Find the inverse Laplace Transform for F(s)=1/ s(s^2+6s+5)

Question

Please help me sir/madam, question : Find the inverse Laplace Transform for F(s)=1/ s(s^2+6s+5)

thank you.

Answer 1

F(s) = 1/[s(s + 1)(s + 5)] = a/s + b/(s + 1) + c/(s + 5); a(s² + 6s + 5) + b(s² + 5s) + c(s² + s) = 1;

a + b + c = 0; 6a + 5b + c = 0; 5a = 1; a = 1/5; b + c = - 1/5; 5b + c = - 6/5; 4b = - 1; b = - 1/4; c = 1/20

F(s) = 1/(5s) - 1/[4(s + 1)] + 1/[20(s + 5)];

f(t) = 1/5 - 1/4e^{- t} + 1/20e^{- 5t}

Answer 2

Factor the denominator as s(s + 1)(s + 5), and perform partial fraction decomposition to obtain

1/[s(s+1)(s+5)] = 1/(5s) -1/[4(s+1)] + 1/[20(s+5)]

Recall that the inverse Laplace transform of 1/(s - a) is e^(at). Thus

L^(-1)(1/(5s)) = 1/5

L^(-1)(1/[4(s+1)]) = (1/4)e^(-t)

L^(-1)(1/[20(s+5)]) = (1/20)e^(-5t)

Finally,

L^(-1)[1/(s(s+1)(s+5))] = L^(-1)(1/(5s)) - L^(-1)(1/[4(s+1)]) + L^(-1)(1/[20(s+5)])

....................................= 1/5 - (1/4)e^(-t) + (1/20)e^(-5t)