Chapter 7: The Laplace Transform

In problemsfind either f(S)or f(t),as indicated$\mathit{L}\mathbf{\left\{}{\mathbf{t}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{\right\}}$

In Problems 37–40 find $L\left\{f\left(t\right)\right\}$by first using a trigonometric identity.

40.$f\left(t\right)=10.\cos \left(t-\frac{\pi }{6}\right)$

$L\left\{\right(3t+1\left)U\right(t-1\left)\right\}$

In Problems 35–42 use the Laplace transform to solve the given

equation.

40. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{e}$

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{12}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{x}}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}\mathbf{\times }\mathbf{t}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{k}\mathbf{)}\mathbf{.}$

We have encountered the gamma function$\Gamma \left(\alpha \right)$in our study of Bessel functions in Section 6.4 (page 263). One definition of this function is given by the improper integral

$\Gamma \left(\alpha \right)={\int }_0^{\infty }{t}^{\alpha -1}{e}^{-t}dt,\alpha >0.$

Use this definition to show that $\Gamma (\alpha +1)=\alpha \Gamma \left(\alpha \right).$ When $\alpha =n$ is a positive integer the last property can be used to show that$\Gamma (n+1)=n!$. See Appendix A.

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

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\sqrt{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\sqrt{\mathbf{2}\mathbf{t}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{10}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}$

$L\left\{\cos 2tU\right(t-\pi \left)\right\}$

In Problems 35–42 use the Laplace transform to solve the given

equation.

$\mathbf{41}\mathbf{.}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{t}\mathbf{+}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathbf{y}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{\tau }\mathbf{)}\mathbf{d}\mathbf{\tau }\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Use Problem 41 and the change of variable $u=st$to obtain the generalization

$\mathcal{L}\left\{t^{\alpha }\right\}=\frac{\Gamma (\alpha +1)}{s^{\alpha +1}},\alpha >-1,$

of the result in Theorem 7.1.1(b).

$L\left\{\sin tU\left(t-\frac{\pi }{2}\right)\right\}$