Question

Inverse laplace of L^-1 {5s/s^2 + 3s - 4}​

Answer 1

Answer:

$e^t+e^{-4t}$

Step-by-step explanation:

We have to simplify the original function using partial fraction, hence:

$\frac{5s}{s^2+3s-4} =\frac{5s}{(s-1)(s+4)}\\\\=\frac{A}{s-1}+\frac{B}{s+4}\\\\Therefore:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{5s}{(s-1)(s+4)}\\\\Eliminating\ the \ denominator:\\\\A(s+4)+B(s-1)=5s\\\\substitute\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\tsubstitute\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Therefore\ substituting\ A\ and\ B\ gives:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}$

$From\ Laplace\ inverse:\\\\But\ L^{-1}[\frac{1}{s-a} ]=e^{at}\\\\Hence:\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]=e^{t}+4e^{-4t}\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=e^{t}+4e^{-4t}$