Question

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{5s-2}\right\}$

Short answer

Therefore, the inverse Laplace transform is $\frac{1}{5}{\mathrm{e}}^{2\mathrm{t}/5}$.

Step-by-step solution

Definition of Laplace Transform

Let $\mathbf{f}$be a function defined for. Then the integral $\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$ is said to be the Laplace transform of $\mathbf{f}$, provided that the integral converges.

Consider the formula for inverse Laplace given below:

1. $\mathit{1}\mathit{=}{\mathit{ℒ}}^{\mathit{-}\mathit{1}}\left\{\frac{1}{s}\right\}$
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 transform

From the formulas the inverse Laplace transform is calculated as:

$$\begin{gathered} L^{-1}\left\{\frac{1}{5s-2}\right\}=L^{-1}\left\{\frac{1}{5}·\frac{1}{s-\frac{2}{5}}\right\} \\ =\frac{1}{5}{e}^{\frac{2t}{5}} \end{gathered}$$