# First order autonomous equations

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• Sep 21st 2013, 07:54 PM
AnotherGeek
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!
• Sep 21st 2013, 10:48 PM
chiro
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).