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anonymous
 5 years ago
how would I do the laplace transform of sin(2t)e^(5t)?
anonymous
 5 years ago
how would I do the laplace transform of sin(2t)e^(5t)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0L(sin(2t)e^5t)=2/((s5)^2+2^2) \[2 \over (s5)^2+4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you show me the steps, because I got the answer from a website, but I can't actually get there on my own with integration by parts. I keep on ending up with sin and cos on the top.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see.. I actually just apply it to the formula, you can find it in any laplace table.. I will try the integration now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess.. just gimme a minute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you so much for your help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to do integration by parts twice for the integration \[I=\int\limits_{0}^{\infty}e ^{(5s)t}\sin(2t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}udv=uv\int\limits_{}^{}vdu\] u=e^(5s)t , dv=sin(2t) ... I think you know how to proceed. you will get another integration with cos this time, do integration by parts once more with the substitution for u and proceed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you can substitute the new result in the previous one to get the final integration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got all of that, but I still end up with wonky sin and cos stuff on the top, even when using the indicator function with integration by parts twice.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you're right.. the final step is the trick, if I may say, you will notice that the integral you got at last is the same one you have in the beginning, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me write, directly the result of the two integrations by parts: (1) \[\int\limits_{0}^{\infty}e ^{(5s)t}\sin(2t)dt=1/2e ^{(5t)s}\cos(2t)1/2\int\limits_{0}^{\infty}e ^{(5s)t}\cos(2t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep, that's what I got as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think that I also got the next part, but when I do the algebra for the last step and get I on its own again, I end up with those wonky sin and cos bits in the numerator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now integrate the new integral: (2) \[\int\limits_{0}^{\infty}e ^{(5s)t}\cos(2t)dt=1/2e ^{(5s)t}\sin(2t)+\int\limits_{0}^{\infty}e ^{(5s)t}\sin(2t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now just substitute (2) in (1) to, and take the part with integration in one side..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try it out and see if it works with you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you talking about the substitution in the end of infinity and zero?
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