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Inv Laplace Tranforms

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    (4s^2+2)/(s^3+6s-20)


    3. 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
     
  2. jcsd
  3. Mar 2, 2009 #2
    anyone?
     
  4. Mar 2, 2009 #3
    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...
     
  5. Mar 2, 2009 #4
    plz anyone?
     
  6. Mar 2, 2009 #5
    anyone know wat to do?
     
  7. Mar 3, 2009 #6
  8. Mar 3, 2009 #7
    surely someone knows how to do this
     
  9. Mar 3, 2009 #8
    cmon plz someone help
     
  10. Mar 3, 2009 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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 [tex]\frac{3s+4}{(s+1)^2+ 9}[/tex], separate the two parts of the numerator: write it as
    [tex]3\frac{s}{(s+1)^2+ 9}+ 4\frac{1}{(s+1)^2+ 9}[/tex]
    You should be able to find both of those in any table of Laplace transforms.
     
  11. Mar 3, 2009 #10
    ok thankyou very much.. and sorry wont happen again.
     
  12. Mar 3, 2009 #11
    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
     
  13. Mar 3, 2009 #12
    i have no idea how the 4... turns into a 9....
     
  14. Mar 3, 2009 #13
    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?
     
  15. Mar 3, 2009 #14
    plz someone?
     
  16. Mar 3, 2009 #15
    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
     
  17. Mar 3, 2009 #16
    nope.... im close i think.. tho
     
  18. Mar 3, 2009 #17
    back in couple of hrs, feel free to help = )
     
  19. Mar 4, 2009 #18
    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?
     
  20. Mar 4, 2009 #19
    dang...
     
  21. Mar 4, 2009 #20
    nope its right lol
     
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