35E - Fundamentals Of Differential Equations And Boundary Value Problems

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+9}{s^2+1}\right)$

Step-by-Step Solution

Answer

${\mathcal{L}}^{-1}\left(F\right)=\frac{2cost-2cos3t}{t}$

Step 1: Simplify the function and find derivative

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

$$\begin{gathered} F\left(s\right)=\ln \left(\frac{s^2+9}{s^2+9}\right) \\ =\ln \left(s^2+9\right)-\ln \left(s^2+1\right) \end{gathered}$$

Find the derivative of F with respect to s:

$$\begin{gathered} \frac{dF}{ds}=\frac{d}{ds}\left(\ln \left(s^2+9\right)-\ln \left(s^2+1\right)\right) \\ \frac{dF}{ds}=\frac{d}{ds}\left(\ln \left(s^2+9\right)\right)-\frac{d}{ds}\left(\ln \left(s^2+1\right)\right) \\ \frac{dF}{ds}=\frac{2s}{s^2+9}-\frac{2s}{s^2+1} \end{gathered}$$

Step 2: 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:

$$\begin{gathered} {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{\left(-t\right)}{\mathcal{L}}^{-1}\left\{\frac{dF}{ds}\right\} \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{\left(-t\right)}{\mathcal{L}}^{-1}\left(\frac{2s}{s^2+9}-\frac{2s}{s^2+1}\right) \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{t}\left(-{\mathcal{L}}^{-1}\left(\frac{2s}{s^2+9}\right)+{\mathcal{L}}^{-1}\left(\frac{2s}{s^2+1}\right)\right) \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{1}{t}\left(-2\cos 3t+2\cos t\right) \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{2\cos t-2\cos 3t}{t} \\ Therefore, \\ {\mathcal{L}}^{-1}\left(F\right)=\frac{2\cos t-2\cos 3t}{t} \end{gathered}$$