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Differential Equations Problem

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    This is actually a question we went over in class, but I kind of spaced out when my teacher was explaining it. I have since solved it myself while I was reviewing for my calculus test, but when I compared my answer with the answer key my teacher provided to our class, I found out my answer was wrong. Here's the question.

    "Write and solve the differential equation that models the statement in the following problem.

    The rate of change of P with respect to t is proportional to 10-t."

    2. Relevant equations
    I should add that this is part of our chapter involving exponential growth and decay.

    3. The attempt at a solution
    Now, I understand how to do this particular problem. You essentially start off with the following differential equation:

    dP/dt = K/(10-t)

    You then multiply the dt over to isolate all the t variable expressions on the right side.

    dP = [K/(10-t)]dt

    Then you would take an indefinite integral of both sides respectively. The problem is, my teacher says that the answer should be:

    P = -K*ln|10-t| + C

    When I solved this on my own, I produced the same result essentially, just without a negative sign in front of the K constant. Am I missing something here? Could someone please explain to me why there should be a negative sign or did my teacher make a mistake?
  2. jcsd
  3. Mar 17, 2013 #2


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    Did you integrate the right side using u-substitution? After you use that method, the negative will appear in the answer.
  4. Mar 17, 2013 #3
    Ah, no. I didn't. I tried integrating using substitution just now, and it worked. >.<
    Thank you for pointing that out to me. It makes sense now.
  5. Mar 18, 2013 #4


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    When you say ##y## is proportional to ##x## that means there is a constant ##k## such that ##y=kx##. So if the rate of change of ##P## is proportional to ##10-t##, then ##\frac{dP}{dt}= k(10-t)##. So you either are solving the problem wrong or stated the problem you are solving wrong.
    Last edited: Mar 18, 2013
  6. Mar 18, 2013 #5
    That's my mistake. I stated the problem incorrectly. It should say "inversely proportional". But regardless, eumyang pointed out what I was missing. I didn't use the substitution method to integrate the differential equation, so I never produced that negative sign.
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