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anonymous
 3 years ago
\[ y\ddot{y}\dot{y}^2=1 \]
Which method would I use here? The boundary conditions are \[y(a)=y(a)=1 \]
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think reduction of order with\[p=\dot y\]will lead us to something maybe?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just note that\[\ddot y=p\dot p \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \dot{p}=\frac{ dp }{ dx}= \frac{d^2y}{dx^2}=\ddot{y} \]Surely?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0[yy']'=yy''+y'^2 so just difference in sign. Maybe it is a derivative of some quotient.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0make 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually that is\[\ddot y=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p \dot p_y\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, I got confused by the notation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large yy''  \left( y' \right)^2 = \left(y'\right)^2\left( \frac{y}{y'} \right)'\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Very nice, but how does that help?
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