Question

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

${\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}\left(t\right)={\left(-t\right)}^{n}f\left(t\right),$

where,$f={\mathcal{L}}^{-1}\left\{F\right\}$.Use this equation in Problems 33-36 to compute${\mathcal{L}}^{-1}\left\{F\right\}.$$F\left(s\right)=ln\left(\frac{s+2}{s-5}\right)$

Short answer

${\mathcal{L}}^{-1}\left(F\right)=\frac{1}{t}\left(e^{5t}-e^{2t}\right)$

Step-by-step solution

Simplify the function and find derivative

Using the property of function $\ln t$ we get:

$$\begin{gathered} F\left(s\right)=\ln \left(\frac{s+2}{s-5}\right) \\ F\left(s\right)=\ln \left(s+2\right)-\ln \left(s-5\right) \end{gathered}$$

Find the derivative of F with respect to s:

$$\begin{gathered} \frac{dF}{ds}=\frac{d}{ds}\left(\ln \left(s+2\right)\right)-\frac{d}{ds}\left(\ln \left(s-5\right)\right) \\ \frac{dF}{ds}=\frac{1}{s+2}-\frac{1}{s-5} \end{gathered}$$

Find the Laplace inverse

From the given condition, we have ${\mathcal{L}}^{-1}\left(F\right)=\frac{1}{{\left(-t\right)}^n}{\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}$,apply this to find the Laplace inverse as:

$$\begin{gathered} {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{{\left(-t\right)}^1}{\mathcal{L}}^{-1}\left\{\frac{dF}{ds}\right\} \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{\left(-t\right)}{\mathcal{L}}^{-1}\left(\frac{1}{s+2}-\frac{1}{s-5}\right) \\ =\frac{1}{t}\left(-{\mathcal{L}}^{-1}\left(\frac{1}{s+2}\right)+{\mathcal{L}}^{-1}\left(\frac{1}{s-5}\right)\right) \\ =\frac{1}{t}\left(e^{5t}-e^{2t}\right) \\ Therefore, \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{t}\left(e^{5t}-e^{2t}\right) \end{gathered}$$