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Math Help - Separable Equations Application

  1. #16
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    Your two equations should be

    A+B=0,

    Ab+Ba=1.

    You got a minus sign in there that doesn't belong.

    So, what is the solution to this system?
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  2. #17
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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?
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  3. #18
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    Quote Originally Posted by CluelessSenior View Post
    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?
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  4. #19
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    B=-A
    Ab+(-A)a=1
    right?
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  5. #20
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    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 05:38 PM. Reason: Corrected.
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  6. #21
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    Does A= 1/(b-a) then?
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  7. #22
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    Looks good. Then B is what?
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  8. #23
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    B= -(1/(b-a))


    So I now have to integrate [(1/(b-a))/(a-x)]-[(1/(b-a))/(b-x)]?
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  9. #24
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    That's correct. What do you get?
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  10. #25
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    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?
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  11. #26
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    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\int\frac{1}{x-a}\,dx=\ln|x-a|+C, but

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

    How does that change things?
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  12. #27
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    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)?
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  13. #28
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    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:

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

    Now exponentiate.
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  14. #29
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    [(b-x)/(a-x)]/(b-a)?

    Or do I have to do that bottom as well?
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  15. #30
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    Well, what you really have is the equation

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

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

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

    Now exponentiate.
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