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Homework Help: 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


    after partial fraction decompistion and comparing coeffcients I found these equations

    0= 9c

    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.

  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

  5. Apr 14, 2012 #4

    Ray Vickson

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    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.

  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):
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