- #1

ThomasMagnus

- 138

- 0

## Homework Statement

Model for learning in the form of a differential equation:

[itex]\frac{dP}{dt}[/itex]= k(M-P)

Where P(t) measures the performance of someone learning a skill after 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?

## Homework Equations

## The Attempt at a Solution

I think I am doing this problem the correct way. However, my textbook uses a different method. Would you be able to confirm if I am do this correctly?

dP=k(M-P)dt

[itex]\frac{dP}{M-P}[/itex] = kdt

[itex]\int\frac{dP}{M-P}[/itex] = k [itex]\int dt [/itex]

for [itex]\int\frac{dP}{M-P}[/itex] let u=M-P, du=-dP

-[itex]\int\frac{1}{u}[/itex] du= -ln|M-P|

-ln|M-P|=kt+C

ln|M-P|=-kt-c

e^(-kt-c)=|M-P|

[itex]\pm[/itex]e^(-kt)*e^(-c) =M-P

[itex]\pm[/itex] e^(-c) is a constant so call it A

**M-Ae^(-kt)=P(t)**

as t→∞ P→M

The book does something different. At the very start they say: [itex]\int\frac{dP}{P-M}[/itex] = [itex]\int -kdt [/itex] and get a final answer of: P(t)= M+Ae^(-kt). Are these answers equivalent because we can just say -A is another constant?