AMath350.F13.A3.Sol

# 05y 365 dt the initial condition is y 0 1000 the de is

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Unformatted text preview: on is y (0) = 1000. The DE is linear, and an integrating factor is µ=e R 1 dt 20 t/20 =e . Factoring this into the equation, we obtain e t/20 dy 1 e 20 dt t/20 y = 365e t/20 which must be equivalent to d e dt t/20 = 365e t/20 (and we can quickly check that it is, so we have calculated our integrating factor correctly). Integrating, we ﬁnd e so t/20 y= t/20 7300e y= +C 7300 + Cet/20 . Apply the initial condition: =) y (0) = 1000 =) =) 1000 = 7300 + C C = 8300 y = 8300et/20 7300. Therefore, after 10 years, the account balance should be approximately y (10) = 8300e1/2 = \$6384.39. 7300 AMATH 350 Page 5 Assignment #3 - Fall 2013 4. (Consider a commodity (such as wheat) where the supply ﬂuctuates on an annual basis, i.e., S = 5(1 + cos(2⇡ t)), where t is measured in years. Suppose that the rate of increase of price is equal to the rate of decrease of quantity and the initial price is p0 . ) a) (If the demand for this commodity is given by D = 20 p/10, ﬁnd the price as a function of time. What is the long term behaviour of the price? How does this depend on p0 ? Are there any equilibrium solutions? Use Maple or Matlab to sketch the solution for diﬀerent values of p0 . ) Solution: dp = D S = 20 p/10 5(1 + cos(2⇡ t)), p(0) = p0 . dt 1 dp + p = 15 5 cos(2⇡ t). Linear DE with standard form: dt 10 Integrating factor: e 1/10 dt = et/10 dp = D S = 20 p/10 5(1 + cos(2⇡ t)), p(0) = p0 . 5. (a) dt dp et/10 t/10 1 et/10 + = 15edp 5et/10 cos(2⇡ t) + p = 15 5 cos(2⇡ t). Linear DE with standard form: dt 10p dt 10 d e1 Integrating factor: t/10 p/10 dt =15t/1010 5et/10 cos(2⇡ t) [e = e et/ dt t/10 e t/10 dp Integrate by parts twice e + = 15et/10 5et/10 cos(2⇡ t) on RHS (details below) dt 10p 50 t/10 et/10 et/10 cos(2⇡ t) + 20⇡ et/10 sin(2⇡ t) + C d e 10 p = 150t/10 t/ 1t/10400⇡ 2⇡ t) + cos(2 [e p] = 15e 5e dt 50 t [cos(2⇡ t + RHS (details below)/10 General Solution: p(t) = 150 Integrate 400parts twice )on 20⇡ sin(2⇡ t)] + Ce 1 + by ⇡ 2 50 et/10 p = 150et/10 et/10 cos(2⇡50) + 20⇡ et/10 sin(2⇡ t) + C t 50 IC: p(0) = p0 ) p0 = 150 1+400 2 + C 1 +C =⇡ 20 150 + 1+400 2 ) 400 p 50 50 50 [cos(2⇡ t + 20⇡ sin(2 0 150+ t/10 General Solution: (t = Solution of IVP: p(t)p= )150 150 1 2 [cos(22 t) + 20⇡)sin(2⇡ t)]+(p⇡...
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## This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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