Show that the nonlinear oscillator $y" + f(y) =0$ is equivalent to the system

$y'= -z $,

$z'= f(y)$

and that the solutions of the system lie on the family of curves

$2F(y)+ z^2 = constant $

where $F_y= f(y)$. verify that if $f(y)=y$ the curves are circle.

=>

nonlinear oscillator $y''+ f(y) =0$

where

$y'= -z $,

$z'= f(y)$

so that means

$z''+z =0$

for the solution of the system lie on the family of curves, i was thinking

$\frac{d}{dt}[2F(y(t))+z^2(t)]= 2F \frac{dy}{dt} + 2z \frac{dz}{dt}$

$=-2Fz +2zf(y)$

$=-2f(y)z+2zf(y)$

$\frac{d}{dt}[2F(y)+z^2]=0$

$2F(y)+ z^2 = constant $

for last part

if $f(y)=y$ , then the differential equation is $y'' + y =0$, meaning that

$y=A cosx +B Sinx$ and $z=-y'= - A sinx +B cosx$.

And I also know $z'' + z =0$

im trying to connect the above equations.

So, I can get $cos^2x + sin^2x =1$ which show the curves on the circles.

can someone please check my first,second and last part of answer.