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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    L(sin(2t)e^5t)=2/((s-5)^2+2^2) \[2 \over (s-5)^2+4\]

  2. anonymous
    • 5 years ago
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    would 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.

  3. anonymous
    • 5 years ago
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    I see.. I actually just apply it to the formula, you can find it in any laplace table.. I will try the integration now

  4. anonymous
    • 5 years ago
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    thanks

  5. anonymous
    • 5 years ago
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    any luck?

  6. anonymous
    • 5 years ago
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    I guess.. just gimme a minute

  7. anonymous
    • 5 years ago
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    thank you so much for your help

  8. anonymous
    • 5 years ago
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    you need to do integration by parts twice for the integration \[I=\int\limits_{0}^{\infty}e ^{(5-s)t}\sin(2t)dt\]

  9. anonymous
    • 5 years ago
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    \[\int\limits_{}^{}udv=uv-\int\limits_{}^{}vdu\] u=e^(5-s)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.

  10. anonymous
    • 5 years ago
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    then you can substitute the new result in the previous one to get the final integration.

  11. anonymous
    • 5 years ago
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    I 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.

  12. anonymous
    • 5 years ago
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    you'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?

  13. anonymous
    • 5 years ago
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    let me write, directly the result of the two integrations by parts: (1) \[\int\limits_{0}^{\infty}e ^{(5-s)t}\sin(2t)dt=-1/2e ^{(5-t)s}\cos(2t)-1/2\int\limits_{0}^{\infty}e ^{(5-s)t}\cos(2t)dt\]

  14. anonymous
    • 5 years ago
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    yep, that's what I got as well

  15. anonymous
    • 5 years ago
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    I 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.

  16. anonymous
    • 5 years ago
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    now integrate the new integral: (2) \[\int\limits_{0}^{\infty}e ^{(5-s)t}\cos(2t)dt=1/2e ^{(5-s)t}\sin(2t)+\int\limits_{0}^{\infty}e ^{(5-s)t}\sin(2t)dt\]

  17. anonymous
    • 5 years ago
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    now just substitute (2) in (1) to, and take the part with integration in one side..

  18. anonymous
    • 5 years ago
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    try it out and see if it works with you

  19. anonymous
    • 5 years ago
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    are you talking about the substitution in the end of infinity and zero?

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