First-Order differential Equation

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SUMMARY

The discussion centers on solving the first-order differential equation for a chemical kinetics problem, specifically the equation d[A]/dt=-(k+k`)[A]+k`[A]0. The solution derived is [A]= (k`+ke^(k`+k)t)*[A]0/(k`+k). The transformation from the differential equation to its solution involves rewriting it in standard form and applying an integrating factor, u=e^{\int P(t) dt}, where P(t) is (k+k`). This method is essential for understanding the behavior of reactants over time in chemical reactions.

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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
 
Last edited:
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If I am reading it correctly, your DE is

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

Which can be rewritten as

\frac{dA}{dt}+(k+k`)A=k'A_0

Since this is in the form

\frac{dA}{dt}+P(t)A= Q(t)

an integrating factor 'u' is given by

u=e^{\int P(t) dt}

see http://en.wikipedia.org/wiki/Integrating_factor" for more info.
 
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Thank you! :D
 

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