## henpen Group Title $y\ddot{y}-\dot{y}^2=1$ one year ago one year ago

1. henpen Group Title

$y\ddot{y}-\dot{y}^2=1$

2. henpen Group Title

The boundary conditions are $y(a)=y(-a)=1$ . $\large \ddot{y}=\frac{ \dot{y}^2+1}{y}$ Let $p=\dot{y}$ . $\large \dot{p}=\frac{dp}{dy}p=\frac{ p^2+1}{y}$ $\large\int \frac{p}{p^2+1}dp= \int \frac{1}{y}dy$ Let $u=p^2$ $\large \frac{1}{2}ln|u+1| =ln|\sqrt{u+1}|=ln|y|+c$ $\large \sqrt{u+1} =ky$ $\large \dot{y}^2+1 =by^2$ What do I do now?

3. matricked Group Title

after this dy/dx = sqrt(by2-1) and then proceed ...