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Math Help - 2yy''-[(y')^2]+1=0

  1. #1
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    2yy''-[(y')^2]+1=0

    2yy''-[(y')^2]+1=0

    [(x+c2)c1/4]-1/c1 ~?
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  2. #2
    MHF Contributor chisigma's Avatar
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    The way to 'attack' this type of second order DE is explained here...

    http://www.mathhelpforum.com/math-he...iliar-ode.html

    In particular the formula to be used is...

    y^{''} = y^{'}\cdot \frac{dy^{'}}{dy} (1)

    Kind regards

    \chi \sigma
    Last edited by chisigma; March 22nd 2010 at 02:44 AM. Reason: clarification...
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  3. #3
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    what is the answer ~?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by sainimu78 View Post
    2yy''-[(y')^2]+1=0

    [(x+c2)c1/4]-1/c1 ~?
    Use the chain rule and the substituion

    u=y' \implies y''=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=u\fra  c{du}{dy}

    This gives the first order ODE

    2yu\frac{du}{dy}-u^2+1=0 \iff \frac{du}{dy}-\frac{1}{2y}u=\frac{1}{2y}

    This can be solved via an integrating factor
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    Use the chain rule and the substituion

    u=y' \implies y''=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=u\fra  c{du}{dy}

    This gives the first order ODE

    2yu\frac{du}{dy}-u^2+1=0 \iff \frac{du}{dy}-\frac{1}{2y}u=\frac{1}{2y}

    This can be solved via an integrating factor

    I solved this equation two days ,
    this is my [(x+c2)c1/4]-1/c1 answer

    [(x+c2)c1/4]-1/c1 is different from the key at the textbook
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  6. #6
    MHF Contributor chisigma's Avatar
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    The second order DE is...

    2\cdot y\cdot y^{''} - y^{' 2} + 1 = 0 (1)

    ... that with the substitution...

    y^{''} = y^{'}\cdot \frac{dy^{'}}{dy} (2)

    ... becomes...

    2\cdot y\cdot y^{'}\cdot \frac{dy^{'}}{dy} = y^{' 2} -1 \rightarrow \frac{y^{'}}{y^{' 2} -1}\cdot dy^{'} = \frac{dy}{2y} \rightarrow

    \rightarrow \ln (y^{' 2} -1) = \ln y + \ln c_{1} \rightarrow y^{' 2} = 1+ c_{1}\cdot y \rightarrow

    \rightarrow y^{'} = \pm \sqrt {1 + c_{1}\cdot y} (3)

    The first order DE we have obtained has separable variables so that it can be solved in 'standard form'...

    \frac{dy}{\sqrt {1 + c_{1}\cdot y}} = \pm dx \rightarrow \frac{2}{c_{1}}\cdot \sqrt {1 + c_{1}\cdot y } = \pm x + c_{2} \rightarrow

    \rightarrow \frac{4}{c_{1}}\cdot y = x^{2} \pm 2\cdot c_{2}\cdot x + c_{2}^{2} - \frac{4}{c_{1}^{2}} \rightarrow \dots (4)

    Kind regards

    \chi \sigma
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