(due: April 15 in class)

1) Show that the following initial value problems are well posed:

a) dy/dt = y cos(t); 0 <= t <= 1; Y(0)=1

b) dy/dt = gy; 0<= t<= t_{Max}; y(0)=y_{0}

c) dy/dt = 4t^{3}y/(1+t^{4}); 0 <= t <= 1; y(0)=1

**Bonus Problem** (10% extra)

Give a Proof for Theorem 5.10 in Burdon and Faires

(the proof is similar to the one for Theorem 5.9).

Population growth is described by an ODE of the form dN/dt = gN, where g is the growth rate (this was discussed
in class). In a typical population, the growth rate is not a constant, but is density dependent. For example, as
the population grows, there might be less food available, and as a result the growth rate decreases. A simple form
to describe this is the following equation:

dN/dt = (g - bN) N; 0 <= t <= 50;
N(0) = N_{0}

which is also called the Logistic Eqation.

The analytic solution is given by N(t) = [g/b]/[1+((g-b*N_{0})/(b*N_{0}))*exp(-g*t)].

Solve this equation numerically using Euler's method. Assume the following values: N_{0}=1000, g=0.2, b=0.00005.
Solve this problem for 0 <= t <= 50. Choose the timestep h=10, 5, 1, 0.5, 0.1.

You may use the code that has been discussed in class and modify it.

a) Compare the solutions to the analytic solution.

b) Discuss the behaviour of the solutions as a function of h. What happens for very large timesteps ?

c) Can you interpret the asymptotic behaviour of the solution in biological terms ? What happens if you change
b to b=0.0002 ? Can you expplain this change ? (it might be useful to think in terms of the total, effective growth
rate!)

You should turn in plots of population vs. time, and also tables that show the difference between the numerical and the exact solution.