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Math Help - First order diff eqns, the -ive and +ive signs?

  1. #1
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    Question First order diff eqns, the -ive and +ive signs?

    Hey guys im stuck as to the origins of the + and minus signs of the following question? (It is attached via a jpeg)

    The answer tells me that u

    int/ dy/y(y+1) = int/ dx/ x(x-1)

    int/ (1/y - 1/y+1) dy = int/ (-1/x + 1/x-1)

    ok i dont get this part, where do they get the -ive sign for the 1/x and how do they decide that it is -ive plus sign for the int of the y side and +ive sign for the x side of the eqn. Am i missing a really obvious point? Please help me, im fine with the eqns but the signs are giving completely diff answers
    Attached Thumbnails Attached Thumbnails First order diff eqns, the -ive and +ive signs?-first-order-diff-question1.jpg  
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  2. #2
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    What you are doing here is simply splitting the equation into its y part and its x part. So you have
    \frac{dy}{dx}=\frac{y(y+1)}{x(x-1)}

    [\math]\frac{1}{y(y+1)}dy=\frac{1}{x(x+1)}dx[/tex]

    Integrate both sides. Fiddle about to get y. And then use y(2)=whatever to solve for the constant.
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  3. #3
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    Quote Originally Posted by Latkan View Post

    int/ dy/y(y+1) = int/ dx/ x(x-1)

    int/ (1/y - 1/y+1) dy = int/ (-1/x + 1/x-1)
    To get from here

    \int \frac{dy}{y(y+1)} = \int\frac{ dx}{ x(x-1)}

    to here

    \int \frac{1}{y} - \frac{1}{y+1}~ dy = \int \frac{-1}{x} + \frac{1}{x-1}~dx

    The denominators have been broken up using partial fractions. Do you know this method?

    This might be worth a read

    Partial fraction - Wikipedia, the free encyclopedia


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