Modeling Population Growth [dP/dt = k P - A P2 - h]

  • Thread starter NZBRU
  • Start date
  • #1
20
0
Does anyone know how to solve dP/dt = k P - A P2 - h for P. I understand partial fractions are needed and I have already solved dP/dt = k P - A P2. Is anyone able to solve it, Cheers NZBRU.
 

Answers and Replies

  • #2
1,225
75
Hi.

dt = dP/(k P - A P^2 - h) = -1/A dP/[ (P- k/A)^2-{(k/A)^2+h} ] = -1/A dp/[p^2 - {(k/A)^2+h}) ] , p= P- k/A

= -1/A dp/[p - sqrt{(k/A)^2+h} ][p +sqrt {(k/A)^2+h} ]

now you can integrate.
 
  • #3
20
0
If I typed that in correctly the line would be [-1/A [dp]/[[p - sqrt{(k/A)^2+h} ][p +sqrt {(k/A)^2+h} ]]]=dt or would it be: [-1/A [dp] [p +sqrt {(k/A)^2+h} ]/[p - sqrt{(k/A)^2+h} ]]=dt? (I have not used ASCIIMath extensively). Thank you for the fast response.
 
  • #4
798
34
Hi !
integration gives t(P)
The inverse fonction is P(t)
 

Attachments

  • EDO.JPG
    EDO.JPG
    2.1 KB · Views: 331

Related Threads on Modeling Population Growth [dP/dt = k P - A P2 - h]

  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
21K
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
7K
  • Last Post
Replies
2
Views
20K
Top