Your two equations should be You got a minus sign in there that doesn't belong. So, what is the solution to this system?
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That's where I am confused. What do I do about the a and b. Or do I just treat them like constants?
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?
B=-A Ab+(-A)a=1 right?
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.
Last edited by Ackbeet; February 18th 2011 at 06:38 PM. Reason: Corrected.
Does A= 1/(b-a) then?
Looks good. Then B is what?
B= -(1/(b-a)) So I now have to integrate [(1/(b-a))/(a-x)]-[(1/(b-a))/(b-x)]?
That's correct. What do you get?
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?
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: but How does that change things?
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)?
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: Now exponentiate.
[(b-x)/(a-x)]/(b-a)? Or do I have to do that bottom as well?
Well, what you really have is the equation I'd recommend throwing the to the RHS thus: Now exponentiate.
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