Math / CalculusQuestionFind the inverse Laplace transform of F(s)=8e−9ss2+49F(s)=\frac{8 e^{-9 s}}{s^{2}+49}F(s)=s2+498e−9s f(t)=1f(t)=1f(t)=1 ( Use step (t−c)(t-c)(t−c) for uc(t)u_{c}(t)uc(t).)Studdy Solutionf(t)=87u9(t)sin(7(t−9))f(t) = \frac{8}{7}u_9(t)\sin(7(t-9))f(t)=78u9(t)sin(7(t−9))View Full Solution - FreeWas this helpful?
