Question

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

Answer 1

`e^t+e^(-4t)`

Explanation:

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

`(5s)/(s^2+3s-4) =(5s)/((s-1)(s+4))\\\\=(A)/(s-1)+(B)/(s+4)\\\\Therefore:\\\\(A(s+4)+B(s-1))/((s-1)(s+4))=(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:\\\\(5s)/(s^2+3s-4)=(1)/(s-1)+ (4)/(s+4)\\\\`

`From\ Laplace\ inverse:\\\\But\ L^(-1)[(1)/(s-a) ]=e^(at)\\\\Hence:\\\\L^(-1) [(5s)/(s^2+3s-4)]=L^(-1)[(1)/(s-1) ]+L^(-1)[(4)/(s+4) ]=e^(t)+4e^(-4t)\\\\L^(-1) [(5s)/(s^2+3s-4)]=e^(t)+4e^(-4t)`