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henpen

  • 2 years ago

\[ y\ddot{y}-\dot{y}^2=1 \] Which method would I use here? The boundary conditions are \[y(a)=y(-a)=1 \]

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  1. mukushla
    • 2 years ago
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    i think reduction of order with\[p=\dot y\]will lead us to something maybe?

  2. mukushla
    • 2 years ago
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    just note that\[\ddot y=p\dot p \]

  3. henpen
    • 2 years ago
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    \[ \dot{p}=\frac{ dp }{ dx}= \frac{d^2y}{dx^2}=\ddot{y} \]Surely?

  4. myko
    • 2 years ago
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    [yy']'=yy''+y'^2 so just difference in sign. Maybe it is a derivative of some quotient.

  5. mukushla
    • 2 years ago
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    make p a function of y\[\ddot y=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p \dot p\] and what @myko mentioned can be a good start too

  6. mukushla
    • 2 years ago
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    actually that is\[\ddot y=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p \dot p_y\]

  7. henpen
    • 2 years ago
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    Sorry, I got confused by the notation

  8. micahwood50
    • 2 years ago
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    \[\Large yy'' - \left( y' \right)^2 = -\left(y'\right)^2\left( \frac{y}{y'} \right)'\]

  9. henpen
    • 2 years ago
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    Very nice, but how does that help?

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