**1.**-
- (a)
- Just integrate to obtain and use the
initial data to find
**C = 3**. - (b)
- Separate and integrate (using partial fractions) to
obtain
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**.

**2.**-
- (a)
- The balance law behind the equation is
Rearrange, divide by and let go to zero to obtain

The growth rate is equal to divided by the doubling time, so .

- (b)
- The maximum (constant) harvest rate that does not wipe out the population
is 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.
- (c)
- Just let the yeast grow for eight hours and then harvest it all at once. You have kg per day, which is a much larger yield per day than the maximum for constant harvesting.

**3.**- 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**.

**4.**-
- (a)
- The characteristic equation is
**r + 3 = 0**so a homogeneous solution is . - (b)
- Look for a particular solution of the form
. 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**. - (c)
- Solve
and obtain
**C= 7**, hence . - (d)
- The exponential term is transient and so the solution looks
like the particular solution for large
**t**.

**5.**-
- (a)
- Solve to obtain two steady states
**y=1**and**y=3**; these are constant solutions for the differential equation. - (b)
- 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. - (c)
- The sketch indicates that
**y = 1**is stable because trajectories which start nearby converge to**y=1**. The steady state**y=3**is unstable; trajectories starting at any**y> 3**increase and do not stay close to**3**. - (d)
- Use with ,
and . You obtain
(whatever units you want).

These solutions are provided without warranty, implied or otherwise. Use at your own risk.

© 1996 by Will Brother. All rights Reserved. File last modified on March 22, 1996.