Chapter 7: The Laplace Transform

(a) Assume that Theorem 7.3.1 holds when the symbol a is replaced by ki, where k is a real number and ${\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$. Show that $\mathbf{L}\left\{{\mathrm{te}}^{\mathrm{kti}}\right\}$ can be used to deduce

$$\begin{gathered} \mathcal{L}\mathbf{\left\{}\mathbf{tcoskt}\mathbf{\right\}}\mathbf{=}\frac{{\mathbf{s}}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}{{\left(s^2+k^2\right)}^{\mathbf{2}}} \\ \mathcal{L}\mathbf{\left\{}\mathbf{tsinkt}\mathbf{\right\}}\mathbf{=}\frac{\mathbf{2}\mathbf{ks}}{{\left(s^2+k^2\right)}^{\mathbf{2}}} \end{gathered}$$

(b) Now use the Laplace transform to solve the initial value problem

${\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{cos\omega t}\mathbf{,}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Appropriately modify the procedure of Problem 72 In this problem you are led through the commands in Mathematica that enable you to obtain the symbolic Laplace transform of a differential equation and the solution of the initial-value problem by finding the inverse transform. In Mathematica the Laplace transform of a function y(t) is obtained using Laplace Transform [y[t], t, s]. In line two of the syntax we replace Laplace Transform [y[t], t, s] by the symbol Y. (If you do not have Mathematica, then adapt the given procedure by finding the corresponding syntax for the CAS you have on hand.)

Consider the initial-value problem

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{tsin}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}$

Load the Laplace transform package. Precisely reproduce and then, in turn, execute each line in the following sequence of commands. Either copy the output by hand or print out the results.

diffequat$\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left[}\mathbf{t}\mathbf{\right]}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\left[}\mathbf{t}\mathbf{\right]}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{\left[}\mathbf{t}\mathbf{\right]}\mathbf{=}\mathbf{=}\mathbf{tsin}\mathbf{\left[}\mathbf{t}\mathbf{\right]}$transformdeq 5 Laplace Transform,$\mathbf{[}\mathbf{}\mathbf{diffequat}\mathbf{,}\mathbf{t}\mathbf{,}\mathbf{s}\mathbf{]}\mathbf{}\mathbf{/}\mathbf{.}$$\left\{\mathbf{y}\mathbf{\left[}\mathbf{0}\mathbf{\right]}\mathbf{-}\mathbf{>}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left[}\mathbf{0}\mathbf{\right]}\mathbf{-}\mathbf{>}\mathbf{-}\mathbf{1}\right.$ Laplace Transform$\mathbf{\left[}\mathbf{y}\mathbf{\right[}\mathbf{t}\mathbf{]}\mathbf{,}\mathbf{t}\mathbf{,}\mathbf{s}\mathbf{]}\mathbf{-}\mathbf{>}\mathbf{Y}\mathbf{]}$soln = Solve[transformdeq, Y]//Flatten Y 5 Y/.soln

Inverse Laplace Transform[Y, s, t] to find a solution of

${\mathbf{y}}^{\mathbf{w}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$

$\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$

$\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$

Use the Laplace transform to find the charge $q\left(t\right)$on the capacitor in an RC series circuit subject to the given conditions.

$q\left(0\right)=0,R=2.5\Omega ,C=0.08f,E\left(t\right)$given in figure.

In Problems 3-24 fill in the blanks or answer true or false.

If fis not piecewise continuous on $\left[0,\infty \right)\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathbf{}\mathcal{L}\left\{f\left(t\right)\right\}$will not exist.

In Problems 9-14use the Laplace transform to solve the given initialvalue problem. Use the table of Laplace transforms in Appendix C as needed.

$\mathbf{10}\mathbf{.}\mathbf{}{\mathbf{y}}^{\mathbf{t}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{te}}^{\mathbf{t}}\mathbf{sint}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

In Problems 9–14 use the Laplace transform to solve the given initial value problem. Use the table of Laplace transforms in Appendix C as needed

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$

In Problems 9–14 use the Laplace transform to solve the given initial value problem. Use the table of Laplace transforms in Appendix C as needed

_{${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{sint}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$}

In Problems 9–14 use the Laplace transform to solve the given initial value problem. Use the table of Laplace transforms in Appendix C as needed

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}$$\begin{array}{l}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\mathbf{cos}\mathbf{4}\mathbf{t}\right\}\mathbf{,}\mathbf{0}\mathbf{}\mathit{\epsilon }\mathbf{}\mathbf{t}\mathbf{<}\mathbf{\pi }\\ \left\{\mathbf{0}\right\}\mathbf{,}{\mathbf{t}}^{\mathbf{3}}\mathbf{\pi }\end{array}$

In Problems 9–14 use the Laplace transform to solve the given initial value problem. Use the table of Laplace transforms in Appendix C as needed

_{${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$}

Where$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\begin{array}{l}1,0\le t<\frac{\pi }{2}\\ sint,t\ge \frac{\pi }{2}\end{array}\right\}$

In Problems 15 and 16 use a graphing utility to graph the indicated

solution.

y (t) of Problem $\mathbf{0}\le \mathbf{t}<\mathbf{2}\pi$