• Support PF! Buy your school textbooks, materials and every day products Here!

Differential Equations model

  • #1
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?
 

Answers and Replies

  • #2
138
0
So then it would be: let -A=D

M+De^(-kt)=P(t)
 
  • #3
33,505
5,191
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.
 

Related Threads on Differential Equations model

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
12
Views
770
Replies
1
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
1K
Top