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Math Help - simple question about autonomous first-order DEs

  1. #1
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    simple question about autonomous first-order DEs

    consider the autonomous first-order equation dy/dx = y - y^3. and the initial condition y(0)=y_0. How do you go about drawing the graph of a typical solution y(x) when
    a) y_0>1
    b) 0<y_0<1
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  2. #2
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  3. #3
    MHF Contributor chisigma's Avatar
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    The DE is of 'separable variables type' and can be written as...

    \frac{dy}{y\cdot (1-y^{2})} = dx (1)

    Because is...

    \frac{1}{y\cdot (1-y^{2})} = \frac{1}{y} + \frac{y}{1-y^{2}} (2)

    ... is also...

    \int \frac{dy}{y\cdot (1-y^{2})} = \ln y - \frac{1}{2} \cdot \ln (1-y^{2}) + \ln c (3)

    ... so that the solution of (1) is...

    \frac{y}{\sqrt{1-y^{2}}} = c \cdot e^{x} (4)

    ... that can be 'explicitate' as function of x as...

    y= \pm \frac{c\cdot e^{x}}{\sqrt{1+c^{2}\cdot e^{2x}}} (5)

    If now we intend to impose the initial condition y(0)= y_{0} we 'discover' that we can do it only if is |y_{0}|<1. To overcome the handycap in case |y_{0}|>1 we write (2) as...

    \frac{1}{y\cdot (1-y^{2})} = \frac{1}{y} - \frac{y}{y^{2}-1} (6)

    ... and repeat the procedure obtaining...

    y= \pm \frac{c\cdot e^{x}}{\sqrt{c^{2}\cdot e^{2x}-1}} (7)

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 13th 2010 at 12:56 AM.
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