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Math Help - Homogeneous and Inhomogeneous differential equations

  1. #1
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    Homogeneous and Inhomogeneous differential equations

    My question is:

    "Let x(t) represent the number of rabbits and y(t) represent the number of foxes in a particular population at time t months. In a simplified model, these satisfy the differential equations:

    dx/dt = 90 - y dy/dt = x - 40

    Initially there are 90 foxes and 70 rabbits.

    Obtain and solve differential equation for y in terms of x. Deduce the point (x,y) moves around the circle of radius 30. Sketch this circle showing which way the point moves.

    Differentiate the 1st equation with respect to t. Substitute for dy/dx. You now have a second order differential equation for x. You already know the initial value of x. Show that the initial value of dx/dt is 0. Find x as a function of t. Use your answer and one of the given differential equations to find y as a function of t. Confirm that (x,y) does lie on the circle. Find the time before there are 90 foxes and 70 rabbits again."


    For the first bit, I got dy/dx = (x - 40) / (90 - y).

    I solved this and got y + 180y + x + 80x = C. By completing the square you get (y - 90) + (x - 40) = C. By putting in the x = 70 and y = 90, you get C = r = 900. So radius is 30. I drew the circle with origin at (70,90).

    I didn't know how to show which way it turned. I got that when x = 40, dy/dt = 0 meaning that it was at the top of the circle. But then I didn't know how to find out what happened to dx/dt. How do you do this?

    By differentiating the first equation, you get (dy)/(dx) = -(dy/dt) = 40 - x. Therefore (dy)/(dx) + x = 40 shows us that when x = 40, dy/dx = 0.

    The next bit is to find x as a function of t. I know that this is an inhomogeneous differential equation but I don't know how to solve it. I know the complementary solution will be 40, but I don't know what the general solution ad particular solution will be and how to use these to find the functions. I haven't got a clue how to do the bit for y either.

    Please can you help me.

    Thank you
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  2. #2
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    If the center of the circle is at (40,90), and the radius of the circle is 30, then the point (70,90) is on the circle. Plug that point into the original DE, and see which way x and y are changing.
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  3. #3
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    You have some errors in your solution of the DE

    dy/dx = (x - 40)/(90 - y)

    \int[(90 - y)dy] = \int[(x - 40)dx]

    90y - (1/2)y^2 + C_1 = (1/2)x^2 - 40x + C_2

    C_1 - C_2 = (1/2)x^2 - 40x + (1/2)y^2 - 90y

    2(C_1 - C_2) = x^2 - 80x + y^2 - 180y... It appears that it merely a typo though because your completion of the square is correct.

    2(C_1 - C_2) + (-40)^2 + (-90^2) = x^2 - 80x + (-40)^2 + y^2 - 180y + (-90)^2

    r^2 = (x - 40)^2 + (y - 90)^2, where r^2 = 2(C_1 - C_2) + (-40)^2 + (-90)^2.


    BTW why would you draw the circle with the centre at (70, 90)? Surely you can see that the circle's centre is (40, 90).
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  4. #4
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    Quote Originally Posted by Ackbeet View Post
    If the center of the circle is at (40,90), and the radius of the circle is 30, then the point (70,90) is on the circle. Plug that point into the original DE, and see which way x and y are changing.
    What do you mean? dx/dt = 90 - 90 = 0. dy/dt = 70 - 40 = 30.

    So the change in x is 0 and the change in y is positive. So what does this show? If change in y is positive, then the whole thing moves up because change in x is 0 right?

    Quote Originally Posted by Prove It View Post

    BTW why would you draw the circle with the centre at (70, 90)? Surely you can see that the circle's centre is (40, 90).
    Yeah, I think that was a just a typo. I checked with my sheet of paper and I put the center as (40, 90).

    Can anyone help me with the second part of the question?
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  5. #5
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    Quote Originally Posted by ironz View Post
    What do you mean? dx/dt = 90 - 90 = 0. dy/dt = 70 - 40 = 30.

    So the change in x is 0 and the change in y is positive. So what does this show? If change in y is positive, then the whole thing moves up because change in x is 0 right?
    Yes, and since the point (40, 90) is on the right of the circle, that means the motion is counter-clockwise (or if you are British, "anti-clockwise".



    Yeah, I think that was a just a typo. I checked with my sheet of paper and I put the center as (40, 90).

    Can anyone help me with the second part of the question?[/QUOTE]
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