Chapter 7: The Laplace Transform

Use the Laplace transform to solve the given initial-value problem

${\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{6}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{{\left(\frac{2}{s}-\frac{1}{s^3}\right)}^2\right\}$

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{{(s+1)}^3}{s^4}\right\}$

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{{\left(s+2\right)}^2}{s^3}\right\}$

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

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

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

${\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{\mathbf{4}}{\mathbf{s}}\mathbf{+}\frac{\mathbf{6}}{{\mathbf{s}}^{\mathbf{5}}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{s}\mathbf{+}\mathbf{8}}\right\}$

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{4s+1}\right\}$

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

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

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,localid="1663927220755" $\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/.solnInverse Laplace Transform[Y, s, t]

If the impressed force $f\left(t\right)5sint$acts on the system for $0⩽t<2\pi$and is then removed.

L{f(t)}.In Problems 1 and 2 use, the definition of the Laplace transform to find $\mathcal{L}\mathbf{\left\{}\mathit{f}\mathbf{\right(}\mathit{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{.}$

$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{cc}\mathbf{0}& \mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{<}\mathbf{2}\\ \mathbf{1}& \mathbf{2}\mathbf{\le }\mathbf{t}\mathbf{<}\mathbf{4}\\ \mathbf{0}& \mathbf{t}\mathbf{\ge }\mathbf{4}\end{array}$