First-Order differential Equation

sparkle123
Messages
172
Reaction score
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:
Physics news on Phys.org
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.
 
Last edited by a moderator:
Thank you! :D
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top