1. Your two equations should be

$\displaystyle A+B=0,$

$\displaystyle Ab+Ba=1.$

So, what is the solution to this system?

2. That's where I am confused.
What do I do about the a and b. Or do I just treat them like constants?

3. Originally Posted by CluelessSenior
That's where I am confused.
What do I do about the a and b. Or do I just treat them like constants?
Yes, little a and b are just constants, and you're solving for the capital A and B. What do you get?

4. B=-A
Ab+(-A)a=1
right?

5. See post # 16. I think you're off by a minus sign in your second equation there.

[EDIT]: I see you corrected the sign error.

6. Does A= 1/(b-a) then?

7. Looks good. Then B is what?

8. B= -(1/(b-a))

So I now have to integrate [(1/(b-a))/(a-x)]-[(1/(b-a))/(b-x)]?

9. That's correct. What do you get?

10. I get [ln(a-x)-ln(b-x)]/(b-a)=kt.

Then I multiplied both sides by e.

Getting [(a-x)-(b-x)]/(b-x) = e^kt, right?

11. I'm not sure your integrations are correct. I think that, yet again, you're off by a minus sign. The rule goes like this:

$\displaystyle \displaystyle\int\frac{1}{x-a}\,dx=\ln|x-a|+C,$ but

$\displaystyle \displaystyle\int\frac{1}{a-x}\,dx=-\ln|x-a|+C.$

How does that change things?

12. The signs always seem to get me...

So I get [-ln(a-x)+ln(b-x)]/(b-a)=kt

-a+b-2x=(b-a)(e^kt)?

13. No, no. The logarithms don't simplify that way. Exponentiation does NOT distribute over addition (most functions don't). I would use the logarithm identities to help here:

$\displaystyle \dfrac{-\ln(a-x)+\ln(b-x)}{b-a}=\dfrac{\ln((b-x)/(a-x))}{b-a}.$

Now exponentiate.

14. [(b-x)/(a-x)]/(b-a)?

Or do I have to do that bottom as well?

15. Well, what you really have is the equation

$\displaystyle \dfrac{\ln((b-x)/(a-x))}{b-a}=kt.$

I'd recommend throwing the $\displaystyle b-a$ to the RHS thus:

$\displaystyle \ln\left(\dfrac{b-x}{a-x}\right)=kt(b-a).$

Now exponentiate.

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# integrate dx/dt = k (a-x) (b-x)

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