Question 1185370

It is known that {{{L^(-1) (1/(s^2-9)) = (1/6)*(e^(3t) - e^(-3t)) = (1/3)*sinh(3t)}}}.


===> {{{L^(-1) (e^(-9s)/(s^2-9)) = (1/6)*H(t-9)*(e^(3(t-9)) - e^(-3(t-9))) = (1/3)*H(t-9)*sinh(3(t-9))}}}, where H(t) is the Heaviside step function.