AMath350.F13.A3.Sol

400 1 400 1 400 2 longterm behaviour as t 1 the last

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Unformatted text preview: 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)]+(p 150+ Solution of IVP: p(t) = 150 [cos(2 )e t/10 . 0 2 1 + 400⇡ equilibrium solutions. 1 + 400⇡ 2 By inspection of the DE there are no Longterm behaviour: as t ! 1, the last term ! 0, thus 50 p(t) ! 150 [cos(2⇡ t) + 20⇡ sin(2⇡ t)] . 1 + 400⇡ 2 By inspection of the DE there are no equilibrium solutions. 5. (a) Integration by parts: let u = cos(2⇡ t) and dv = et/10 : Z et/10 cos(2⇡ t) dt = 10et/10 cos(2⇡ t) + 20⇡ Z et/10 sin(2⇡ t) dt = 10et/10 cos(2⇡ t) + 200⇡ et/10 sin(2⇡ t) Integration by parts: let u = cos(2⇡ t) and dv = et/10 : Z Z 400⇡ 2 Z et/10 cos(2⇡ t) dt t/10 t/ cos(2 20⇡ et/ e sin(2⇡ t) dt Thus t/10t/10 cos(2⇡ t) dt = 10e10 102cos(2⇡ t) + ⇡ t) + 20⇡10t/10 sin(2⇡ t)]. e e cos(2⇡ t) dt = 1+400 [e dp Z (b) (Suppose S = (10 p)(1 + cos(2⇡ t)) 5(1 + cos(2⇡an = (5 p)(1 +2cos(2⇡ tD p(0) = p0 . = D that the demand 10 b) t/ also fluctuates on10 )) annual basis, i.e., ), = t/ t et/10 cos(2⇡ t) dt = 10e cos(2⇡ t) + 200⇡ e sin(2⇡ t) 400⇡ dt (10 be )(1 + cos(2⇡ t))or Repeat part (a) with this demand function.) Can psolved as linear . separable DE. Linear approach is shown here. 10 Thus et/10 cos(2⇡ t) dt = 1+400 2 [et/10 cos(2⇡ t) + 20⇡ et/10 sin(2⇡ t)]. Solution: 3 dp (b) = D S = (10 p)(1 + cos(2⇡ t)) 5(1 + cos(2⇡ t)) = (5 p)(1 + cos(2⇡ t), p(0) = p0 . dt Can be solved as linear or separable DE. Linear approach is shown here. 3 AMATH 350 Page 6 Assignment #3 - Fall 2013 dp + (1 + cos(2⇡ t))p = 5(1 + cos(2⇡ t)). dt 1 Integrating Factor: e 1+cos(2 t) dt = et+ 2 sin(2 t) . Standard form: 1 et+ 2 sin(2 t) dp dt 1 + (1 + cos(2⇡ t))et+ 2 sin(2 t) 1 p = 5(1 + cos(2⇡ t))et+ 2 1 d t+ 1 sin(2 t) e2 p = 5(1 + cos(2⇡ t))et+ 2 dt 1 1 et+ 2 sin(2 t) p = 5et+ 2 sin(2...
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