18.03SC - Part I Problems and Solutions

Part I Problems and Solutions Problem 1: Find the general solution by separation of variables: dy = 2 − y, dx Solution: dy = dx → 2−y y (0) = 0 dy = 2−y − ln |2 − y| = x + c → |2 − y| = Ce−x y = 2 − Ce− x dx → (with C = ec ) → (with C any number) IC: y(0) = 2 − C = 0 → C = 2 → y = 2(1 − e− x ) Problem 2: Find the general solution by separation of variables: ( y − 1)2 dy = dx ( x + 1)2 Solution: dy dx = → ( y − 1)2 ( x + 1)2 − dy = ( y − 1)2 dx → ( x + 1)2 1 1 =− +c x+1 y−1 Extra: solve for y as a function of x: Answer: y = 1 + x +1 1− c ( x +1) Problem 3: The rate of change of a certain population is proportional to the square root of its size. Model this situation with a differential equation. Solution: Let P = P(t) be the size of the population as a function of time t. Then √ dP =k P dt where k > 0 is the constant of proportionality. Problem 4: The rate of change of the velocity of an object is proportional to the square of the velocity. Model this situation with a differential equation. Solution: dv = kv2 dt with k > 0, constant. Problem 5: In a population of fixed size S, the rate of change of the number N of persons who have heard a rumor is proportional to the number of those who have not yet heard it. Model this situation with a differential equation. Solution: dN = k(S − N ) dt with k > 0, constant. Problem 6: The amount of a certain medicine in the bloodstream decays exponentially with a half-life of 5 hours. In order to keep a patient safe during a one-hour procedure, there needs to be at least 50 mg of medicine per kg of body weight. How much medicine should be administered to a 60kg patient at the start of the procedure? Solution: Let x (t) be the amount of the medicine in mg present in the bloodstream at time t in hours. Given information: x (t) = x0 e−kt with k = ln 2 5 1 hr , since half-life is given as 5 hours. Since x (t) = x0 e−kt is decreasing, x (t) ≤ x (1) for t ≥ 1 hrs. So the patient will be (just) safe if x (1) = x0 e−k·1 = 50 · 60 = 3000 mg = 3 g where k = ln52 . Thus, x0 = x (1) · ek = e(ln 2)/5 · 3000 ≈ 3446 mg (or about 3.446 g). Problem 7: Early one morning it starts to snow. At 7AM a snowplow sets off to clear the road. By 8AM, it has gone 2 miles. It takes an additional 2 hours for the plow to go another 2 miles. Let t = 0 when it begins to snow, let x denote the distance traveled by the plow at time t. Assuming the snowplow clears snow at a constant rate in cubic meters/hour: a) Find the DE modeling the value of x. b) When did it start snowing? Solution: a) One approach: let k1 be the rate (in height/hour) of snowfall and k2 be the rate of snow clearance. The height of snow is k1 t → Δx · k1 t ≈ k2 Δt → This is then dx dt Δx Δt ≈ k2 k1 t = kt , where k is a constant. b) Solving by separation of variables, x = k ln t + C. Let t = t1 at 7AM, so t = t1 + 1 at 8AM and t = t1 + 3 at 10AM. 2 miles between 7 and 8AM → 2 = x (t1 + 1) − x (t1 ) = k ln ((t1 + 1) /t1 ) 4 miles between 7 and 10AM → 4 = x (t1 + 3) − x (t1 ) = k ln ((t1 + 3) /t1 ) Thus, ln t1 +3 t1 = 2 ln t1 +1 t1 → t1 +3 t1 = t1 +1 t1 2 After a little algebra, t1 = 1, so the snow started at 6AM Problem 8: A tank holds 100 liters of water which contains 25 grams of salt initially. Pure water then flows into the tank, and salt water flows out of the tank, both at 5 liters/minute. The mixture is kept uniform at all times by stirring. a) Write down the DE with IC for this situation. b) How long will it take until only 1 gram of salt remains in the tank? Solution: minutes. Let x = x (t) be the amount of salt in the tank in grams, with t the time in a) DE: net rate of change of salt. x 0 − 5 · 100 . DE: dx dt = −.05x, dx dt = salt rate in - salt rate out = IC x (0) = 25 b) Solution to DE: x (t) = Ce−.05t . IC x (0) = 25 = c, so x (t) = 25e−.05t . x (t) = 1 when 25e−.05t = 1 → t = ln 25 .05 ≈ 64.38 min