## henpen Group Title $y\ddot{y}-\dot{y}^2=1$ Which method would I use here? The boundary conditions are $y(a)=y(-a)=1$ one year ago one year ago

1. mukushla Group Title

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

2. mukushla Group Title

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

3. henpen Group Title

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

4. myko Group Title

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

5. mukushla Group Title

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 Group Title

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

7. henpen Group Title

Sorry, I got confused by the notation

8. micahwood50 Group Title

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

9. henpen Group Title

Very nice, but how does that help?