Consider the direction field below for a differential equation y' = y(y-3)(y-5)

Question

1.Find the equilibrium solutions for the differential equation.

where I have found y = (0,3,5)

2.If the initial condition is y(0) = c, for what values of c is lim_t→∞ y(t) finite?

Answer:(as an interval): ??

Answer 1

The three equilibria y=0, y=3, and y=5 divide the plane into four regions. From the equation we can see that the derivative y' is positive in the regions 0<y<3 and y>5, and y' is negative in y<0 and 3<y<5. By observing the direction (slople) field of the equation, we can see that a solution with y(0)=c is increasing in the regions where y' is positive and decreasing in the regions where y' is negative. Such a solution will approach −∞ as x approaches ∞ when c<0; or it will approach ∞ as x approaches ∞ when c>5. On the other hand, when 0<c<5, y will approach 3 as x approaches ∞. So the conclusion is that the the limit of the solution approaches a finite value if and only if c belons to the interval [0,5]. In addition, the only stable equilibrium is y=3.