Question

(with steps please) Find the inverse Laplace transform, f(t), of the function: 16/(s-4)^3

Answer 1

Answer:  The required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

Step-by-step explanation:  We are given to find the inverse Laplace transform, f(t), of the following function :

$F(s)=\dfrac{16}{(s-4)^3}.$

We have the following Laplace formula :

$L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\\\\\\\Rightarrow L^{-1}\{\dfrac{1}{(s-a)^{n+1}}\}=\dfrac{t^ne^{at}}{n!}.$

Therefore, we get

$f(t)\\\\=L^{-1}\{\dfrac{16}{(s-4)^3}\}\\\\\\=16\times\dfrac{t^{3-1e^{4t}}}{(3-1)!}\\\\\\=\dfrac{16}{2}t^2e^{4t}\\\\\\=8t^2e^{4t}.$

Thus, the required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$