Need use the laplace transform of te^(3t) cos(3t) So I started out with the property L{t^n f(t)}=(-1)^n (d)/(ds) L{f(t)} Which yielded -s^2-b^2+2s^2/(s^2+b^2)^2. If I did it correctly?

madeeha1d8

madeeha1d8

Answered question

2022-09-20

I have to take the laplace transform of t e 3 t cos ( 3 t ) So I started out with the property L { t n f ( t ) } = ( 1 ) n d d s L { f ( t ) } Which yielded s 2 b 2 + 2 s 2 / ( s 2 + b 2 ) 2 . If I did it correctly.
Then I would be using L { e a t f ( t ) } = L { f } ( s a )
When I plug in s,a,b, I get in the end 3 ( s 3 ) 2 + 49 / ( ( s 3 ) 2 + 49 ) 2
The denominator is correct, but the book says the numerator is ( s 3 ) 2 49. What did I do wrong?

Answer & Explanation

Davian Nguyen

Davian Nguyen

Beginner2022-09-21Added 9 answers

F ( s ) = L { t e 3 t cos 3 t }
F ( s ) = L { t cos 3 t } s s 3
F ( s ) = d d s L { c o s 3 t } s s 3
F ( s ) = d d s { s s 2 + 9 } s s 3
F ( s ) = [ 1 ( s 2 + 9 ) s ( 2 s ) ( s 2 + 9 ) 2 ] s s 3
F ( s ) = [ s 2 + 9 ( s 2 + 9 ) 2 ] s s 3 = ( s 3 ) 2 9 ( ( s 3 ) 2 + 9 ) 2
F ( s ) = s 2 6 s ( s 2 6 s + 18 ) 2

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