Q9RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3-10, determine the Laplace transform of the given function.${\mathit{t}}^{\mathbf{2}}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{4}\mathbf{)}$

Step-by-Step Solution

Answer

Therefore, the solution is.$\text{L}\left\{{\text{t}}^{\text{2}}\text{u(t-4)}\right\}{\text{=e}}^{\text{-4s}}\left(\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{8}}{{\text{s}}^{\text{2}}}\text{+}\frac{\text{16}}{\text{s}}\right)$

 

Step 1: Given Information

The given function value is$t^{2}u(t-4)$

Step 2: Determining the Laplace transform

Using the following Laplace transform property, to find the Laplace of given integral:

$\text{L}\left\{\text{f}\left(\text{t}\right)\text{u}\left(\text{t-a}\right)\right\}{\text{=e}}^{\text{at}}\text{L}\left\{\text{f}\left(\text{t+a}\right)\right\}$

Let. $f\left(t\right)=t^2$On applying linearity of Laplace and we get:

$\begin{array}{c}\text{L}\left\{{\text{t}}^{\text{2}}\text{u(t-4)}\right\}{\text{=e}}^{\text{-4s}}\text{L{f(t+4)}}\\ {\text{ =e}}^{\text{-4s}}\text{L}\left\{{\text{(t+4)}}^{\text{2}}\right\}\\ {\text{ =e}}^{\text{-4s}}\text{L}\left\{{\text{t}}^{\text{2}}\text{+8t+16}\right\}\\ {\text{ =e}}^{\text{4s}}\left(\text{L}\left\{{\text{t}}^{\text{2}}\right\}\text{+8L}\left\{t\right\}\text{+16L{1}}\right)\end{array}$

 Simplify further as follows:

$\begin{array}{c}\text{L}\left\{{\text{t}}^{\text{2}}\text{u(t-4)}\right\}{\text{=e}}^{\text{-4s}}\left(\frac{\text{2!}}{{\text{s}}^{\text{3}}}\text{+8}\frac{\text{1!}}{{\text{s}}^{\text{2}}}\text{+16}\frac{\text{1}}{\text{s}}\right)\\ {\text{ =e}}^{\text{-4s}}\left(\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{8}}{{\text{s}}^{\text{2}}}\text{+}\frac{\text{16}}{\text{s}}\right)\end{array}$

Therefore$\text{L}\left\{{\text{t}}^{\text{2}}\text{u(t-4)}\right\}{\text{=e}}^{\text{-4s}}\left(\frac{{\displaystyle \text{2}}}{{\displaystyle {\text{s}}^{\text{3}}}}\text{+}\frac{{\displaystyle \text{8}}}{{\displaystyle {\text{s}}^{\text{2}}}}\text{+}\frac{{\displaystyle \text{16}}}{{\displaystyle \text{s}}}\right)$