Question

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPdt represents the rate at which performance improves. If M is the maximum level of performance of which the learner is capable, then a model for learning is given by the differential equation dPdt=k(M−P(t)) where k is a positive constant. a) First solve this differential equation for P(t) using C as your final (simplified) constant parameter introduced by integrating.

Answer 1

`P(t)=M+Ce^(-kt)`

Explanation:

Given the differential model

`(dP)/(dt)=k[M-P(t)]`

We are required to solve the equation for P(t).

`(dP)/(dt)=kM-kP(t)\\$Add kP(t) to both sides\\(dP)/(dt)+kP(t)=kM\\$Taking the integrating factor\\e^(\int k dt) =e^(kt)\\$Multiply all through by the integrating factor\\(dP)/(dt)e^(kt)+kP(t)e^(kt)=kMe^(kt)\\(dP)/(dt)e^(kt)=kMe^(kt)\\(Pe^(kt))'=kMe^(kt) dt\\$Take the integral of both sides with respect to t\\\int (Pe^(kt))'=\int kMe^(kt) dt\\Pe^(kt)=kM \int e^(kt) dt\\Pe^(kt)=(kM)/(k) e^(kt) + C_0, C_0$ a constant of integration`

`Pe^(kt)=Me^(kt) + C\\$Divide both side by e^(kt)\\P(t)=M+Ce^(-kt)\\P(t)=M+Ce^(-kt)\\$Therefore:\\P(t)=M+Ce^(-kt)`