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Math Help - Differential Equation Substitution

  1. #1
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    Differential Equation Substitution

    Hey guys I'm a bit stuck on this problem, any help would be greatly appreciated!

    For, \frac{dy}{dx}=\frac{2cos^{2}x-sin^{2}x+y^2}{2cos^{2}x}

    Substitute, y(x)=sinx+\frac{1}{u(x)}

    to get, \frac{du}{dx}=-utanx-\frac{1}{2}secx

    and then find the general solution for y(x)


    So far I've got to u^{2}=\frac{2cosx}{sin^{2}x+y^{2}}

    Am I on the right track, and where do I go from here?
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  2. #2
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    Quote Originally Posted by jwade456 View Post
    Hey guys I'm a bit stuck on this problem, any help would be greatly appreciated!

    For, \frac{dy}{dx}=\frac{2cos^{2}x-sin^{2}x+y^2}{2cos^{2}x}

    Substitute, y(x)=sinx+\frac{1}{u(x)}

    to get, \frac{du}{dx}=-utanx-\frac{1}{2}secx

    When I make that substitution, I get:

    u'=-1/2\sec^2(x)-\sec(x)\tan(x)u+(\cos(x)-1)u^2

    The DE however, looks like a Riccati equation:

    y'+Q(x)y+R(x)y^2=P(x)

    So for:

    \frac{dy}{dx}=1-1/2\tan^2(x)+1/2\sec^2(x)y^2

    The standard approach for these is to let y=\frac{u'}{Ru}

    In that case I get:

    u''+2\tan(x)u'-1/8\sec^2(x)(1/2\tan^2(x)-1)u=0

    Second order I know, but we've changed a non-linear equation to a linear one.
    Last edited by shawsend; October 2nd 2008 at 01:58 PM.
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