Final answer:
The equilibrium solutions of the given differential equation are y = 1 and y = 5. The equilibrium solution y = 1 is asymptotically stable, while the equilibrium solution y = 5 is unstable. The phase line shows that between y = 1 and y = 5, the solutions decrease.
Step-by-step explanation:
To determine the equilibrium solutions of the given differential equation, we set dt/dy equal to zero and solve for y:
0 = -(y-1)(y-5)
Setting each factor equal to zero, we find two equilibrium solutions: y = 1 and y = 5.
To classify each equilibrium solution, we can analyze the sign of dt/dy around each solution. Plugging in y = 1 in the original differential equation, we get dt/dy = -(1-1)(1-5) = 0. Since dt/dy is zero at y = 1, it is an asymptotically stable equilibrium solution. Plugging in y = 5, we get dt/dy = -(5-1)(5-5) = 0. Since dt/dy is zero at y = 5, it is an unstable equilibrium solution.
The phase line is a number line with the equilibrium solutions marked as points and the signs of dt/dy indicated in between. Between y = 1 and y = 5, dt/dy is negative, indicating that the solutions decrease. Outside this interval, dt/dy is positive, indicating that the solutions increase.
To sketch a few solution curves in the ty-plane, we can choose some initial conditions and integrate the differential equation. For example, starting at y = 0, the solution curve will approach y = 1 as t approaches negative infinity. Starting at y = 2, the solution curve will approach y = 5 as t approaches positive infinity. We can continue this process to sketch more solution curves.