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Math Help - desperate seconds

  1. #16
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    I've found the same result as in yours. Is it weird or is that the true result? y= Ax+ B/x + C x ln|x|
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  2. #17
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    Quote Originally Posted by freakymath View Post
    3) ty''+(1-2t)y'-2y=0 y(0)=1 , y'(0)=2 Laplace
    For the Laplace question, this should also be straight forward. Just apply the laplace transformation to each term. You can look up the transformations on a table.
    Once you get the solution in the transformed space, use the table to get the inverse laplace transform of your solution.
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  3. #18
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    Quote Originally Posted by freakymath View Post
    I've found the same result as in yours. Is it weird or is that the true result? y= Ax+ B/x + C x ln|x|
    You should actually have a value for C (I just left it as a variable since I didn't do the work).
    The C x ln|x| is the particular solution.

    Why is it weird? You can plug it back in to check.
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  4. #19
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    Maybe i mustn't force my limits! Does that laplace question have a long answer?
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  5. #20
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    Quote Originally Posted by freakymath View Post
    Maybe i mustn't force my limits! Does that laplace question have a long answer?
    If you know how to do it, it is straightforward, but requires a bit of work.

    For your test at least know how to setup the question.
    ty''+(1-2t)y'-2y=0
    Apply laplace denoted by L
    L(ty'' + (1-2t)y' - 2y) = L(0)
    L(ty'' + y' - 2ty' - 2y) = 0
    L(ty'') + L(y') - 2L(ty') - 2L(y) = 0

    If you have time, you look up the laplace transforms of these individual terms on a table (probably provided with the exam) that looks like this: http://en.wikipedia.org/wiki/Laplace...s_and_theorems
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  6. #21
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    Thx for ur all kind replies sir -> Take care
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