# Thread: Trajectories in phase space

1. ## Trajectories in phase space

the functions x(t) and y(t) satisfy:

x'=x+ 2xy
y' =-3y + xy

a) show that the solution trajectories in x,y phase space are given by

ln|(x^3)y| = x +2y + constant

b) determine the critical points of the system. solve the linear comparison system corresponding to each critical point(i.e. determine a relationship between x and y (or between u and v for the translated critical points)) - determine the type and stability of each critical point, and sketch the trajectories in the vicinity of each.

honestly, i'm lost, and any help would be much appreciated

2. Originally Posted by thatgirlrocks
the functions x(t) and y(t) satisfy:

x'=x+ 2xy
y' =-3y + xy

a) show that the solution trajectories in x,y phase space are given by

ln|(x^3)y| = x +2y + constant

honestly, i'm lost, and any help would be much appreciated
Edit, snap read it incorrectly. 3 posts in a row that i've had to edit (geeze, i need a break!) 1 sec

This system is of the form

$x=f(x(t),y(t))$ and $y=g(x(t),y(t))$

Find

$\frac{dy}{dx}$

And see what we have, if it is seperable great! If it is not, more then likely it will be exact or something else that we can evaluate.

Of course, once we figure out which type we're dealing with we go ahead accordingly and find the general solution. This will then form our trajectories.

3. My issue is i have tried to solve the differential equations, and i'm just getting nowhere. been attempting it for 3 weeks now and still lost. i don't even understand the trajectories bit!! :S

It is an issue :P thank you for the hints though

4. Originally Posted by thatgirlrocks
My issue is i have tried to solve the differential equations, and i'm just getting nowhere. been attempting it for 3 weeks now and still lost. i don't even understand the trajectories bit!! :S

It is an issue :P thank you for the hints though
Why don't you walk us through your solutions? Where do you get and why do you stop?

5. Originally Posted by thatgirlrocks
the functions x(t) and y(t) satisfy:

x'=x+ 2xy
y' =-3y + xy

a) show that the solution trajectories in x,y phase space are given by

ln|(x^3)y| = x +2y + constant

honestly, i'm lost, and any help would be much appreciated
So I decided to go ahead and attempt this and see if there were any major problems. I will show you the solution to part A, you must follow with Part B.

What we have here is

$\frac{dy}{dt} = =-3y + xy$

$\frac{dx}{dt} = x+ 2xy$

$\frac{dy}{dx} = \frac{-3y + xy}{x+ 2xy}$

$\frac{dy}{-3y+xy}=\frac{dx}{x+2xy}$

$\frac{dy}{(-3+x)} \frac{1}{y} =\frac{dx}{(1+2y)} \frac{1}{x}$

$\frac{(1+2y)dy}{y} = \frac{(-3 +x)dx}{x}$

$\int \frac{(1+2y)dy}{y} = \int \frac{(-3 +x)dx}{x}$

$ln(y) + 2y = -3ln(x) + x + c$

Of course then we can simplify to get to what you have there

$ln(y) + ln(x^3) = 2y + x + c$

$ln(yx^3) = 2y + x + c$

Did you get this far or no?