Respuesta :
Answer:
The laplace transform is [tex]\frac{2}{(s+6)^3}[/tex]
Step-by-step explanation:
We will solve this problem by applying the laplace transform properties (their proofs are beyond the scope of this explanation).
Consider first the function f(t) = 1. By definition of the laplace transform, we have
[tex]F(s) = \int_{0}^{\infty}f(t)e^{-st}dt[/tex]
when f(t) = 1 we get
[tex]F(s) = \int_{0}^{\infty}e^{-st}dt = \left.\frac{-e^{-st}}{s}\right|_{0}^{\infty} = \frac{1}{s}[/tex]
We will apply the following properties: Define L(f) as applying the laplace transform
[tex]L(e^{at}f(t)) = F(s-a)[/tex] (this means, multiplying by an exponential corresponds to a shift in the s parameter of the transform of f)
[tex]L(t^nf(t)) = (-1)^n\frac{d^n F}{ds^n}[/tex] (this is, multypling by [tex]t^n[/tex] is equivalent to taking the n-th derivative of the transform.
We are given the function [tex]g(t) = t^2e^{-6t}\cdot 1 [/tex]. Since the transform of the constant function 1 is 1/s, by applying the first property we get
[tex]L(e^{-6t}\cdot 1 ) = \frac{1}{s+6}[/tex]
By applying the second property we get
[tex]L(g(t)) = L(t^2 e^{-6t} \cdot 1) = (-1)^2\frac{d^2}{ds^2}(\frac{1}{s+6}) = \frac{2}{(s+6)^3}[/tex]
