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July 21st, 2014, 03:17 AM   #1
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Inverse Laplace

Hello, is it good?

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I dont wanna use complex numbers.
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July 24th, 2014, 03:01 PM   #2
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Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
According to the tables $\displaystyle\frac{1}{s^3-1}=\frac{-s-2}{3 \left(s^2+s+1\right)}+\frac{1}{3 (s-1)}$, $\displaystyle f(t)=\frac{1}{3} e^{-\frac{t}{2}} \left(e^{\frac{3 t}{2}}-\sqrt{3} \sin \left(\frac{\sqrt{3} t}{2}\right)-\cos \left(\frac{\sqrt{3} t}{2}\right)\right)$.

Last edited by ZardoZ; July 24th, 2014 at 03:04 PM.
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