UBC Math 215/255, Fall 2009, Solutions to assignment 1
1.1 (pa g e 7): 6
For y > -2, the slopes are positive, and hence the solutions increase. For y < -2,
the slopes are negative, and hence the solutions decrease. All solutions diverge away
from the equilibrium solution y(t) = -2.
1.3 (pa g e 24): 17
2.1 (pa g e 39): 14, 16, 31
2.2 (pa g e 47): 4, 9, 22
2.3 (pa g e 59): 4, 10, 24
10.
Since we are assuming continuity, either convert the monthly payment into an annual
payment or convert the yearly interest rate into a monthly interest rate for 240 months. We
will choose the first option and convert everything yearly. Let S(t) be the amount of the load
remaining at year t, then dS/dt = rS – k, where r=0.09 and k=800x12 = 9600 is the
maximal annual payment. The equation can be rewritten as (S exp(-rt))' = - k exp(-rt).
Integrating,
S(t) exp(-rt) - S(0) = (k/r) [ exp (-rt) - 1]
We want S(20)=0 when S(0) is the maximal amount this buyer can afford to borrow. Thus
S(0) = (k/r)[1-exp(-20r)] = $89,034.79.
The total interest paid is k x 20 – S(0) = $102,965.21
2.5 (pa g e 88): 15