and use the
initial data to find C = 3.
Rearrange to find
and use the initial data to find 
.
Solve finally for y,
and note that this solution is valid only for 
if
you start at t = 0.
Rearrange, divide by 
and let 
go to zero to obtain
The growth rate is equal to
divided by the doubling time, so 
.
where 
is the initial population.
You can get this from the solution for the initial value problem or
just by setting the righthand side of the differential
equation equal to zero. You have 
, so the
population does not decay, if and only
if 
. If, on the other hand,
, then 
for
all time and the yeast population will reach zero.
The amount harvested in a day with 
kg and 
is then 
, which is approximately 
kg per day.
kg per day,
which is a much larger yield per day than the maximum for
constant harvesting.
is a homogeneous solution and 
is a particular
solution so, because the equation is linear, the general solution
is 
. The initial data determine C
from 
, or C= 3.
.
. Plug it in to obtain
the equations 2A + 3B = 13 and 3A - 2B = 0 for A and B.
The solutions are A=2 and B=3.
and obtain C= 7, hence
.
for large t.
to obtain two steady states
y=1 and y=3; these are constant solutions for the differential
equation.
is positive and so y is increasing for y<1
and y > 3. 
is negative and so y is decreasing
for 1 < y < 3.
The solution starting at 0 increases and approaches y=1.
The solution starting at 1 remains y=1.
The solution starting at 5 increases and does not approach anything.
with 
,
and 
. You obtain
(whatever units you want).
© 1996 by Will Brother. All rights Reserved. File last modified on March 22, 1996.