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
$\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