Question

15. In Problems 1-30 use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform.

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{2s-6}{s^2+9}\right\}$

Short answer

The inverse Laplace transform is $2\cos \left(3\mathrm{t}\right)-2\sin \left(3\mathrm{t}\right)$.

Step-by-step solution

Define the formulas for inverse Transform

Consider the formula for inverse Laplace given below:

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}$
   
2. $t^n={\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\},n=1,2,3,\dots$
3. $e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}$
4. $\sin kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}$
5. $\cos kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2+k^2}\right\}$
6. $\sinh kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2-k^2}\right\}$
7. $\cosh kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2-k^2}\right\}$
   

Determine the inverse Laplace transform:

From the formulas, simplify the expression and solve for the inverse Laplace transform as:

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{2s-6}{s^2+9}\right\}=2{\mathcal{L}}^{-1}\left\{\frac{s}{s^2+9}\right\}-L^{-1}\left\{\frac{2·3}{s^2+9}\right\} \\ =2\cos \left(3t\right)-2\sin \left(3t\right) \end{gathered}$$

Therefore, the inverse Laplace transform is $2\cos \left(3t\right)-2\sin \left(3t\right)$.