Question

Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. to find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = t2e−6t

Answer 1

Answer:

The laplace transform is $\frac{2}{(s+6)^3}$

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

$F(s) = \int_{0}^{\infty}f(t)e^{-st}dt$

when f(t) = 1 we get

$F(s) = \int_{0}^{\infty}e^{-st}dt = \left.\frac{-e^{-st}}{s}\right|_{0}^{\infty} = \frac{1}{s}$

We will apply the following properties: Define L(f) as applying the laplace transform

$L(e^{at}f(t)) = F(s-a)$ (this means, multiplying by an exponential corresponds to a shift in the s parameter of the transform of f)

$L(t^nf(t)) = (-1)^n\frac{d^n F}{ds^n}$ (this is, multypling by $t^n$ is equivalent to taking the n-th derivative of the transform.

We are given the function $g(t) = t^2e^{-6t}\cdot 1$. Since the transform of the constant function 1 is 1/s, by applying the first property we get

$L(e^{-6t}\cdot 1 ) = \frac{1}{s+6}$

By applying the second property we get

$L(g(t)) = L(t^2 e^{-6t} \cdot 1) = (-1)^2\frac{d^2}{ds^2}(\frac{1}{s+6}) = \frac{2}{(s+6)^3}$