hw1solns

# hw1solns - will choose the first option and convert...

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UBC Math 215/255, Fall 2009, Solutions to assignment 1 1.1 (page 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 (page 24): 17 2.1 (page 39): 14, 16, 31

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2.2 (page 47): 4, 9, 22
2.3 (page 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

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Unformatted text preview: 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 (page 88): 15...
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## This note was uploaded on 01/29/2011 for the course MATH 215 taught by Professor Keqinliu during the Fall '08 term at UBC.

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hw1solns - will choose the first option and convert...

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