What is Laplace transform of (e^(-t)/(t) sin 3t sin 2t

skystasvs

skystasvs

Answered question

2022-09-13

What is Laplace transform of e t t sin 3 t sin 2 t

Answer & Explanation

nirosoh9

nirosoh9

Beginner2022-09-14Added 16 answers

e t t sin ( 3 t ) sin ( 2 t ) = e t 2 t cos ( t ) e t 2 t cos ( 5 t )
Let f ( t ) = e t cos ( t ) and g ( t ) = e t cos ( 5 t ), we get
e t t sin ( 3 t ) sin ( 2 t ) = 1 2 ( f ( t ) t g ( t ) t )
Using the property that
L ( f ( t ) t ) = s F ( u )   d u
we get
L ( e t t sin ( 3 t ) sin ( 2 t ) ) = 1 2 [ s F ( u )   d u s G ( u )   d u ]
where F ( s ) , G ( s ) are Laplace Transforms of f ( t ) , g ( t ). Using L ( e a t cos ( w t ) ) = s + a ( s + a ) 2 + w 2 , we can say
F ( u ) = s + 1 ( s + 1 ) 2 + 1
G ( u ) = s + 1 ( s + 1 ) 2 + 25
Computing the integrals:
s F ( u )   d u = s u + 1 ( u + 1 ) 2 + 1   d u = 1 2 [ lim u ln ( ( u + 1 ) 2 + 1 ) ln ( ( s + 1 ) 2 + 1 ) ]
s G ( u )   d u = s u + 1 ( u + 1 ) 2 + 25   d u = 1 2 [ lim u ln ( ( u + 1 ) 2 + 25 ) ln ( ( s + 1 ) 2 + 25 ) ]
Denote the limits as A,B, then:
L ( e t t sin ( 3 t ) sin ( 2 t ) ) = 1 4 [ ln ( ( s + 1 ) 2 + 25 ) ln ( ( s + 1 ) 2 + 1 ) + A B ]
which is
L ( e t t sin ( 3 t ) sin ( 2 t ) ) = 1 4 [ ln ( s + 1 ) 2 + 25 ( s + 1 ) 2 + 1 + lim u ln ( u + 1 ) 2 + 25 ( u + 1 ) 2 + 1 ] = 1 4 ln ( s + 1 ) 2 + 25 ( s + 1 ) 2 + 1
Makayla Reilly

Makayla Reilly

Beginner2022-09-15Added 3 answers

With integration of the formula L ( e c t ) = 1 s c respect to c we find
L ( e c t t ) = ln 1 s c
then by sin α = e i α e i α 2 i we write
L ( e t sin 3 t sin 2 t t ) = 1 4 L ( e ( 5 i 1 ) t e ( i 1 ) t e ( i 1 ) t + e ( 5 i 1 ) t t ) = 1 4 L ( ln 1 s ( 5 i 1 ) ln 1 s ( i 1 ) ln 1 s ( i 1 ) + ln 1 s ( 5 i 1 ) ) = 1 4 ln ( s + 1 ) 2 + 1 ( s + 1 ) 2 + 25

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