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

Question

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

anonymous · Answer

i think reduction of order with$$p=\dot y$$will lead us to something maybe?

anonymous · Answer

just note that$$\ddot y=p\dot p $$

anonymous · Answer

$$ \dot{p}=\frac{ dp }{ dx}= \frac{d^2y}{dx^2}=\ddot{y} $$Surely?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

actually that is$$\ddot y=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p \dot p_y$$

anonymous · Answer

Sorry, I got confused by the notation

anonymous · Answer

$$\Large yy'' - \left( y' \right)^2 = -\left(y'\right)^2\left( \frac{y}{y'} \right)'$$

anonymous · Answer

Very nice, but how does that help?