Find the inverse Laplace transform of F(s)=3s/(s^2-s-6).
Answer: f(t)=(9/5)e^3t+(6/5)e^-2t
Find the inverse Laplace transform of F(s)=3s/(s^2-s-6).
Answer: f(t)=(9/5)e^3t+(6/5)e^-2t
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
Comments
If without using the table. Do you know that the inverse Laplace transform is called Mellin's transform and takes countur integration in the complex plane, finding singular points of the function? How much did you study about this and waht is required to be done? Just use of tables for inverse Lapalce transform?