(with steps please) Find the inverse Laplace transform, f(t), of the function: 16/(s-4)^3
Accepted Solution
A:
Answer: The required inverse transform of the given function is[tex]f(t)=8t^2e^{4t}.[/tex]Step-by-step explanation: We are given to find the inverse Laplace transform, f(t), of the following function :[tex]F(s)=\dfrac{16}{(s-4)^3}.[/tex]We have the following Laplace formula :[tex]L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\\\\\\\Rightarrow L^{-1}\{\dfrac{1}{(s-a)^{n+1}}\}=\dfrac{t^ne^{at}}{n!}.[/tex]Therefore, we get[tex]f(t)\\\\=L^{-1}\{\dfrac{16}{(s-4)^3}\}\\\\\\=16\times\dfrac{t^{3-1e^{4t}}}{(3-1)!}\\\\\\=\dfrac{16}{2}t^2e^{4t}\\\\\\=8t^2e^{4t}.[/tex]Thus, the required inverse transform of the given function is[tex]f(t)=8t^2e^{4t}.[/tex]