Question

The table in Appendix C does not contain an entry for

${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{8k^{3}s}{{(s^2+k^2)}^3}\right\}$

(a) Use (4) along with the results in (5) to evaluate this inverse transform. Use a CAS as an aid in evaluating the convolution integral.

(b) Re-examine your answer to part (a). Could you have obtained the result in a different manner?

Short answer

The laplace transformation is$L^{-1}\left\{\frac{8k^{3}s}{{(s^2+k^2)}^3}\right\}=t\sin kt-k{t}^2\cos kt$.

Step-by-step solution

Applying inverse transformation property;

(a)$L^{-1}\left\{\frac{8k^{3}s}{{\left(s^2+k^2\right)}^3}\right\}$as the product of F(s)G(s), where

F(s) =$\frac{2k^3}{{\left(s^2+k^2\right)}^2}$and G(s) =$\frac{4s}{ s^2+k^2}\frac{1}{2}$

Using inverse Laplace transform; 

According to the inverse Laplace transformation property

$\begin{array}{l}\Rightarrow f\left(t\right)=L^{-1}\left\{F\left(s\right)\right\}\\ =L^{-1}\left\{\frac{2k^3}{{\left(s^2+k^2\right)}^2}\right\}\\ =\sin kt-kt\cos kt\\ \\ \Rightarrow g\left(t\right)=L^{-1}\left\{G\left(s\right)\right\}\\ =4L^{-1}\left\{\frac{s}{s^2+k^2}\right\}\\ =4\cos kt\end{array}$

Using convolution;

Use convolution to get

$L^{-1}\left\{\frac{8k^{3}s}{{\left(s^2+k^2\right)}^3}\right\}=f\left(t\right)*g\left(t\right)$

$=4\underset{0}{\overset{t}{\int }}\left[\sin k\tau -k\tau \cos k\tau \right]\cos k(t-\tau )d\tau$

$L^{-1}\left\{\frac{8k^{3}s}{{(s^2+k^2)}^3}\right\}=t\sin kt-k{t}^2\cos k$

Using a CSA to evaluate the integral we get

$L^{-1}\left\{\frac{8k^{3}s}{(s^2+k^2)3}\right\}=t\sin kt-k{t}^2\cos kt$

If n=1, then

$\begin{array}{l}{L}^{-1}\left\{t\left[\sin kt-k{t}^2\cos kt\right]\right\}=(-1)\frac{d}{ds}\left\{\frac{2k^3}{{(s^2+k^2)}^2}\right\}\\ =\frac{8k^{3}s}{{(s^2+k^2)}^3}\end{array}$

Hence the final answer is $L^{-1}\left\{\frac{8k^{3}s}{{(s^2+k^2)}^3}\right\}=t\sin kt-k{t}^2\cos kt$