# Inv Laplace Tranforms

## Homework Statement

(4s^2+2)/(s^3+6s-20)

## The Attempt at a Solution

ok from an earlier question, i used partial frac, to re write it as (1/(s-2))+((3s+4)/(s^2+2s+10))
which i can re write to, by completing square to (1/(s-2))+((3s+4)/((s+1)^2+9))

now i think ive done all that ok.. so my inv laplace is:

e^(2t)+invlaplace((3s+4)/((s+1)^2+9))....

now im not completely sure on how to use s and t shifts properly(i just started this stuff today), and im quite unsure what to do from now but the answer should be
e^(2t)+(1/3)e^(-t)(9cos(3t)+sin(3t)) and i have no idea how to get that from that... what happens to the 4?? can anyone help plz

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anyone?

i could do..
e^(2t)+(1/3)e^(-t)invlaplace((9s+12)/(s^2+9)) but in my examples i have no idea.. how to get cos and sin inside the invlaplace brackets.. coz then i would need to
e^(2t)+(1/3)e^(-t)invlaplace(9s/(s^2+9).... which would get me 9cos(3t) but then the sin........something but then i have a 12 and would need to subtract 9 to get to 3????.. idk im confused...

plz anyone?

anyone know wat to do?

?????

surely someone knows how to do this

cmon plz someone help

HallsofIvy
Homework Helper
Every ten minutes you complain that people are not answering you? We are not sitting here waiting for some to post a question that we will answer immediately! Have patience.

To find an inverse transform of $$\frac{3s+4}{(s+1)^2+ 9}$$, separate the two parts of the numerator: write it as
$$3\frac{s}{(s+1)^2+ 9}+ 4\frac{1}{(s+1)^2+ 9}$$
You should be able to find both of those in any table of Laplace transforms.

ok thankyou very much.. and sorry wont happen again.

ok i know what you said to do is right, but i still dont quite understand, form what ive learnt. i have e^(2*t)+invlaplace(3*s/((s+1)^2+9)+4/((s+1)^2+9))
=e^(2 t)+(3e^(-t)*invlaplace(s/(s^(2)+9))+4e^(-t)*invlaplace(1/(s^(2)+9))
=e^(2*t)+3*e^(-t)*cos(3*t)+(4/3)*invlaplace(3/(s^2+9))
=e^(2*t)+3*e^(-t)*cos(3*t)+(4*e^(-t)*1/3)*sin(3*t)
im pretty sure this is wrong.... plz can u tell me where stuffing up

i have no idea how the 4... turns into a 9....

ok its the (3*s/(s+1)^2+9) bit im stuffing up... i know its 3e^-t*cos(3t) but i think there also has to be a -sin(3t) somewhere... but where does it come from?

plz someone?

omg did sleep thinking actually help is 3s/(s+1)^2+9=e^(-t)(3cos(3t)-sin(t) because i need 3s+1 on top so i subtract 1. then its

3*(s+1)/((s+1)^2+9)-(1/3)(3/(s+1)^2+9)???? coz i think that works

nope.... im close i think.. tho

back in couple of hrs, feel free to help = )

ok back i been thinking again and is it..
invlaplace((3(s+1))/(s+1)^2+9)-3/(s+1)^2+9))?
then it would be e^(-t)3cos(3t)-e^(-t)sin(st)

so then + (4/3)sin(3t) would be

e^(-t)3cos(3t)+e^(-t)(1/3)sin(3t)
=(1/3)e^(-t)(9cos(3t)+sin(3t) yes i think that is right, plz can someone confirm?

dang...

nope its right lol