Help with solving Laplace Transform problem

  • Thread starter Ric-Veda
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  • #1
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Homework Statement


Solve Laplace Transform L{tsin(2t)sin(5t)}

Homework Equations


cos(bt)=s/s^2+b^2
trig identity (product identity): sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)
t^nf(t)=(-1)^nd^n/ds^nF(S)
(the template is complicated for me to use. Srry for the inconvinience)

The Attempt at a Solution


So far I have 1/2L{cos(-3)-1/2L{cos(7)}
1/2(s/s^2+4)-1/2(s/s^2+49)

Am I right?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Solve Laplace Transform L{tsin(2t)sin(5t)}

Homework Equations


cos(bt)=s/s^2+b^2
trig identity (product identity): sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)
t^nf(t)=(-1)^nd^n/ds^nF(S)
(the template is complicated for me to use. Srry for the inconvinience)

The Attempt at a Solution


So far I have 1/2L{cos(-3)-1/2L{cos(7)}
1/2(s/s^2+4)-1/2(s/s^2+49)

Am I right?

No, not even close.

BTW: you wrote
$$(1)\hspace{1.5cm}1/2 \left(\frac{s}{s^2}+4 \right) - 1/2 \left( \frac{s}{s^2} + 49 \right),$$
which equals ##(1/2)(-45)##. If you mean
$$(2) \hspace{1.5cm}1/2 \left(\frac{s}{s^2+4} \right) - 1/2 \left( \frac{s}{s^2 + 49} \right),$$
then you need to use parentheses, like this: 1/2 s/(s^2+4), etc. (When I said your answer was wrong I assumed you meant (2), not (1).)
 
  • #3
LCKurtz
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Homework Statement


Solve Laplace Transform L{tsin(2t)sin(5t)}

Homework Equations


cos(bt)=s/s^2+b^2
This equation makes no sense, with the varable ##t## on one side and ##s## on the other. They aren't equal.
trig identity (product identity): sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)
t^nf(t)=(-1)^nd^n/ds^nF(S)

Same comment

(the template is complicated for me to use. Srry for the inconvinience)

The Attempt at a Solution


So far I have 1/2L{cos(-3)-1/2L{cos(7)}
What happened to the ##t## variable in that last line?
 

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