MATH 308 Lecture 21

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Lecture Notes

Theorem 6.3.1

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

${\mathcal {L}}\left\{u_{c}(t)h(t-c)\right\}=\mathrm {e} ^{-c\,s}\,{\mathcal {L}}\left\{f(t)\right\}=\mathrm {e} ^{-c\,s}\,F(s)\quad s>a$

Exercise 4

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

Rewrite $f(t)=2t+u_{2}(t)\left(t^{2}-2t\right)+u_{5}(t)\left(-t^{2}\right)$

${\mathcal {L}}\left\{f(t)\right\}=2{\mathcal {L}}\left\{t\right\}+{\mathcal {L}}\left\{u_{2}(t)\,(t^{2}-2t)\right\}-{\mathcal {L}}\left\{u_{5}(t)\,t^{2}\right\}$

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

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

Theorem 6.3.2

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

${\mathcal {L}}\left\{\mathrm {e} ^{c\,t}\,f(t)\right\}=F(s-c)\quad s>a+c$

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

$\mathrm {e} ^{c\,t}\,f(t)={\mathcal {L}}^{-1}\left\{F(s-c)\right\}$

Exercise 5

Find the inverse Laplace transform of the functions:

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

${\begin{aligned}{\mathcal {L}}^{-1}\left\{G(s)\right\}&={\mathcal {L}}^{-1}\left\{{\frac {\mathrm {e} ^{-s}}{s}}-{\frac {\mathrm {e} ^{-3s}}{s}}+{\frac {3}{s}}\right\}\\&=u_{1}(t)-u_{3}(t)+3\end{aligned}}$

Exercise 7

Find solution of initial value problems $y''+y=u_{3\pi }(t)$, where $y(0)=0$ and $y'(0)=1$

${\begin{aligned}{\mathcal {L}}\left\{y''\right\}+{\mathcal {L}}\left\{y\right\}&={\mathcal {L}}\left\{1-u_{3\pi }(t)\right\}\\s^{2}{\mathcal {L}}\left\{y\right\}-y'(0)-s\,y(0)+{\mathcal {L}}\left\{y\right\}&={\frac {1-\mathrm {e} ^{-3\pi \,s}}{s}}\\(s^{2}+1){\mathcal {L}}\left\{y\right\}-1&={\frac {1-\mathrm {e} ^{-3\pi \,s}}{s}}\end{aligned}}$