1. ## Conserved quantity

I have read a paper recently in which the authors consider a system of differential equations show below.

$\displaystyle x'=x+a-y$
$\displaystyle y'=x(b-2y)/c$
Where

a,b,c are constants.

They claim that the quantity:

$\displaystyle y^2/2-y(a+x)^2+bx^2/2$

is conserved. After trying to derive this quanity from the governing equations, I can't work out where the 'c' disappears. Any thoughts would much much appreciated.

2. Very sorry, I've posted the question wrong. My problems remain though...

The actual question should have read:

$\displaystyle x'=x^2+a-y$
$\displaystyle y'=b(x-2y)/c$

The conserved quantity is:

$\displaystyle bx^2/2-y(a+x^2)+y^2/2$

Here's what I did with it though, maybe this will help...
If you ignore the 'c'.

Then you can do the following.

$\displaystyle x'=x^2+a-y$

so
$\displaystyle x'-x^2-a+y=0$

then multiply through by [tex] y' [\math] to get

$\displaystyle x'y'-x^2y'-ay'+yy'=0$

then substitute (ignoring c) $\displaystyle y'=x(b-2y)$ to get

$\displaystyle bxx'-2xx'y-x^2y'-ay'+yy'=0$

Which you can integrate to get:

$\displaystyle bx^2/2-ay-x^2y+y^2/2=constant$

Which gives the result. Don't know how to do it with the c though.

3. Originally Posted by kyz1024
Very sorry, I've posted the question wrong. My problems remain though...

The actual question should have read:

$\displaystyle x'=x^2+a-y$
$\displaystyle y'=b(x-2y)/c$ (*)

The conserved quantity is:

$\displaystyle bx^2/2-y(a+x^2)+y^2/2$

Here's what I did with it though, maybe this will help...
If you ignore the 'c'.

Then you can do the following.

$\displaystyle x'=x^2+a-y$

so
$\displaystyle x'-x^2-a+y=0$

then multiply through by [tex] y' [\math] to get

$\displaystyle x'y'-x^2y'-ay'+yy'=0$

then substitute (ignoring c) $\displaystyle y'=x(b-2y)$ (*) to get

$\displaystyle bxx'-2xx'y-x^2y'-ay'+yy'=0$

Which you can integrate to get:

$\displaystyle bx^2/2-ay-x^2y+y^2/2=constant$

Which gives the result. Don't know how to do it with the c though.
I see a conflict. See the above in red (*)