Question

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

Answer 1

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

Answer 2

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

Answer 3

Or just a separate problem?

Answer 4

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

Answer 5

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

Answer 6

s*

Answer 7

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

Answer 8

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

Answer 9

for which one? ugh

Answer 10

\[\large \frac{1}{s(s-3)} \quad = \quad \frac{A}{s}+\frac{B}{s-3}\]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

Answer 11

then once i have A and B?

Answer 12

i can use the laplace table to determine inverse?

Answer 13

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.

Answer 14

okay sweet thanks