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Laplace transform

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Problem: laplace transform of x"+2x'+x=sin(t) x=x'=0

    2. Relevant equations

    3. The attempt at a solution
    Attempt at problem: I was able to get Y(s)=1/(s^2+1)(s^2+2s+5) and then i expanded it to get [s/10(s^2+2s+5)]-[s/10(s^2+1)]+[1/5(s^2+1)]. The last two terms are easy to transform the first term i completed the square and got 1/10*s/(s+1)^2+4

    Question: from maple i know it will be a exp times a sin and cos function of some sort. I don't know how they got that even though i see it in the denominator.
  2. jcsd
  3. Nov 18, 2008 #2
    Hi Jumpman,

    Firstly, if you did mean to write the DE as x"+2x'+x=sin(t) (with no 5 in front of the x) then you should restart from there. Otherwise, I'll assume you meant x"+2x'+5x=sin(t) with initial conditions x(0)=x'(0)=0.

    To find the inverse laplace of [tex]\frac{s}{(s+1)^2+4}[/tex] using the formulae

    [tex]\frac{s-a}{(s-a)^2+b^2}\mapsto e^{at}\cos{bt}[/tex]


    [tex]\frac{b}{(s-a)^2+b^2}\mapsto e^{at}\sin{bt}\, ,[/tex]

    begin by rewriting it as

    [tex]\frac{s}{(s+1)^2+4} = \frac{s+1}{(s+1)^2+4} - \frac{1}{(s+1)^2+4}[/tex].

    Can you proceed from there?
  4. Nov 18, 2008 #3
    yes I do see it now...thank you
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