Laplace transform of derivatives: Given the following ODE: x′′−x′−6x=0. In addition we are given the following definitions: Lf′(t)=F(s)−f(0) and Lf′'=s^2−sf(0)−f′(0)

Harrison Mills 2022-09-26 Answered
Given the following ODE: x x 6 x = 0. In addition we are given the following definitions:
L f ( t ) = s F ( s ) f ( 0 ) and
L f = s 2 s f ( 0 ) f ( 0 )
The part I am confused about is what is the transformation of −6x?
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Answers (2)

vidovitogv5
Answered 2022-09-27 Author has 10 answers
x x 6 x = 0
Apply the Laplace Transform:
L ( x x 6 x ) = 0
L ( x ) L ( x ) L ( 6 x ) = 0
Since you have that: L [ 6 x ] ( s ) = 6 0 e t s x ( t ) d t = 6 X ( s ) Hence:
L ( x ) L ( x ) 6 X ( s ) = 0
You know what to do with the first two terms.
L ( x ) = s 2 X ( s ) s x ( 0 ) x ( 0 )
L ( x ) = s X ( s ) x ( 0 )
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shaunistayb1
Answered 2022-09-28 Author has 4 answers
By definition,
L [ 6 x ] ( s ) = 0 e t s 6 t d t = I B P [ 1 s e t s 6 t ] t = 0 + 6 s 0 e t s d t
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