Finding the Inverse Laplace Transform of (4s^2+2)/(s^3+6s-20)

[fredrick08]

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 I've done all that ok.. so my inv laplace is:

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

now I am not completely sure on how to use s and t shifts properly(i just started this stuff today), and I am 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 please

 

[fredrick08]

anyone?

 

[fredrick08]

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 I am confused...

 

[fredrick08]

please anyone?

 

[fredrick08]

anyone know wat to do?

 

[fredrick08]

?

 

[fredrick08]

surely someone knows how to do this

 

[fredrick08]

cmon please someone help

 

[HallsofIvy]

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.

 

[fredrick08]

ok thankyou very much.. and sorry won't happen again.

 

[fredrick08]

ok i know what you said to do is right, but i still don't quite understand, form what I've 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... please can u tell me where stuffing up

 

[fredrick08]

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

 

[fredrick08]

ok its the (3*s/(s+1)^2+9) bit I am 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?

 

[fredrick08]

please someone?

 

[fredrick08]

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

 

[fredrick08]

nope... I am close i think.. tho

 

[fredrick08]

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

 

[fredrick08]

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, please can someone confirm?

 

[fredrick08]

dang...

 

[fredrick08]

nope its right lol