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  1. Top | #1

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    2010
    y x ( ) ...
    5 !!!!!

    ; 05-26-2016 09:20 PM

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    $ \frac {2x}{dx} = \frac {1+y^2}{2ydy}$
    $ \frac { dx}{ x} = \frac{2y}{1+y^2}dy$
    $ \int\frac { dx}{ x} =\int\frac{2y }{1+y^2}dy $
    $ \ln|x|=\ln(1+y^2 ) +lnc $
    $ \ { cx} = 1+y^2 $
    $when x=2, y=3$


    $c=5 $

    $ 5x = 1+y^2 $





    \
    \frac {2x}{dx} = \frac {1+y^2}{2ydy}
    \frac { dx}{ x} = \frac{2ydy}{1+y^2}
    \int\frac { dx}{ x} =\int\frac{2ydy}{1+y^2}
    \ln|x|=\ln(1+y^2 ) +lnc
    \ { cx} = 1+y^2
    when x=2, y=3
    c=5
    \ 5 = 1+y^2 x
    ; 06-02-2016 01:57 PM

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  7. Top | #5

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    ln c

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  11. Top | #7

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    $2xy\frac{dy}{dx}=1+y^2$
    $\frac{2ydy}{1+y^2}=\frac{dx}{x}$
    $\int {\frac{2ydy}{1+y^2}}=\int {\frac{dx}{x}}$
    $ln|1+y^2|=ln|x|+c$
    $ when x=2 , y=3 $
    $ln|1+9|=ln|2|+c$
    $c=ln5$
    $ln|1+y^2|=ln|x|+ln5$
    $y^2=5x-1$



    $ln|1+y^2|=ln|x|+c$
    $e^{ln|1+y^2|}=e^{ln|x|+c}$
    $|1+y^2|=e^{c}|x|$
    $|1+y^2|=C|x|$
    $x=\frac{1}{C}(1+y^2)$
    $ where\ C=-+e^{c}$
    $C=5$
    $x=\frac{1}{5}(1+y^2)$
    ; 05-31-2016 07:51 PM

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    ln ...
    ln

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    ; 05-30-2016 07:15 AM

  16. Top | #12

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