Answer:
[tex]e^{t}+4e^{-4t}[/tex].
Step-by-step explanation:
To find the Laplace inverse of the above problem, we have to first solve the problem as a partial fraction, hence:
[tex]\frac{5s}{s^2+3s-4}=\frac{5s}{s^2+4s-s-4} \\\\=\frac{5s}{s(s+4)-1(s+4)}\\\\=\frac{5s}{(s-1)(s+4)}=\frac{A}{s-1}+\frac{B}{s+4}\\\\= \frac{A(s+4)+B(s-1)}{(s-1)(s+4)}\\\\Hence:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{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:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}\\\\[/tex]
[tex]L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]\\\\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}\\\\=e^{t}+4e^{-4t}[/tex]
