Question

use the translation(or shifting) theorem to find the laplace transform of f(t)=sinh(kt)sin(kt).
also show that this answer is algebraically equivalent to F(s)=(2(k^2)s)/(s^4+4k^4)

Answer 1

The hint that I was given was to remember that sinh(kt)=((e^kt)-(e^-kt))/2

Answer 2

You can split sinh(kt) into two different exponential terms, and the translation theorem of Laplace Transforms tells you that

\[L(e^{at}f(t)) = F(s-a).\] Break your function into two different terms with sin(kt) multiplying an exponential, and apply the above rule.

Answer 3

Don't forget: \[L(\sin(at)) = \frac{a}{s^2+a^2}.\]

Answer 4

just solved in on paper, lmk if you have any follow-up questions after solving it how Quandtum did it. (proper way)

Answer 5

well, I'm still learning this stuff and I'm not really sure how to apply the formula. I know it, but I'm still a little confused.

Answer 6

I have the two functions but I'm not sure where to go from there.

Answer 7

You can look up any old derivation of L(sin(at)) online, but the important part of how to use the formula above is to notice that F(s) is the Laplace transform of f, the function multiplying your exponential. If f(t) = t (or the unit step function), then F(s) = 1/s^2. However, if you want L(e^(at))*t), then you just replace the "s" in F(s) with "s-a" = 1/(s-a)^2.

TL;DR: Whenever you shift your original function by an exponential e^(at), just replace every "s" in the original transform with "s-a".

Answer 8

Now, you can take those two functions in the original problem, and split them into two terms: \[f(t) = \sinh(kt)\sin(kt) = \frac{1}{2}e^{kt}\sin(kt) - \frac{1}{2}e^{-kt}\sin(kt)\] and proceed as I described.

Answer 9

so say in the first one, you take the laplace of sin(kt) to get k/(s^2+k^2) then substitute and that's your answer?

Answer 10

Yep, you replace the s^2 term with (s-k)^2. The opposite goes for the next term.

Answer 11

so then the answer to the first half is something like\[\frac{ k }{2 (s-k)^2+k^2 }\] after including the 1/2

Answer 12

yes?

Answer 13

I think so, just make sure to be careful about that factor of 1/2. You want it multiplying the whole fraction, not just that term of the denominator.

Answer 14

okay, and my last question(I think) on this one is how do I account for the e^(-kt) in the second half? I'm not sure if I'm over thinking this one, but I'm not sure what to do with it

Answer 15

You would have (s - (-k)), or (s + k)..

Answer 16

The general formula for the translation theorem works in whatever way is dictated by the sign of your constant, "a". With a negative number in the exponent, the shift will be s+a, with a positive number (as in the first term), the shift will be s-a.

Answer 17

so the final answer for this part is \[\frac{ k }{ 2(s-k)^2+2k^2 }-\frac{k}{2(s+k)^2+2k^2}\] Correct?

Answer 18

I do believe so.

Answer 19

for the second part then, just solve the equation and that should prove it then.

Answer 20

It's not an equation that you have to solve, but rather that you have to manipulate one side until it becomes the other (to show that they're algebraically equivalent). If you set the two expressions equal to each other, you will ultimately reach 1=1 or something like that, and it won't be in good form.

Answer 21

right. sorry, I meant simplify

Answer 22

Yep. I'm getting something slightly off from the expression that you're trying to get to in the end, but it's late and my algebra is probably shot.

Answer 23

same here but I haven't come up with an answer quite yet

Answer 24

http://www.wolframalpha.com/input/?i=simplify+1%2F2*%28k%2F%28%28s-k%29%5E2%2Bk%5E2%29+-+k%2F%28%28s%2Bk%29%5E2%2Bk%5E2%29%29

a.k.a: it works. :P Just takes some grunt work.

Answer 25

I love that website... It does everything

Answer 26

and thank you for your help. I more or less understand it. Thanks

Answer 27

No problem! I recommend MIT's OCW lectures in 18.03 for future reference, if you need some more help, or simply another reference.

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/