Q16RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3-10, determine the Laplace transform of the given function.$\frac{\mathbf{1}}{{\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{)}}^{\mathbf{2}}}$

Step-by-Step Solution

Answer

Therefore, the solution is${\mathcal{L}}^{-1}\left\{\frac{1}{{(s^2+9)}^2}\right\}=\frac{\sin 3t-3t\cos 3t}{54}$

Step 1: Given Information

The given function value in s domain is $\frac{1}{{(s^2+9)}^2}$.

Step 2: Use convolution theorem

Let $f\left(t\right)=\sin 3t$Then

$\begin{array}{c}F\left(s\right)={\mathcal{L}}^{-1}\left\{f\right(t\left)\right\}\left(s\right)\\ ={\mathcal{L}}^{-1}\left\{\sin 3t\right\}\left(s\right)\\ =\frac{3}{s^2+3^2}\\ \\ \end{array}$

We can write the given function as

$\frac{1}{{(s^2+9)}^2}=\frac{1}{9}\frac{3}{s^2+3^2}\frac{3}{s^2+3^2}$

Now, using convolution theorem we get

$\begin{array}{c}{\mathcal{L}}^{-1}\left\{\frac{1}{{(s^2+9)}^2}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{9}\frac{3}{s^2+3^2}\frac{3}{s^2+3^2}\right\}\\ =\frac{1}{9}{\mathcal{L}}^{-1}\left\{F\right(s\left)F\right(s\left)\right\}\\ =\frac{1}{9}\left[f\right(t)*f(t\left)\right]\\ =\frac{1}{9}[\sin 3t*\sin 3t]\end{array}$

Simplify further as:

$\begin{array}{c}{\mathcal{L}}^{-1}\left\{\frac{1}{{(s^2+9)}^2}\right\}=\frac{1}{9}{\int }_0^t\sin 3(t-v)\sin 3tdv\\ =\frac{1}{18}{\int }_0^t\left[\cos \right(6v-3t)-\cos 3t]dv\\ =\frac{1}{18}\left({\int }_0^t\cos \right(6v-3t)dv-{\int }_0^t\cos 3tdv)\\ =\frac{1}{18}({\left[\frac{\sin (6v-3t)}{6}\right]}_0^t-\cos 3t{\left[v\right]}_0^t)\\ =\frac{1}{18}(\frac{\sin 3t}{6}+\frac{\sin 3t}{6}-t\cos 3t)\end{array}$

Further simplify as:

$\begin{array}{c}{\mathcal{L}}^{-1}\left\{\frac{1}{{(s^2+9)}^2}\right\}=\frac{1}{18}\frac{\sin 3t-3t\cos 3t}{3}\\ =\frac{\sin 3t-3t\cos 3t}{54}\end{array}$

Therefore, the required inverse Laplace transform is

${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle 1}}{{\displaystyle {(s^2+9)}^2}}\right\}=\frac{\sin 3t-3t\cos 3t}{54}$