Integrating we nd e so t20 y t20 7300e y c 7300

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = +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 Solutions 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⇡ t)] + Ce )e t/10 . + 400⇡ ⇡ 1 + 400⇡ 1 + 400⇡ 2 Longterm behaviour: as t ! 1, the last term ! 0, thus 50 50 IC: p(0) = p0 ) p0 = 150 1+400 2 + 50 ) C = p0 150 + 1+400 2 C p(t) ! 150 50 [cos(2⇡ t) + 20⇡ sin(2⇡ t)] . 50 1 + 400⇡ 2 ⇡ t) + 20⇡ sin(2⇡ t)]+...
View Full Document

Ask a homework question - tutors are online