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MATH 307 I (Introduction to Differential Equations), Winter 2017 Solution 2 Problem 1. Section 2.3, problems 1, 6(c), 9, 16, 21(a)(b) Sol. 1. Let Q(t) be the amount of dye in the tank at time t. Clearly Q(0) = 200g. The differential equation governing the amount of dye is Q0 (t) = −2Q(t)/200, The solution of this separable equation is Q(t) = Q(0)e−t/100 = 200e−t/100 . We need the time T such that Q(T ) = 2g. This means we have to solve 2 = 200e−T /100 and we obtain T = −100 ln(1/100) = 100 ln(100) ≈ 460.5 min. 6.(c) With A(h) = π, a = 0.01π, α = 0.6, the differential equation for the water level h is p dh = −0.006π 2gh, π dt p with solution h(t) = 0.000018gt2 − 0.006 2gh(0)t + h(0). Setting h(0) = 3 and g = 9.8, h(t) = 0.0001764t2 − 0/046t + 3, resulting in h(t) = 0 for t ≈ 130.4s. 9. Let S(t) be the amount of the loan remaining at time t. Then dS/dt = 0.1S −k, S(0) = $8, 000. Solving for S(t) gives S(t) = 8000e0.1t −10k(e0.1t − 1). Setting S = 0 and substituting t = 3 gives k = $3, 086.64 per year. For 3 years this totals $9, 259.92, so $1, 259.92 has been paid in interest. 16. If T is the temperature of the coffee at any time t, then dT /dt = −k(T − 70), T (0) = 200, TR(1) = 190. Then dT /dt + kT = 70k. Using integrating factors, µ(t) = e kdt = ekt , and the integrating equations is d (T ekt ) = 70kekt . dt Integrating, we obtain T ekt = 70ekt + C, so T (t) = 70 + Ce−kt . Since T (0) = 200, then C = 130 and T (t) = 70 + 130e−kt . To find k we use the fact that T (1) = 190. Then 190 = 70 + 130e−k , so −k = ln(12/13) and T (t) = 70 + 130eln(12/13)t . We want to find t such that 150 = T (t) = 70 + 130eln(12/13)t . Solving for t we find t = ln(8/13)/ ln(12/13). 21.(a) The differential equation for the motion is mdv/dt = −v/30 − mg. Given the initial condition v(0) = 20 m/s, the solution is v(t) = −44.1 + 64.1e−t/4.5 . Letting v(t0 ) = 0, the ball reaches the maximum height at 1 2 t0 = 1.683 s. Integrating v(t), the position is given by x(t) = 318.45 − 44.1t − 288.45e−t/4.5 . Hence the max height is x(t0 ) = 45.78. 21.(b) Setting x(t1 ) = 0, the ball hits the ground at t1 = 5.128s. Problem 2. A reservoir on a farm initially contains 10000 liters of water, in which 200kg nitrate fertilizer is dissolved. The owner of the reservoir decides that the amount of dissolved nitrate needs to be increased, and therefore starts pumping in a 1 kg nitrate per liter water solution at a rate of 100 liters per minute. However, the reservoir simultaneously develops a leak, and starts draining at a rate of 200 liters per minute. (a) Assuming the solution remains perfectly mixed at all times, set up a differential equation for y(t), the mass of nitrate in the reservoir at time t. Sol. Let V (t) be the volume of nitrate water in the reservoir at time t. Note that V (t) = 10000 + (100 − 200)t = 10000 − 100t. The differential equation for y is dy y y = 1(100) − 200 = 100 − 200 , dt V 10000 − 100t or, in standard form, 2 dy + y = 100. dt 100 − t (b) Find the solution y(t). Sol. This a linear first-order ODE, and the integrating factor is R 1 e 2/(100−t) dt = . (100 − t)2 The integrating factor equation becomes y 100 d = 2 dt (100 − t) (100 − t)2 and integrating with respect to t and simplifying gives y = 100(100 − t) + C(100 − t)2 . The initial condition y(0) = 200 becomes 200 = 100(100) + C(100)2 so that C = −0.98, and we get y = 100(100 − t) − 0.98(100 − t)2 = −0.98t2 + 96t + 200 3 (c) What is the maximum amount of nitrate in the reservoir, and when does it occur? The maximal value of the function y is found when y 0 = 0, which gives t ≈ 48.9796. Problem 3. The corpse of King Joffrey is discovered in a room at midnight and its temperature is 80◦ F, while the temperature of the room is kept constant at 70◦ F. Two hours later, the temperature of the corpse dropped to 75◦ F. Joffrey’s mother, Cersei, asks you to investigate the crime scene. Even if you don’t like Joffrey and Cersei (who does?), you accept. Assuming that the temperature of King Joffrey’s body just before dying was a healthy 98.6◦ F, can you find the time of death? Sol. Let T (t) denote the temperature of Joffrey’s corpse t hours from midnight, so that T (0) = 80 and T (2) = 75. Newton’s law of cooling tells us that dT = k(T − 70), dt a separable equation whose solution is T = 70 + Cekt . Using T (0) = 80 gives C = 10, so that T = 70 + 10ekt . Using T (2) = 75 then gives k = 0.5 ln 0.5, so that T = 70 + 10e0.5t ln 0.5 . We want t so that T = 98.6, i.e. 98.6 = 70 + 10e0.5t ln 0.5 . Solving for t gives t ≈ −3.0320, so that Joffrey died approximately three hours before midnight.