First-Order differential Equation

[sparkle123]

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

 

[rock.freak667]

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.

 

[sparkle123]

Thank you! :D