Question

For each differential equation, determine (with reasons) whether or not graphs of solution functions will ever touch any and all equilibrium solution functions (consider both positive and negative values of t ).

(a) dt/dL​ =0.5(1−L)
(b) dt/dy​ =0.3y(1− 10y​ )
(c) dt/dy​ =−t+1

Answer 1

Differential equations (a) and (b) have equilibrium solutions that the solution functions approach without touching, while for (c), the linear nature of the equation means that solutions can intersect with the equilibrium line.

Step-by-step explanation:

For differential equation (a) dt/dL = 0.5(1−L), we need to determine if the graph of solution functions will ever touch the equilibrium solution. An equilibrium solution occurs when the derivative is zero, so by setting the right-hand side to zero, we find L=1. Therefore, as L approaches 1 from any initial condition L < 1, the solutions will approach the equilibrium solution without ever touching it, assuming positive values of t.

For (b) dt/dy = 0.3y(1 - 10y), we look for equilibrium solutions by setting the expression equal to zero. This gives us y=0 or y=0.1 as equilibrium solutions. In general, the solutions will tend to approach these equilibria without crossing them, just like in the first equation.

The case (c) dt/dy = -t + 1 is different because here the equilibrium is determined by setting t = 1. In this linear differential equation, the solutions can indeed intersect with the equilibrium line since it's not affected by the y-value and is instead a horizontal line.