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    October 27, 2001


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{dy \over dx} = f(x) g(y)\end{displaymath}

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{1\over g(y)} {dy \over dx} = f(x)\end{displaymath}

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    \begin{displaymath}
\int {1\over g(y)} {dy \over dx} dx = \int f(x) dx
\to \int {1\over g(y)} dy = \int f(x) dx\end{displaymath}

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    $B$5$F!"Nc$H$7$F\begin{displaymath}
{dy \over dx} = {y \over 2x}\end{displaymath}

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    \begin{displaymath}
\int {dy \over y} = \int {dx \over 2x}
\to \log \vert y\vert = {1\over 2} \log \vert x\vert + C\end{displaymath}

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y = \pm \exp C \sqrt{x} = A \sqrt{x}\end{displaymath}

    $\pm \exp C$ $B$r2~$a$F(B A $B$H$*$-D>$7$F$$$^$9!#(B

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    Akira OHNISHI
    10/27/2001