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Math Help - Trajectories in phase space

  1. #1
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    Exclamation 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.



    i would be sooo grateful if someone could please help me with this!!
    honestly, i'm lost, and any help would be much appreciated
    Last edited by mr fantastic; April 15th 2010 at 03:40 PM. Reason: Changed post title
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by thatgirlrocks View Post
    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

    i would be sooo grateful if someone could please help me with this!!
    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.
    Last edited by AllanCuz; April 15th 2010 at 11:08 AM.
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  3. #3
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    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
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by thatgirlrocks View Post
    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?
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by thatgirlrocks View Post
    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

    i would be sooo grateful if someone could please help me with this!!
    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?
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