First-Order differential Equation

  • Thread starter sparkle123
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  • #1
sparkle123
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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
 
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Answers and Replies

  • #2
rock.freak667
Homework Helper
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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.
 
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  • #3
sparkle123
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Thank you! :D
 

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