Question

In Problems 3–10, determine the Laplace transform of the given function.${\mathit{t}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{9}\mathbf{t}}$

Short answer

$L\left\{t^2{e}^{-9t}\right\}\left(s\right)=\frac{2}{{(s+9)}^3}$

Step-by-step solution

Step 1:Given Information.

The given function is .$t^2{e}^{-9t}$

Determining the Laplace transform:

Using the Laplace transform definition,$L\left\{t^n{e}^{at}\right\}=\frac{n!}{{(s-a)}^{n+1}}$we get

$\begin{array}{c}L\left\{t^2{e}^{-9t}\right\}=\frac{2!}{{(s+9)}^3}\\ =\frac{2}{{(s+9)}^3}\end{array}$

Determining the Result

Thus, the required Laplace transform is .$L\left\{t^2{e}^{-9t}\right\}\left(s\right)=\frac{2}{{(s+9)}^3}$