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

1. mukushla

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

2. mukushla

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

3. henpen

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

4. myko

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

5. mukushla

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

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

7. henpen

Sorry, I got confused by the notation

8. micahwood50

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

9. henpen

Very nice, but how does that help?

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