Question

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

Answer 1

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

Explanation:

To find the Laplace inverse of the above problem, we have to first solve the problem as a partial fraction, hence:

`(5s)/(s^2+3s-4)=(5s)/(s^2+4s-s-4) \\\\=(5s)/(s(s+4)-1(s+4))\\\\=(5s)/((s-1)(s+4))=(A)/(s-1)+(B)/(s+4)\\\\= (A(s+4)+B(s-1))/((s-1)(s+4))\\\\Hence:\\\\(A(s+4)+B(s-1))/((s-1)(s+4))=(5s)/((s-1)(s+4))\\\\A(s+4)+B(s-1)=5s\\\\to\ find \ A,put\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\to\ find \ B,put\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Hence:\\\\(5s)/(s^2+3s-4)=(1)/(s-1)+ (4)/(s+4)\\\\`

`L^(-1) [(5s)/(s^2+3s-4)]=L^(-1)[(1)/(s-1) ]+L^(-1)[(4)/(s+4) ]\\\\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)\\\\=e^(t)+4e^(-4t)`