Question

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)).

Answer 1

Answer:

Step-by-step explanation:

a) The function is:

$f(t)=e^{at}t^n$

Is is necessary to demonstrate that:

$L[f(t)] = \frac{n!}{(s-a)^{n+1}}$

you can use two identities of Laplace's transforms:

$L[t^n]=\frac{n!}{s^{n+1}}\\\\L[e^{at}g(t)]=G(s-a)$

By using the previous transforms you obtain:

$L[f(t)] = \frac{n!}{(s-a)^{n+1}}$

b) The other way is to use the following formula:

$L[t^nf(t)]=(-1)^n\frac{d^n}{ds^n}F(s)$

Then, you use the following identities:

$L[e^{at}]=\frac{1}{s-a}\\\\\frac{d^n}{ds^n}[\frac{1}{s-a}]=(-1)^n\frac{n!}{(s-a)^{n+1}}$

then, by replacing you obtain:

$L[f(t)] = \frac{n!}{(s-a)^{n+1}}$