Differential equations

1. Apr 11, 2014

wel

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.

Last edited: Apr 11, 2014
2. Apr 11, 2014

HallsofIvy

Staff Emeritus
This shows that the single equation $y''+ f(y)= 0$ can be written as the two equations $y'= -z$ and $z'= f(y)$. To show they are equivalent you must also show the other way: that if $y'= -z$ and $z'= f(y)$ then $y''+ f(y)= 0$.

where did you get that $y''= -y$? That would be true ony iin the very simple case that $f(y)= -y$. If this is a "nonlinear oscillator then $f$ must be nonlinear.

This isn't true. You are missing a factor of $2$. Further you have "lost" F on the right. $\frac{dF}{dy}$ is not necessarily equal to $y$.

Where did the $-yz+ zy$ come from? What you had before was a single expression that was not equal to anything. Why do

No, if $y= A \cos(x)+ B \sin(x)$ then $z= -y'= A sin(x)- B cos(x)$.

Last edited by a moderator: Apr 11, 2014
3. Apr 11, 2014

micromass

Staff Emeritus
4. Apr 11, 2014

wel

i have edited my answers for the first and second part. Would you someone don't mind checking it please.

5. Apr 11, 2014

wel

correction for last part of answer
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?

Last edited: Apr 11, 2014