2 with2013 y assignment 3 winter f x a the 1 dv t

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: 2.2 with2013 y ) = Assignment #3 Winter f (x, (a) The 1 dV t + 20 dt 1 V (t + 20)2 1 d V dt t + 20 1 V t + 20 General solution: V y (1+y ) , x 1 t + 20 1 =k t + 20 =k = k ln(|t + 20|) + C = k (t + 20) ln |t + 20| + C (t + 20) Apply IC: V (0) = V0 ) 20k ln(20) + 20C = V0 ) C = V0 /20 k ln(20) V0 k ln(20) (t + 20) Solution of IVP: V (t) = k (t + 20) ln(t + 20) + 20 Interval of existence: t > 20. Given that V0 = 1000, need to find k so that V (45) = 500000. 500000 = k (65) ln(65) + (50 Thus k = 500000/65 50 ln(65) ln(20) k ln(20))65 6483.92. So the person must contribute $6483.92 per year. 2 AMATH 350 Page 6 Assignment #3 - Winter 2013 5. (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 I...
View Full Document

This document was uploaded on 02/09/2014.

Ask a homework question - tutors are online