Question

Como determinar a transformada inversa de Laplace de: F(s) = 1/(s - 1) (s + 3)(s + 5)
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Answer 1

$F(s)=\dfrac{1}{(s-1)(s+3)(s+5)}\\\\\\\dfrac{1}{(s-1)(s+3)(s+5)}=\dfrac{A}{s-1}+\dfrac{B}{s+3}+\dfrac{C}{s+5}\\\\\\A(s+3)(s+5)+B(s-1)(s+5)+C(s-1)(s+3)=1\\\\As^2+8As+15A+Bs^2+4Bs-5B+Cs^2+2Cs-3C=1\\\\(A+B+C)s^2+(8A+4B+2C)s+(15A-5B-3C)=0s^2+0s+1\\\\\\\left\{\begin{matrix}A+B+C=0\\8A+4B+2C=0\\ 15A-5B-3C=1\end{matrix}\right.\\\\\\A=\dfrac{1}{24},\ B=-\dfrac18,\ C=\dfrac{1}{12}$


$F(s)=\dfrac{1}{24}\left ( \dfrac{1}{s-1} \right )-\dfrac18\left ( \dfrac{1}{s+3} \right )+\dfrac{1}{12}\left ( \dfrac{1}{s+5} \right )\\\\\\\boxed{f(t)=\dfrac{1}{24}e^t-\dfrac18e^{-2t}+\dfrac{1}{12}e^{-5t}}$