Help with solving Laplace Transform problem

[Ric-Veda]

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?

 

[Ray Vickson]

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).)

 

[LCKurtz]

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?