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Laplace transform

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation using laplace transforms. Assume zero intial conditions and that the forcing functions are zero prior to t=0

    x''(t) + 6x'(t) +8x(t)=sin3t

    2. Relevant equations



    3. The attempt at a solution

    I took the laplace transform which I found to be after simplyfing

    X(s)=-8s^2-57/(s^2+3^2)(s^2+6s)

    after partial fraction decompistion and comparing coeffcients I found these equations

    0=a+b+c+d
    -8=6a+6b+6c
    0= 9c
    -57=54c+9d

    When I plug that into Ax=b it gives me free variables. i am not sure if my logic is wrong or I just made some mistakes along the way. Any help would be gratly appreciated.

    MEM33
     
  2. jcsd
  3. Apr 14, 2012 #2
    I just made it. Try to write the two polynomials as a factor of the roots like (s-r1)(s-r2)(s-r3)(s-r4).
    I also think you made the transform wrong.
    Btw, if when you factor the poly in the denominator you can get the part.frac coefficients like this:
    [itex] \lim\limits_{s\rightarrow r_1}\{X(s)(s-r_1)\}[/itex]
    Can you see why?
     
    Last edited: Apr 14, 2012
  4. Apr 14, 2012 #3
    I found my mistake

    Thanks!
     
  5. Apr 14, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    Is your expression supposed to be [tex]X(s) = -8s^2 -\frac{57}{(s^2+3^2)(s^2+6s)} \text{ or }
    X(s) = \frac{-8s^2 - 57}{(s^2+3^2)(s^2+6s)}?[/tex]
    Using standard rules the expression, as written, means the first one.

    RGV
     
  6. Apr 14, 2012 #5
    Sorry after re-working it my X(s) looks like 15/(s^2 + 3^2)(s +4)(s +2)

    and my solution is X(t) = -.4cos(3t) +.5sin(3t) - .3exp(-4t) +.47exp(-2t)

    Can anyone tell me if that is the correct solution?
     
  7. Apr 15, 2012 #6
    i get (evaluated as decimal numbers):
    [itex]-0.06\cos(3t)-0.003\sin(3t)+0.12e^{-2t}-0.06e^{-4t}[/itex]
     
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