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Math Help - Substitution when doing Laplace transforms

  1. #1
    uhm
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    Substitution when doing Laplace transforms

    Hi,

    I'm doing a coursework on Laplace transforms and a lecturer mentioned that we would have to use substitution for it. I'm not asking for help with answering the question but I don't understand how you're supposed to do substitution.

    Any help would be much appreciated!
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  2. #2
    Grand Panjandrum
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    Re: Substitution when doing Laplace transforms

    Quote Originally Posted by uhm View Post
    Hi,

    I'm doing a coursework on Laplace transforms and a lecturer mentioned that we would have to use substitution for it. I'm not asking for help with answering the question but I don't understand how you're supposed to do substitution.

    Any help would be much appreciated!
    Use substitution to do some of the integrals maybe?

    Other than that your question is too vague.

    CB
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  3. #3
    uhm
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    Re: Substitution when doing Laplace transforms

    Yea I think so, an example of the type of question being asked is:

    ℒ{t^3 e^6t}
    = 3!/S^4 1/S-5

    Then to get the final transformation my lecturer said to use substitution to get one single answer.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Substitution when doing Laplace transforms

    Is...

    \mathcal{L} \{t^{3}\ e^{6 t}\}= \int_{0}^{\infty} t^{3}\ e^{6 t}\ e^{-s\ t}\ dt (1)

    ... and using the substitution s-6= p You obtain...

    \mathcal{L} \{t^{3}\ e^{6 t}\}= \int_{0}^{\infty} t^{3}\ e^{-pt}\ dt= \frac{3!}{p^{4}} = \frac{3!}{(s-6)^{4}} (2)

    Kind regards

    \chi \sigma
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  5. #5
    uhm
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    Re: Substitution when doing Laplace transforms

    Thank you for replying!

    what I dont understand about it is where you have gotten s-6=p?
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  6. #6
    Grand Panjandrum
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    Re: Substitution when doing Laplace transforms

    Quote Originally Posted by uhm View Post
    Thank you for replying!

    what I dont understand about it is where you have gotten s-6=p?
    chisigma is chosen the substitution as it reduces the exponentials to a simpler form:

    e^{6t}e^{-st}=e^{(6-s)t}=e^pt

    and the integral becomes what he gave.

    CB

    CB
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