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AMath350.W13.A3.Sol

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

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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...
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