Is the Nonlinear Oscillator Equivalent to a System of Equations?

[ra_forever8]

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

$y=A cosx +B Sinx$ and $z=-y'= - A sinx +B cosx$are the rotate axes.$pA^2+qAB+rB^2=1$$p,q,r$ depends on $x$choose $x$ such that $q=0$$pA^2+rB^2=1$what can i do after that?

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

 

[jvicens]

Sure, let's take a look at your answer and see if we can provide some additional clarification and help.

First, you correctly identified the system of equations that is equivalent to the nonlinear oscillator: $y'=-z$ and $z'=f(y)$. This can be seen by substituting the second equation into the first, giving us $y''+f(y)=0$, which matches the original equation.

Next, you correctly calculated the derivative of $2F(y)+z^2$ and showed that it is equal to $0$. This means that this quantity is constant, which is what we want to show in order to prove that the solutions lie on a family of curves.

However, in your next step, you wrote $2F(y)+z^2=0$ instead of $2F(y)+z^2=constant$. This is a small error, but it is important to make sure that we are correctly representing the quantity we are trying to prove is constant.

Finally, you correctly identified that if $f(y)=y$, then the differential equation becomes $y''+y=0$, which has the solutions $y=A\cos x+B\sin x$ and $z=-y'=-A\sin x+B\cos x$. However, in your last step, you wrote $pA^2+qAB+rB^2=1$ without explaining where these coefficients come from. It is not clear how this relates to the problem at hand.

Instead, we can use the solutions for $y$ and $z$ to rewrite the equation $2F(y)+z^2=constant$ as follows:

$2F(A\cos x+B\sin x)+(-A\sin x+B\cos x)^2=constant$

$2(A\sin x+B\cos x)+A^2\sin^2 x+B^2\cos^2 x -2AB\sin x\cos x=constant$

$A^2\sin^2 x+B^2\cos^2 x +2AB\cos x\sin x = constant$

$A^2+B^2=constant$

This shows that the solutions for $y$ and $z$ lie on a family of curves defined by $A^2+B^2=constant$. This means that they lie on circles centered at the origin with varying radii, depending on the value of the constant.

Overall, your answer