Mathematics, 05.05.2020 17:30 yasyyas646646
Show that laplace transform StartSet e Superscript at Baseline t Superscript n EndSetℒeattn(s)equals=StartFraction n exclamation mark Over (s minus a )Superscript n plus 1 EndFraction n! (s−a)n+1 in two ways.
(a) Use the translation property for F(s).
(b) Use the formula laplace transform StartSet t Superscript n Baseline f (t ) EndSet (s )ℒtnf(t)(s)equals=(negative 1 )Superscript n Baseline StartFraction d Superscript n Over ds Superscript n EndFraction (laplace transform StartSet f EndSet (s ))(−1)n dn dsn(ℒ{f}(s)).
Answers: 1
![L[f(t)] = \frac{n!}{(s-a)^{n+1}}](/tpl/images/0640/7896/57934.png)
![L[t^n]=\frac{n!}{s^{n+1}}\\\\L[e^{at}g(t)]=G(s-a)\\\\](/tpl/images/0640/7896/b200b.png)
![L[t^nf(t)]=(-1)^n\frac{d^n}{ds^n}F(s)](/tpl/images/0640/7896/57e8e.png)
![L[e^{at}]=\frac{1}{s-a}\\\\\frac{d^n}{ds^n}[\frac{1}{s-a}]=(-1)^n\frac{n!}{(s-a)^{n+1}}](/tpl/images/0640/7896/8c08e.png)
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Show that laplace transform StartSet e Superscript at Baseline t Superscript n EndSetâ„’eattn(s)equals...
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