Critical points of nonlinear second-order ODE

Question

I want to find the critical points of the autonomous nonlinear ODE $y'' = \sin(y) - y'y$. I know these are the points where $0 = y'' = \sin(y) -y'y$.

I'm unsure if I need to solve this system outright. Other than $y = 0$ I'm having a hard time visualizing any other critical points. Any help?

Answer 1

If we rewrite the equation as a system: $$ y'=z,\qquad z'= \sin y - yz, $$ where $z=y'$, then it becomes clear that the equilibrium points are given by $$ \left\{\begin{array}{lll} z&=&0\\ \sin y - yz&=&0 \end{array}\right. $$ or $$ \left\{\begin{array}{lll} y'&=&0\\ \sin y &=&0. \end{array}\right. $$ Hence, the equilibrium points are $y=\pi k$, $k\in\mathbb Z$.