Laplace transform proof

  • Thread starter magnifik
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  • #1
magnifik
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show that L{f(t/b)} = bF(bs), b is not equal to 0

i know that
L{f(t)} = [tex]\int[/tex]e-stf(t) dt = F(s)
so
L{f(t/b)} = [tex]\int[/tex]e-stf(t/b) dt

any tips on how to start? thx
 

Answers and Replies

  • #2
Dick
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Do a u substitution to change the integration variable. Substitute u=t/b.
 
  • #3
magnifik
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do i also have to plug in (t/b) into e-st so that it becomes e-s(t/b) ?
 
  • #4
Dick
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do i also have to plug in (t/b) into e-st so that it becomes e-s(t/b) ?

You need to write e^(-st) in terms of u.
 
  • #5
magnifik
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so..
u = t/b
t = bu
e^(-st)
e^(-sbu)
 
  • #6
Dick
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so..
u = t/b
t = bu
e^(-st)
e^(-sbu)

Sure. Now don't forget to add a factor to change the dt to a du.
 

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