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Differential equation help

  1. Sep 5, 2007 #1
    Solve the Differential Equation

    (1-cos(x))dy + (2ysin(x) - tan(x)) dx = 0

    I have tried this two ways so far and either way that I do it does not look correst.

    First way: As a seperable DE.

    (1-cos(x))dy/dx = 2ysin(x) - tan(x)

    simplifying I eventually get......

    dx - sec(x)dx = [cscx(cosx + 1)/2y]dy

    and more simplifing

    [1- sec(x)]dx/cscx(cosx + 1) = dy/2y


    Then I integrated the right side and am having trouble integrating the left side.

    int.[tan(x) * (cosx - 1)/(cosx + 1)] = (1/2) ln (y ) +c


    is this a reasonable place to keep going from? if so, any tips on the integration?




    The second way I did this is as a first order linear DE.

    I got it in the form

    (dy/dx) + p(x)y = q(x)

    I used e and raised it to the integral of p(x)

    and did some more work

    y[e^[-2ln(1 - cosx)] = integrate.[ -tanx( 1 / (1-cosx))(e^[-2ln(1 - cosx))]



    the right side does not look like other problems I do at the end when I do the final integration.


    any thoughts on this...? Even make any sense?
     
  2. jcsd
  3. Sep 5, 2007 #2

    Dick

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    Science Advisor
    Homework Helper

    The equation isn't separable - so your first approach is just an error. Use the second approach with the integrating factor. But you have a sign error in your integral of p. And note that exp(2ln(1-cosx)) can be written a lot more simply.
     
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