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Math Help - Laplace Transforms to solve Initial-Value problem

  1. #16
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    Quote Originally Posted by TheEmptySet View Post
    I think you need to have

    \displaystyle \frac{A}{(s+2)}+\frac{B}{(s+2)^2}
    So the correct PF expansion would be,

    Y=\dfrac{A}{s+2}+\dfrac{B}{(s+2)^{2}} + \dfrac{C(s+1)+D}{(s+1)^{2}+1}

    ?
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  2. #17
    Behold, the power of SARDINES!
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    Quote Originally Posted by jegues View Post
    So the correct PF expansion would be,

    Y=\dfrac{A}{s+2}+\dfrac{B}{(s+2)^{2}} + \dfrac{C(s+1)+D}{(s+1)^{2}+1}

    ?
    It should be

    \displaystyle \dfrac{A}{s+2}+\dfrac{B}{(s+2)^{2}} + \dfrac{Cs+D}{(s+1)^{2}+1}
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  3. #18
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    I, for one, am not seeing why there has to be a

    \dfrac{B}{(s+2)^{2}}

    term when there isn't one in the LT. Is there one in the LT? If so, why didn't it cancel? Here's what I have from the original DE:

    y'' + 4y' + 4y = e^{-t}(\sin(t) + \cos(t)), y(0) = 0, y'(0) = 0.

    LT gives

    s^{2}Y+4sY+4Y=\dfrac{1}{(s+1)^{2}+1}+\dfrac{s+1}{(  s+1)^{2}+1}=\dfrac{s+2}{(s+1)^{2}+1}. This implies

    Y(s+2)^{2}=\dfrac{s+2}{(s+1)^{2}+1}, or

    Y(s+2)=\dfrac{1}{(s+1)^{2}+1}, and hence

    Y=\dfrac{1}{(s+2)((s+1)^{2}+1)}.

    There's no 1/(s+2)^{2} in there.

    Did I do something wrong?
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  4. #19
    Behold, the power of SARDINES!
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    Quote Originally Posted by Ackbeet View Post
    I, for one, am not seeing why there has to be a

    \dfrac{B}{(s+2)^{2}}

    term when there isn't one in the LT. Is there one in the LT? If so, why didn't it cancel? Here's what I have from the original DE:

    y'' + 4y' + 4y = e^{-t}(\sin(t) + \cos(t)), y(0) = 0, y'(0) = 0.

    LT gives

    s^{2}Y+4sY+4Y=\dfrac{1}{(s+1)^{2}+1}+\dfrac{s+1}{(  s+1)^{2}+1}=\dfrac{s+2}{(s+1)^{2}+1}. This implies

    Y(s+2)^{2}=\dfrac{s+2}{(s+1)^{2}+1}, or

    Y(s+2)=\dfrac{1}{(s+1)^{2}+1}, and hence

    Y=\dfrac{1}{(s+2)((s+1)^{2}+1)}.

    There's no 1/(s+2)^{2} in there.

    Did I do something wrong?
    No, I did something wrong, I have apparently forgotten how to add :/
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  5. #20
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    Well, you know you're a mathematician if you can do vector calculus but not long division.
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