Find the inverse Laplace transform?

Question

Find the inverse Laplace transform of F(s)=3s/(s^2-s-6).

Answer: f(t)=(9/5)e^3t+(6/5)e^-2t

 

Answer 1

Sol/   use a partial fractions to rewrite F(s)in a form.    Please carefully watch the parentheses 

3s/(s^2-s-6) = 3s/(s-3)(s+2) = A/(s-3) + B/(s+2) = (A(s+2)+B(s-2))/((s-3)(s+2))

Therefore we have  A(s+2) +B(s-3)  = 3s plugging  in  s = 3 and s = -2 ,   solve A and B, 

A =9/5,and B =6/5

so that F(s) = 3s/((s^2-s-6))  = 9/5 (1/s-3)  + 6/5(1/(s+2)

L inverse of F(s) = 9/5  L^-1 (1/(s-3))  +6/5 L^-1 (1/(s+2))

using the table  L^-1 (f(s)) = 9/5 e^3t + 6/5 e^-2t