anonymous
  • anonymous
Does anyone know how to find the Inverse Laplace Transform of 6/(s^3+3s)
MIT 18.03SC Differential Equations
katieb
  • katieb
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athe
  • athe
It is combination of simple Laplace transformation:\[\frac{6}{s^3+3s}=\frac{6}{s(s^2+3)}\] Use Method of undetermined coefficients: \[\frac{6}{s(s^2+3)} \rightarrow \frac{2}{s}-\frac{2s}{s^2+3}\] Use well know transformation for this functions: \[\frac{2}{s} \rightarrow 2\cdot H(t)\], H(t) - Heaviside step function \[\frac{2s}{s^2+3}\rightarrow 2\cdot \cos(\sqrt{3}t)\] \[\frac{6}{s^3+3s} \rightarrow 2\cdot H(t) +2\cdot \cos(\sqrt{3}t)\]
anonymous
  • anonymous
we know that L[sin 3^(1/2) t]= (3^(1/2))/(s^2+3) and L[integral f(k) dk (in limits 0 to t)]=L[f(t)]/s. Here we can consider f(t)=sin 3^(1/2) t,then (3^(1/2))/s(s^2 +3) =integral f(k) dk in limits 0 to t .the integral value is 1-(cos3^(1/2) t)/(3^(1/2)). therefore, L[6/((s^3)+3s)] = 2(1-(cos3^(1/2) t)/(3^(1/2)).

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