# Thread: Hard Mathematics C Questions

1. ## 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

2. Originally Posted by Ned Hunter
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:

$\displaystyle \frac{dN}{dt}=0.0004~N~(1000-N),\ \ N(0)=1$

This is of variables seperable type, so:

$\displaystyle \int \frac{1}{N~(1000-N)}~dN = \int 0.0004~dt = 0.0004t+C$

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

RonL

3. 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

4. Originally Posted by Ned Hunter
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
$\displaystyle \frac{1}{N(1000-N)}=\frac{A}{N}+ \frac{B}{1000-N}=\frac{A(1000-N)+BN}{N(1000-N)}$

So $\displaystyle A=B$ and $\displaystyle A=0.001$

Hence:

$\displaystyle \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]$$\displaystyle =0.001 \ln \left(\frac{N}{1000-N}\right)$

So:

$\displaystyle 0.001 \ln \left(\frac{N}{1000-N}\right)=0.0004 t +C$

Now rearrange:

$\displaystyle \left(\frac{N}{1000-N}\right)=A \exp(0.4 t)$

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

RonL