# Homework Help: Differential Equations model

1. Oct 8, 2013

### ThomasMagnus

1. The problem statement, all variables and given/known data

Model for learning in the form of a differential equation:

$\frac{dP}{dt}$= 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?

2. Relevant equations

3. 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
$\frac{dP}{M-P}$ = kdt

$\int\frac{dP}{M-P}$ = k $\int dt$

for $\int\frac{dP}{M-P}$ let u=M-P, du=-dP

-$\int\frac{1}{u}$ du= -ln|M-P|

-ln|M-P|=kt+C
ln|M-P|=-kt-c
e^(-kt-c)=|M-P|
$\pm$e^(-kt)*e^(-c) =M-P
$\pm$ 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: $\int\frac{dP}{P-M}$ = $\int -kdt$ and get a final answer of: P(t)= M+Ae^(-kt). Are these answers equivalent because we can just say -A is another constant?

2. Oct 9, 2013

### ThomasMagnus

So then it would be: let -A=D

M+De^(-kt)=P(t)

3. Oct 9, 2013

### Staff: Mentor

Per PF rules, you shouldn't bump a thread earlier than 24 hours after you post it. If you need to make a change, just edit your post.