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Help doing differential equations and laplace

  1. Dec 1, 2009 #1
    I have to solve and IVP we were given an 2x2 matrix and i used the determinant method to solve for L=s^2 +2s+2, Lx =s-2 and Ly =s. so now i did this LX=Lx so s-2/(s^2+2s+2) and LY=Ly so s/(s^2+2s+2) now i have to Laplace them and im stuck. I think i have to make the (s^2+2s+2) => ((s+1)^2 +1) but im not sure bc it doesnt really help
    Last edited: Dec 1, 2009
  2. jcsd
  3. Dec 2, 2009 #2


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    I have no idea what you mean by "so s-2/(s^2+ 2x+ 2)" or "so s/(s^2+ 2s+ 2)". Are those equal to something? I also have no idea what you mean by "Laplace them". Do you mean "find their inverse Laplace transform"? Are you trying to find the inverse Laplace transform of each separately or are you combining them somehow?
  4. Dec 2, 2009 #3
    If you want to find the inverse Laplace transform of
    [tex]\frac{s-a}{(s+1)^2+1} [/tex]then find the poles which are at [tex]s=-1 \pm i [/tex] so the sum of the residues of
    [tex] e^{sx}\frac{(s-a)}{(s+1-i)(s+1+i)}[/tex] are: [tex] e^{sx}\frac{(s-a)(s+1-i)}{(s+1-i)(s+1+i)}|_{s=-1+i}+ e^{sx}\frac{(s-a)(s+1+i)}{(s+1-i)(s+1+i)}|_{s=-1-i}=


    2*Real (e^{-x}(cos(x)+isin(x))\frac{i(1+a)+1}{2})

    =e^{-x} (cos(x)-(1+a)sin(x))

    So the inverse Laplace transform of [tex]\frac{s}{s^2+2s+2}[/tex] is [tex]e^{-y} (cos(y)-sin(y))[/tex] and of [tex]\frac{s-2}{s^2+2s+2}[/tex] is [tex]e^{-x} (cos(x)-3sin(x))[/tex].

    The reason this is so is that the contour integral for x>0 vanishes at a large semicircle at the real part of s at -infinity and so the inversion formula [tex]f(x)=\frac{1}{2\pi i} \int F(s)e^{sx}ds [/tex] picks out [tex]2\pi i*residues[/tex].
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