(1 2t), y(1) = 2 y. dy dt = t. e t y, y(0) = 1. dr, r(1) = 2 (r = r(θ)) y = t(t2 + 1) 4y 3, y(0) = 1. 2t y + t 2 y, y(0) = 2. 2t 1 + 2y, y(2) = 0

Transcription

1 MATH 307 Due: Problem 1 Text: Solve the following initial value problems (this problem should mainly be a review of MATH 125). 1. y = (1 2t)y 2, y(0) = 1/ y = (1 2t), y(1) = 2 y = t e t y, y(0) = 1 dr dθ = r2, r(1) = 2 (r = r(θ)) θ y = ty 3 (1 + t 2 ) 1/2, y(0) = 1 y = t(t2 + 1) 4y 3, y(0) = 1 2 y = y = 2t y + t 2 y, y(0) = 2 2t 1 + 2y, y(2) = 0 y = 3t2 e t 2y 5, y(0) = 1 y = e t e t 3 + 4y, y(0) = 1 sin 2t =, y(π/2) = π/3 cos 3y = arcsin t, y(0) = 1 y 2 (1 t 2 ) 1/2 Problem 2 Text: Newton s law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of it surroundings. Suppose that the temperature of a cup of coffee obeys Newton s law of cooling. If the coffee has a temperature of 200 F when freshly poured, and 1 minute later has cooled to 190 F in a room at 70 F, determine when the coffee reaches a temperature of 150 F. 1

2 Let T denote the temperature of our cup of coffee, and we know our surrounding s temperature is 70. Our differential equation is given by dt = k(t 70), where k > 0 is our proportionality constant. Solving this we get that T = Ce kt + 70, and since T (0) = 200, we get that T (t) = 130e kt SinceT (1) = 190, we get that and so Thus our equation is To answer the question so 190 = 130e k + 70, k = ln(12/13). T (t) = 130e t ln(12/13) = 130e t ln(12/13) t = ln(8/13) ln(12/13) 6 minutes. Problem 3 Text: (ish) A ball with a mass 0.2kg is thrown upward with initial velocity 20m/s from the roof of a building 30m high. Assume the gravitation constant g = 10m/s. Consider three different models for the trajectory of the ball: i. One that neglects air resistance (that is, the only force acting on the ball is gravity). ii. One where the force due to air resistance is given by v 30 iii. One where the force due to air resistance is given by v

3 Find the maximum height above the ground (not building) that the ball reaches for (i.) and(ii.). For (iii.) just find the velocity as a function of time. Note that the differential equations in case (i.) and (ii.) were solved in class (except I used arbitrary constants), so lacking some details in these cases is fine. a. We have the initial value problem = 10, v(0) = 20. So v(t) = 10t The height, y, is then given by the initial value problem = 10t + 20, y(0) = 30. Hence y(t) = 5t t Maximum height is given when v = 0, that is, t = 2. Hence y(2) = = 50m. b. We have the initial value problem = 10 1 v, v(0) = Separating and integrating so Integrating again gives 6 ln v = t + C, v = 80e t y(t) = 480e t 6 60t

4 v = 0 when t = 6 ln(4/3). Hence y(6 ln(4/3)) = ln(4/3) 46.43m. c. We have the initial value problem Separating and integrating gives v2 = 10, v(0) = v = 50 arctan ( t5 ) + tan(2/5). Problem 4 Text: (ish) A rocket sled having an initial speed of 150 m/hr is slowed by a channel of water. Assume that during the braking process, the acceleration a is given by a(v) = µv 2, where v is the velocity and µ is some constant. a. As in class, use the relation = v dx to write the equation of motion in terms of v and x. b. If it requires a distance of 2000m to slow the sled to 15m/hr, determine the value of µ. c. Find the time, τ, required to slow the sled to 15m/hr. i. We have and so yielding ii. Integrating gives = µv2, v dx = µv2, dx = µv. v = Ce µx. 4

5 Using the initial condition v(0) = 150 yields the solution v(x) = 150e µx. Since we get µ = 15 = 150e µ2000, ln iii. Going back to the original initial value problem which by integrating yields v = µv 2, v(0) = 150, v(t) = 1 µt + 1, 150 where µ was solved for in part (ii.). Setting v = 15, we get that τ = 120 ln 10 52hr. Problem 5 Text: Some diseases (such as typhoid fever) are spread largely by carriers, individuals who can transmit the disease but who exhibit no overall symptoms. Let x and y denote the proportions of susceptibles and carriers, respectively, in the population. Suppose that the carriers are identified and removed from the population at a rate β, so = βy. (1) Suppose also that the disease spreads at a rate proportional to the product of x and y, thus dx = αxy. (2) a. Determine y and at any time t by solving equation (1) subject to the initial condition y(0) = y 0. b. Use the result from part (a.) to find x at any time t by solving equation (2) subject to the initial conditions x(0) = x 0. c. Find the proportion of the population that escapes the epidemic by finding the limiting value of x as t. 5

6 a. y(t) = y 0 e βt. b. Substituting in part (a.), we get the differential equation Separating and integrating then gives dx = αy 0xe βt. ln x = α β y 0e βt + C. So Using x(0) = x 0 then gives to give c. Taking t gives x = Ce α β y 0e βt. C = x 0 e α β y 0 x(t) = x 0 e α β y 0(1 e βt ). lim t x(t) = x 0e α β y 0. 6