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Math 2280-001 Quiz 3 SOLUTIONS January 30, 2015 1) Consider the following differential equation for a function x t , which could be modeling a logistic population with harvesting: x# t =Kx2 C 2 x C 3 . 1a) Draw the phase diagram. Identify the equilibrium solutions and whether or not they are stable. (3 points) We look for roots of the slope function - they will be the equilibrium points corresponding to constant (equilibrium) solutions. Then we analyze whether solutions x t are increasing or decreasing between equilibrium points. Factoring... Kx2 C 2 x C 3 =K x2 K 2 x K 3 =K x K 3 x C 1 . So the equilibrium points are x = 3, x =K1. x !K1 0 x# t = K K K = K 0 x t decreasing; K1 ! x ! 3 0 x# t = K K C = C 0 x t increasing; x O 3 0 x# t = K C C = K 0 x t decreaing. So the phase diagram is ))K1//3)) and x hK1 is unstable; x h 3 is asymptotically stable. 1b) Solve the initial value problem for the differential equation above, with x 0 = 1 : x# t =Kx2 C 2 x C 3 . x 0 =1. Hint: To save time with partial fractions on this very brief quiz you may use the identity 1 1 1 1 =K K . xC1 xK3 4 xC1 xK3 (6 points) 1 Note: On the quiz this partial fractions decomposition was wrong: the minus sign in front of the was 4 missing. You will not be marked down if you used the incorrect partial fractions decomposition correctly. separate variables: x# t =K x K 3 x C 1 dx =Kdt xK3 xC1 dx xK3 1 4 K = Kdt xC1 1 1 K xC1 xK3 dx = Kdt 1 1 K dx = 4 dt = 4 t C C xC1 xK3 ln x C 1 K ln x K 3 = 4 t C C xC1 ln = 4 tCC xK3 exponentiate... xC1 = e4 t C C = eCe4 t = C1 e4 t . xK3 by continuity, xC1 = Ce4 t . xK3 x 0 =10 2 = C =K1. K2 xC1 =Ke4 t xK3 x C 1 = x K 3 Ke4 t =Ke4 t x C 3e4 t x 1 C e4 t =K1 C 3e4 t . K1 C 3e4 t x t =K . 1 C e4 t 1c) For your solution x t to (b), verify that t lim x t does agree with the value implied by your phase /N diagram in part (a). (1 point) 4t It's easier to take the limit if we divide the top and bottom by e : eK4 t C 3 x t =K K4 t . e C1 0C3 lim x t = = 3. t /N 0C1 This agrees with what the phase diagram said - for any initial values x0 OK1, t lim x t = 3. /N