First-Order differential Equation

  • Thread starter sparkle123
  • Start date
  • #1
175
0
d[A]/dt=-(k+k`)[A]+k`[A]0
The solution of this first-order differential equation is:
[A]= (k`+ke^(k`+k)t)*[A]0/(k`+k)

Could you please explain how you get from the first equation to the second? THANKS!

btw: if this helps, this is part of the solution for a chemical kinetics (rates/equilibrium) problem
so A is the reactant, t is the time, and k and k` are the rate constants for the reaction A==B
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
If I am reading it correctly, your DE is

[tex]\frac{dA}{dt}=-(k+k`)A+k'A_0[/tex]

Which can be rewritten as

[tex]\frac{dA}{dt}+(k+k`)A=k'A_0[/tex]

Since this is in the form

[tex]\frac{dA}{dt}+P(t)A= Q(t)[/tex]

an integrating factor 'u' is given by

[tex]u=e^{\int P(t) dt}[/tex]

see http://en.wikipedia.org/wiki/Integrating_factor" [Broken] for more info.
 
Last edited by a moderator:
  • #3
175
0
Thank you! :D
 

Related Threads on First-Order differential Equation

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
822
  • Last Post
Replies
1
Views
497
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
2
Views
905
  • Last Post
Replies
2
Views
964
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
2K
Top