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Math Help - Laplace and Series Question

  1. #1
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    Laplace and Series Question

    I need help on two webwork questions.

    The first is a Laplace


    For this is transformed the equation and ended up with a linear differential eqn of the form:
    Y'(s) +Y(s) (sē-6)/-2s= (8+7sē)/-2sē

    Im not sure if this is the correct method. I'm also stuck after this step as I cannot find a suitable integrating factor.

    Secondly,
    I completed most of this question
    I am just unsure as to what
    " The closed form of solution a" is referring too.




    Thanks !
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  2. #2
    MHF Contributor chisigma's Avatar
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    First question...

    Remembering that if Y(s)= \mathcal{L} \{y(t)\} the following basic properties hold...

    \mathcal{L} \{y^{'}(t)\} = s\cdot Y(s) - y(0)

    \mathcal{L} \{y^{''}(t)\} = s^{2}\cdot Y(s) - s\cdot y(0) - y^{'}(0)

    \mathcal{L} \{t\cdot y(t)\} = - Y^{'} (s) (1)

    ... the DE...

    y^{''} + 2t\cdot y^{'} - 4\cdot y = 8 , y(0)=7, y^{'}(0)=0 (2)

    ... in terms of LT is written as...

    s^{2}\cdot Y(s) -7s - 2\cdot s\cdot Y^{'}(s) - 2\cdot Y(s) -4\cdot Y(s) = \frac{8}{s} (3)

    ... and from (3) ...

    Y^{'}(s)= \frac{s^{2}-6}{2s}\cdot Y(s) - \frac{8 + 7\cdot s^{2}}{2\cdot s^{2} } (4)

    So your solution is correct but the problem now is to solve the first order DE (4) in terms of s...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    In the last post we have shown that the linear second order DE in t...

    y^{''} + 2t\cdot y^{'} - 4\cdot y = 8 , y(0)=7, y^{'}(0)=0 (1)

    ... can be written in terms of LT as the first order DE in s...

    Y^{'}(s)= \frac{s^{2}-6}{2s}\cdot Y(s) - \frac{8 + 7\cdot s^{2}}{2\cdot s^{2} } (2)


    The (2) can be solved in 'standard fashion' and its general solution is...


    Y(s) = c \cdot \frac{s^{\frac{x^{2}}{4}}}{s^{3}} + \frac{36}{s^{3}} + \frac{7}{s} (3)

    For the 'initial value theorem' from the condition y(0)=7 we detrive that it must be...

    \lim_{t \rightarrow 0} y(t) = \lim_{s \rightarrow \infty} s\cdot Y(s) = 7 (4)

    ... and that means that in (3) must be c=0, so that the solution of (1) is...

    y(t)= \mathcal{L}^{-1} \{\frac{36}{s^{3}} + \frac{7}{s}\} = 18 t^{2} + 7 (5)

    Kind regards

    \chi \sigma
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