MATH 308 Lecture 21

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Lecture Notes

Theorem 6.3.1

If $F(s)={ℒ}\{f(t)\}$ exists for $s>a\ge 0$ and if $c$ is a positive constant, then

${ℒ}\{u_c(t)h(t-c)\}={\mathrm{e}}^{-cs}{ℒ}\{f(t)\}={\mathrm{e}}^{-cs}F(s)s>a$

Exercise 4

Find the Laplace transform of $f(t)=\{\begin{array}{ll}2t& 0\le t<2\\ t^2& 2\le t<5\\ 0& 5\le t\end{array}$

Rewrite $f(t)=2t+u_2(t)(t^2-2t)+u_5(t)(-t^2)$

${ℒ}\{f(t)\}=2{ℒ}\left\{t\right\}+{ℒ}\{u_2(t)(t^2-2t)\}-{ℒ}\{u_5(t)t^2\}$

For second term, let $h(t-2)=t^2-2t$, then $h(t)=t^2+2t$

For third term, let $h(t-5)=t^2$, then $h(t)=x^2+10x+25$

Then we have

${ℒ}\{f(t)\}=\frac{2}{s^2}+{\mathrm{e}}^{-2s}(\frac{2}{s^3}+\frac{2}{s^2})-{\mathrm{e}}^{-5s}(\frac{2}{s^3}+\frac{10}{s^2}-\frac{25}{s})$

Theorem 6.3.2

If $F(s)={ℒ}\{f(t)\}$ exists for $s>a\ge 0$ and if $c$ is a constant, then

${ℒ}\{{\mathrm{e}}^{ct}f(t)\}=F(s-c)s>a+c$

Conversely, if $f(t)={{ℒ}}^{-1}\{F(s)\}$, then

${\mathrm{e}}^{ct}f(t)={{ℒ}}^{-1}\{F(s-c)\}$

Exercise 5

Find the inverse Laplace transform of the functions:

• $F(s)=\frac{{\mathrm{e}}^{-2x}}{s^2-2s-3}$
• $G(s)=\frac{{\mathrm{e}}^{-s}-{\mathrm{e}}^{-3s}+3}{s}$
• $H(s)=\frac{2{\mathrm{e}}^{-3s}}{(s-1{)}^2+4}$

$\begin{array}{rl}{{ℒ}}^{-1}\{G(s)\}& ={{ℒ}}^{-1}\{\frac{{\mathrm{e}}^{-s}}{s}-\frac{{\mathrm{e}}^{-3s}}{s}+\frac{3}{s}\}\\ & =u_1(t)-u_3(t)+3\end{array}$

Exercise 7

Find solution of initial value problems $y^{″}+y=u_{3\pi }(t)$, where $y(0)=0$ and $y^{\prime }(0)=1$

$\begin{array}{rl}{ℒ}\left\{y^{″}\right\}+{ℒ}\left\{y\right\}& ={ℒ}\{1-u_{3\pi }(t)\}\\ s^2{ℒ}\left\{y\right\}-y^{\prime }(0)-sy(0)+{ℒ}\left\{y\right\}& =\frac{1-{\mathrm{e}}^{-3\pi s}}{s}\\ (s^2+1){ℒ}\left\{y\right\}-1& =\frac{1-{\mathrm{e}}^{-3\pi s}}{s}\end{array}$