Find the equilibrium points?

Question

Find the equilibrium points of dy/dt=y(y-1)(y-2), yo≥0.

Answer 1

Let's solve the differential equation to come up with an expression for the function. First, let's write it down in the following way:

                                      dy/[y(y-1)(y-2)]   = dt                                    (1)

Now, we have to simplify the rational expression:

                                         1/[y(y-1)(y-2)] = [1/(y-1) - 1/y] 1/(y-2)

and then

                                           1/[(y-1)(y-2)]  = 1/(y-2)  -  1/(y-1)

                                            1/y(y-2)   = (1/2)[1/(y-2) - 1/y]

Now we can integrate the equation:

                                           (1/2) ln [y(y-2)]  - ln (y-1)  + const =  t

or after some transformations

                                                       y(y-2)/(y-1)² = C e ^2t                             (2)

For t = 0 we have                              C = y₀(y₀-2)/(y₀-1)²                       (3)    

but we will keep C letter for furher calculations. By multiplying (2) by (y-1)²  we can obtain a quadratic

equation

                                                     y² - 2y = (y-1)² C e ^2t                             (4)

The solution of this equation is (I am skipping some routin steps)

                                                     y (t) = 1 +(-) 1/sqrt(1-C e^2t)             (5)

Thus, we have two equlibria solutions for t → - ∞ (y = 0 and y = 2). The last solution (y = 1) can be obtained for t →  ∞  if C< 0 which is pssible for  0< y₀ < 2 (look at (3)).

There is no attainable equilibrium state for any finite segment of the timeline ( - ∞ < t < + ∞).

Answer 2

Solving dy/dt=y(y-1)(y-2) = 0 for y, you get the quilibrium points:

y = 0, 1, 2 <==Answer