, B=-1, C=-5, D=-4, and E=-3. The first two terms are unshifted, so you can just look them up in a table. The latter three terms are shifted by 1, so you can apply the theorem to figure out what they correspond to in the time domain.

 

[lpau001]

I see now! Thank you very much, Vela! I think I see the pattern for the Partial Fractions set up.

If you have a factor in the denominator like.. x^3, you break it up into x, x^2, and x^3. Thank you for the help!