find the inverse laplace of 3 - 2s/(s^2 - 3s)^2

Question

anonymous · Answer

F(s) = 1/(s(s - 3))

zepdrix · Answer

What is this second post? Is that what it looks like simplified?

zepdrix · Answer

Or just a separate problem?

anonymous · Answer

I found F'(s) = 3 - 2s/(S^2 - 3s)^2

anonymous · Answer

because f(t) = L^-1[F(s)] = -1/t * L^-1[F'(s)]

anonymous · Answer

s*

anonymous · Answer

The Problem: Find the inverse Laplace of F(s) = 1/(s(s-3)).

zepdrix · Answer

Looks like we need to do Partial Fraction Decomposition.
Remember that business? :O

anonymous · Answer

for which one? ugh

zepdrix · Answer

$$\large \frac{1}{s(s-3)} \quad = \quad \frac{A}{s}+\frac{B}{s-3}$$Multiplying through by the denominator of the original gives us,$$\huge 1=A(s-3)+Bs$$Then you can solve for A and B from there :O
Yes no? :D

anonymous · Answer

then once i have A and B?

anonymous · Answer

i can use the laplace table to determine inverse?

zepdrix · Answer

Yah :) Through this process you'll be able to break up your fractions like this,$$\large \frac{A}{s}+\frac{B}{s-3}$$Where A and B are constants.
And yes you should be able to find those on your table.

anonymous · Answer

okay sweet thanks