Question

In Exercise 9.1.15 we formulated a model for learning in the form of the differential equation

where P(t) measures the performance of someone learning a skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential equation to find an expression for P(t). What is the limit of this expression?

Answer 1

To solve the differential equation given in Exercise 9.1.15, we need to use separation of variables method.

The given differential equation is:

dP/dt = k(M - P)

We can rearrange it as:

dP/(M - P) = k dt

Now, we integrate both sides:

∫ dP/(M - P) = ∫ k dt

ln|M - P| = kt + C

where C is the constant of integration.

We can exponentiate both sides to eliminate the natural logarithm:

|M - P| = e^(kt+C)

Since e^C is just a constant, we can replace it with another constant, A.

|M - P| = Ae^kt

Now, we can separate this equation into two cases:

Case 1: M > P

|M - P| = M - P

Therefore, we can write:

M - P = Ae^kt

Solving for P, we get:

P(t) = M - Ae^kt

Case 2: M < P

|M - P| = P - M

Therefore, we can write:

P - M = Ae^kt

Solving for P, we get:

P(t) = M + Ae^kt

So, the general solution for P(t) is given by:

P(t) =
{ M - Ae^kt, if M > P }
{ M + Ae^kt, if M < P }

Now, let's find the limit of P(t) as t approaches infinity.

When t approaches infinity, the exponential term in the solution goes to infinity. Therefore, the value of Ae^kt becomes very large compared to M, and we can ignore M in the solution.

Now, the solution becomes:

P(t) =
{ - Ae^kt, if M > P }
{ Ae^kt, if M < P }

As t approaches infinity, the exponential term goes to infinity, and the value of P(t) goes to either positive or negative infinity depending on the value of A.

Therefore, the limit of P(t) as t approaches infinity does not exist.