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Math Help - Probably a simple problem.

  1. #1
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    Sep 2009
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    Probably a simple problem.

    It has been awhile since my diffy Q classes. This is my problem:

    dx/dt=1/(1+t)

    x=1 at t=0

    I get:

    x=C1*ln(1+t)

    When I use the intial value of t=0, x=0, which does not agree with my initial conditions. Can you tell me what I'm doing wrong. Yes, I know this is incredibly simple, but what can I say....i'm simple.

    Thanks for any help,
    Jason
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  2. #2
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    May 2009
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    \frac{dx}{dt}=\frac{1}{1+t}

    dx=\frac{dt}{1+t} let's integrate both sides

    \int dx=\int \frac{dt}{1+t}

    x=\ln|1+t|+c and use the condition t=0 x=1

    1=\ln|1+0|+c

    1=0+c so c=1

    x=\ln(1+t)+1 we can actually ignore the absolute value condition
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  3. #3
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    Thanks!

    10 years ago I could have done that....unfortunately it seems I learn one new thing and forget an old thing!
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