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Homework Help: Another Diff Eq's one: am I wrong?

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data

    Sorry to be bringing these in quick succession, but this one really has me perplexed. Is it possible that both the solution manual and I have two different but valid answers?

    I don't want to just go assuming that I'm right...but I think by subsuming certain parts of the solutions manuals constants in it's solution into constants in my solution, that the two can become equivalent...?!

    The problem


    Imgur link to the problem: http://i.imgur.com/c3PgDgU.png

    2. Relevant equations

    3. The attempt at a solution

    [tex]\int \frac{1}{r-kC} dC = \int dt[/tex]

    [tex]\frac{-1}{k} \ln{|r-kC|} = t+K_0[/tex]

    [tex]\ln{|r-kC|} = -kt + K_1[/tex]

    skipping a couple of steps

    [tex]C(t) = \frac{r-Ae^{-kt}}{k}[/tex]

    with init. conditions

    [tex]C(0) = C_0 = \frac{r-A}{k}[/tex]

    [tex]A = r - kC_0[/tex]


    [tex]C(t) = \frac{r - (r-kC_0)e^{-kt}}{k}[/tex]

    [tex]C(t) = C_0 e^{-kt}[/tex]

    The answer in the back is

    [tex]C(t) = \left(C_0 - \frac{r}{k} \right)e^{-kt} + \frac{r}{k}[/tex]

    And the solution given in the solution manual is


    Link to solution: http://i.imgur.com/RVrLe1R.png

    I can see that they've don'e some minor things different algebraically and with subsuming certain constants into other constants...I feel like my solution should be equivalent though. Is it? If it is, are there any particular advantages to the form the solution manual has given it in? Be they practical advantages or purely mathematical advantages?
  2. jcsd
  3. Jul 18, 2015 #2
    You can always check by plugging it back in.
  4. Jul 18, 2015 #3

    [tex]C_0e^{-kt} = \left(C_0 - \frac{r}{k}\right)e^{-kt}+\frac{r}{k}[/tex]

    [tex]0 = \frac{r}{k} - \frac{r}{k}e^{-kt}[/tex]

    [tex]0 = \frac{r}{k}\left(1 - e^{-kt}\right)[/tex]

    So, either r = 0 or [itex]e^{-kt} = 1[/itex]

    That doesn't seem right.
  5. Jul 18, 2015 #4
    I meant double check each in the original diff eq.

    Then double check boundary conditions.
  6. Jul 18, 2015 #5


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    That step is clearly wrong.
  7. Jul 19, 2015 #6
    Thank you. Whenever I do study after a night shift I make stupid mistakes. Cheers :)
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