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Math Help - Hard Mathematics C Questions

  1. #1
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    Hard Mathematics C Questions

    Hey,
    This is an advanced mathematics C problem in Australia. Anyone who can solve the attached question is a GOD!!!!!!! Please gimme a hand, I cant quite get it worked out.

    Cheers,
    Ned
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ned Hunter View Post
    Hey,
    This is an advanced mathematics C problem in Australia. Anyone who can solve the attached question is a GOD!!!!!!! Please gimme a hand, I cant quite get it worked out.

    Cheers,
    Ned

    You have the following initial value problem to solve:

    <br />
\frac{dN}{dt}=0.0004~N~(1000-N),\ \ N(0)=1<br />

    This is of variables seperable type, so:

    <br />
\int \frac{1}{N~(1000-N)}~dN = \int 0.0004~dt = 0.0004t+C<br />

    The integral on the left can be done by using partial fractions.

    RonL
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  3. #3
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    Hey,
    do you think you can show me how to do that? i cannot remember for the life of me and my textbook is very sketchy on it. that would be a great help.

    Cheers
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Ned Hunter View Post
    Hey,
    do you think you can show me how to do that? i cannot remember for the life of me and my textbook is very sketchy on it. that would be a great help.

    Cheers
    <br />
\frac{1}{N(1000-N)}=\frac{A}{N}+ \frac{B}{1000-N}=\frac{A(1000-N)+BN}{N(1000-N)}<br />

    So A=B and A=0.001

    Hence:

    <br />
\int \frac{1}{N(1000-N)}~dN=\int \frac{0.001}{N}+ \frac{0.001}{1000-N}~dN=0.001 \left[ \ln(N) - \ln(1000-N) \right] =0.001 \ln \left(\frac{N}{1000-N}\right)<br />

    So:

    <br />
0.001 \ln \left(\frac{N}{1000-N}\right)=0.0004 t +C<br />

    Now rearrange:

    <br />
\left(\frac{N}{1000-N}\right)=A \exp(0.4 t)<br />

    Now change the subject to get this in the form: N=f(t) and
    choose the constant so that the initial condition is satisfied.

    RonL
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