Inverse LaPlace Transforms With step functions

[PBJinx]

Homework Statement

Find the inverse laplace tansform of this

[URL]http://webwork.math.umass.edu/webwork2_files/tmp/equations/fe/0f481f24ae927e11e80569093b71321.png[/URL]


My question is can i pull out the e^-9s and just find the inverse of

\frac{1}{s²-1s+20}

I broke it up into

\frac{A}{S+4} + \frac{B}{S-5}

i got A=\frac{-1}{9}

and

B= \frac{1}{9}

set the answer as Step(t-9)(\frac{1}{9}(e^5t-e^-4t)

and the answer is wrong

can anyone tell where i am going wrong. I think its my assumption of pulling out the step, but what do i do with the step then?

 

[PBJinx]

alright, i figured out that i was typing it into the homework wrong

the answer is ends up being

(1/9)(step(t-9)(e^(5(t-9))-e^(-4(t-9))))

 

[tedbradly]

Homework Statement

Find the inverse laplace tansform of this

[URL]http://webwork.math.umass.edu/webwork2_files/tmp/equations/fe/0f481f24ae927e11e80569093b71321.png[/URL]


My question is can i pull out the e^-9s and just find the inverse of

\frac{1}{s²-1s+20}

I broke it up into

\frac{A}{S+4} + \frac{B}{S-5}

i got A=\frac{-1}{9}

and

B= \frac{1}{9}

set the answer as Step(t-9)(\frac{1}{9}(e^5t-e^-4t)

and the answer is wrong

can anyone tell where i am going wrong. I think its my assumption of pulling out the step, but what do i do with the step then?

The exponential denotes a time shift for the inverse Laplace, so find the inverse Laplace without the exponential and shift that answer appropriately.
G(s) = \frac{1}{(s+4)(s-5)} = \frac{1}{-9}\frac{1}{s+4} + \frac{1}{9}\frac{1}{s-5}}\leftrightarrows g(t) = \frac{1}{9}(e^{5t}-e^{-4t})u(t)
Then
F(s) = e^{-9s}G(s) \to f(t) = g(t-9) = \frac{1}{9}(e^{5(t-9)}-e^{-4(t-9)})u(t-9)

 

[PBJinx]

The exponential denotes a time shift for the inverse Laplace, so find the inverse Laplace without the exponential and shift that answer appropriately.
G(s) = \frac{1}{(s+4)(s-5)} = \frac{1}{-9}\frac{1}{s+4} + \frac{1}{9}\frac{1}{s-5}}\leftrightarrows g(t) = \frac{1}{9}(e^{5t}-e^{-4t})u(t)
Then
F(s) = e^{-9s}G(s) \to f(t) = g(t-9) = \frac{1}{9}(e^{5(t-9)}-e^{-4(t-9)})u(t-9)

ahhhh

That is very helpful!