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Thread: What are the steps in these equations?

  1. #1
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    What are the steps in these equations?

    What are the steps in these differential equations?
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  2. #2
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    #1 is integration, one side at a time. However, you seem to have changed variables from M to t on the left-hand side. This will make it all more confusing.
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  3. #3
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by fyec View Post
    What are the steps in these differential equations?
    This image is hard to read, but I think the steps were as follows:



    [tex]\mbox{1. }\, \dfrac{dM}{S\, -\, kM}\, =\, dt[/tex]

    [tex]\mbox{2. }\, -\dfrac{1}{k}\, \ln(S\, -\, kM)\,\big|_0^t\, =\, t\,\big|_0^t[/tex]

    [tex]\mbox{3. }\, \left(-\dfrac{1}{k}\right)\, \ln\left(\dfrac{S\, -\, km(t)}{S\, -\, km(0)}\right)\, =\, t[/tex]

    [tex]\mbox{4. }\, M(t)\, =\, M(0)\, e^{-kt}\, +\, \dfrac{S}{k}\, \left(1\, -\, e^{-kt}\right)[/tex]



    When you say that you don't understand these steps (which involve nothing past what you learned back in algebra and in calculus), please clarify what you mean. For instance, you should understand how they got the log expression in step (3), because you understand how log rules work, and how you can convert a subtraction containing two logs into one log containing a division.

    Please be specific. Thank you!

  4. #4
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    Shady physicist notation strikes again! (Note: I am a physicist, not a mathematician). I think a more correct way of writing step one would be

    [tex] \frac{1}{S-kM} \frac{dM}{dt} = 1 [/tex]

    Now, as others have stated, integrate both sides of this equation with respect to t from t = 0 to t = tau (let's say), where tau is the actual time at which we want to evaluate things (t is just a dummy integration variable here).

    Hint: Suppose you have arbitrary function x(t), and then you take the natural log of x, ln(x). Remember from differential calculus that

    [tex]\frac{d}{dt}\left(\ln(x)\right) = \frac{d}{dx}\left(\ln(x)\right)\frac{dx}{dt} = \frac{1}{x}\frac{dx}{dt} [/tex]

    where we derived this by 1) remembering the chain rule, and 2) remembering what the derivative of ln(x) is

    This identity is being used in step 1, except that a second application of the chain rule is also required...

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