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Differential equation: word problem

  1. Feb 12, 2013 #1
    1. In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of δ-glucono lactone into gluconic acid, for example, dy/dt= -.6y when t is measured in hours.

    If there are 90 grams of δ-glucono lactone present when t=0, how many grams will be left after the first hour?
    Answer = ___grams

    2. Alright, I thought this might be worked like the other differential equation problems.

    We are given: dy/dt=-.6y, so I think I have to separate the x's and y's and then integrate.

    dy = -.6ydt

    ∫dy/(-.6y) = ∫dt

    ∫(1/y)(-1/6)dy = t

    ln|y|*-1/6 = t +C

    ...Is this right so far?

    I'm not really sure at all how I would finish the problem, but I will try what I think might be right anyway...

    So, would you substitute t=0 and 90 for x?

    ln|90|*-1/6 = 0 + C

    C = -.75

    ln|y| = -6(t + C)

    eln|y| = e-6(t+C)

    y= e-6(t + C)

    So we want to figure out the grams after 60min?

    y= e-6( 60 + -.75)

    y= 0

    ...Yeah, I don't know if I'm doing this right at all. :/ But I thought I might try!
    Thank you so much for helping! :)
  2. jcsd
  3. Feb 12, 2013 #2


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    Staff: Mentor

    Not that I checked your work thoroughly, but just from skimming:

    - 60 min is 1 hour and 0.6 is given for time unit equal 1 hour.

    - how did you got 1/6 from 0.6?

    Besides, I would integrate [itex]\int \frac {dy}{y} = 0.6\int dt[/itex], makes the rest of the math less clumsy (in the end you moved 6 back to the RHS).

    Edit: typo in the formula corrected.
    Last edited: Feb 12, 2013
  4. Feb 12, 2013 #3


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    Science Advisor

    It sould be simpler to leave the "-.6" on the right: dy/y= -.6.
    ln(y)= -.6t+ C, y= C'e^{-.6t}.

    You have dropped the decimal point: -1/.6, not -1/6.

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