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wel
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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 $$z'' + z =0$$
I am 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.
$$y'= -z $$,
$$z'= f(y)$$
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 $$z'' + z =0$$
I am 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.
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