Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By chiro

Math Help - First order autonomous equations

  1. #1
    Newbie
    Joined
    Sep 2013
    From
    Broken Arrow
    Posts
    4

    First order autonomous equations

    Hello,

    I've been working more out of Christian Constanda's "Differential Equations: A Primer for Scientists and Engineers" ISBN 978-1-4614-7296-4. This is from section 3.3, which is on autonomous equations and their models. In my particular section of Diff-EQ, all homework is 'optional'.

    I'm having what is likely a stupid problem. I've been working later sections all day - second order homogenous differential equations - and have been fine. It's only now that I'm back to first order autonomous equations that I'm hitting a brick wall. This chapter is killing me every time I look at any of its sections!

    The exercise numbers are 1 and 15.

    The instructions are: "Find the critical points and the equilibrium solutions of the given equation and solve the equation with each of the prescribed initial conditions."

    Additionally we're supposed to sketch the graphs of the solutions and comment on the stability/instability of the equilibrium solutions as well as identify what model is governed by the problem (I.E. population with logistic growth, population with ac ritical threshold, chemical reactions, etc). The part in quotes is what I'm looking for help with.

    #1: y'=300y - 2y^2; y(0) = 50; y(0) = 100; and y(0) = 200

    Correct answer: Critical points are at 0, 150. y(t)=0 (unstable, y(t) = 150 (asymptotically stable);
    y=(150y_0)/(y_0-(y_0-150)e^(-300t)).
    This models a population with logistic growth, tau=300 and Beta = 150.

    The given equation in the section for population with logistic growth is y'=((tau)-(alpha)*y)*y OR y'=(tau)*(1-y/(Beta))*y. Alpha is a constant > 0.

    My work is:

    dy/dt = 300y-2y^2

    dy/(y^2-150y) = -2*dt

    1/150 * LN((y-150)/y) = -2T + C

    (y-150)/y = e^(-300T+150C)

    y=(-150*e^(300t))/(e^300t(e^c-1)+1)

    At this point I realized I was swinging my shovel in the air trying to dig myself up out of a hole. Where'd I take a wrong turn/what rule did I break? Is anyone willing to work this problem through to solution?


    The second question I have is for good measure to make certain I get the rest of the section. It's exercise 15, I think I need to see it solved for y.

    y'=y^2+y-6; y(0) = -4; y(0) = -2; y(0) = 1; and y(0) = 3.

    The answer given is:

    -3, 2; y(t) = -3 (asymptotically stable), y(t) = 2 (unstable), and y=[-3(y_0 - 2) - 2(y_0+3)e^(-5t)] / (y_0 - 2 - (y_0 + 3) e^(-5t)]. It does not, by the book's answer, model any specific system.

    Thank you very much for your time!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: First order autonomous equations

    Hey AnotherGeek.

    This line:

    dy/(y^2-150y) = -2*dt

    looks ok but the next one doesn't. You should try completing the square and then solving the integral. It will be in terms of

    Integral 1/[(y-75)^2 - b] for some non-zero b (Complete the square to get b). (Also look up arctanh function).
    Thanks from AnotherGeek
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Autonomous PDE admits a solution of exponential order?
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: December 25th 2011, 08:02 PM
  2. Replies: 2
    Last Post: May 7th 2011, 10:26 AM
  3. first order autonomous eq
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 1st 2010, 06:24 AM
  4. simple question about autonomous first-order DEs
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 12th 2010, 11:39 PM
  5. Replies: 5
    Last Post: March 15th 2010, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum