How do i integrate this? Am i doing it right?

[Kbotz]

dCb/dt = k1Ca-k2Cb
dCb/(k1Ca-k2Cb)=dt
Now integrate.
ln(k1Ca-k2Cb)=t
(k1Ca-k2Cb) = et
Rearranging
k2Cb=-et +k1Ca
Cb= -et +k1Ca/k2

[TylerH]

What is constant and what is not?

[Kbotz]

k1, k2 and Ca are constants

[TylerH]

You forgot your arbitrary constant: ln(k1Ca-k2Cb)=t+C
You forgot you parentheses: Cb= (Cet +k1Ca)/k2

[TylerH]

This may help with differential equations of this form, ie separable, in general: http://tutorial.math.lamar.edu/Classes/DE/Separable.aspx

[Kbotz]

Oh sweet!
Cheers for that!

[Kbotz]

So other than that?
It's all good?

[TylerH]

Yep. You may want to look at http://www.wolframalpha.com/ for checking answers in the future; it's faster.

[ZealScience]

Why not use reverse of Chain Rule?

[TylerH]

He did. That's were the natural logarithm came from.

[ZealScience]

I mean the coefficient of Cb isn't divided by

[TylerH]

It is, but he left out the ()s, see my post above.

[HallsofIvy]

dCb/dt = k1Ca-k2Cb
dCb/(k1Ca-k2Cb)=dt
Now integrate.
ln(k1Ca-k2Cb)=t
You have integrated incorrectly- you should have
-\frac{1}{k}ln(k_1C_a- k_1C_b) = t

(k1Ca-k2Cb) = et
k_1C_a- k_1C_b= e^{-kt}

Rearranging
k2Cb=-et +k1Ca
Cb= -et +k1Ca/k2