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

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

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

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

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