Final answer:
The inverse Laplace transform of the function F(s) = 1 / ((s + a)(s + b)) with a = b is te-at, where a represents a constant.
Step-by-step explanation:
To find the inverse Laplace transform of F(s) = 1 / ((s + a)(s + b)) when a = b, we can use partial fraction decomposition. However, since a and b are equal, this becomes a special case known as repeated linear factors. We can write:
F(s) = A / (s + a)^2
To find A, we would normally differentiate both sides with respect to s, but in this case, we notice that setting A equal to 1 would be sufficient because the derivative of s + a is 1. Thus, F(s) simplifies to:
F(s) = 1 / (s + a)^2
The inverse Laplace transform of 1 / (s + a)^2 is t e-a t, where a represents a constant. Therefore, the inverse Laplace transform of our original function F(s) is:
L-1{F(s)} = t e-a t