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

[Kbotz]

dC_b/dt = k₁C_a-k₂C_b
dC_b/(k₁C_a-k₂C_b)=dt
Now integrate.
ln(k₁C_a-k₂C_b)=t
(k₁C_a-k₂C_b) = e^t
Rearranging
k₂C_b=-e^t +k₁C_a
C_b= -e^t +k₁C_a/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]

dC_b/dt = k₁C_a-k₂C_b
dC_b/(k₁C_a-k₂C_b)=dt
Now integrate.
ln(k₁C_a-k₂C_b)=t

You have integrated incorrectly- you should have
$$-\frac{1}{k}ln(k_1C_a- k_1C_b) = t$$

(k₁C_a-k₂C_b) = e^t

$$k_1C_a- k_1C_b= e^{-kt}$$

Rearranging
k₂C_b=-e^t +k₁C_a
C_b= -e^t +k₁C_a/k₂