(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

Question

7 days ago

10

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

t 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.

Mathematics

1 answer:

tester [11.9K] · Answer 1 · 2026-04-03T07:09:38+03:00

tester [11.9K]7 days ago

4 0

Answer:

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

Step-by-step explanation:

The differential model indicates

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

We need to find P(t) as a solution to the equation.

$\dfrac{dP}{dt}=kM-kP(t)\\$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}$