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Math 315 Homework 1 Solutions § 1.3 #26: A ball is dropped from rest at a height of 200 m. What is the velocity and position of the ball 3 seconds later? Solution: We start with x00 (t) = −g where g = 9.8 m/s2 is the acceleration due to gravity. Integrating, we get x0 (t) = −gt + C1 . Since the ball is dropped from rest, the initial velocity is zero, so x0 (0) = 0. So C1 = 0. So x0 (t) = −gt. Integrating again gives 1 x(t) = − gt2 + C2 2 The ball is dropped from a height of 200 m, so x(0) = 200. So C2 = 200. So 1 x(t) = − gt2 + 200. 2 Thus, the velocity of the ball 3 seconds after dropping it is given by 2 2 x0 (3) = (−9.8 m/s )(3s) = −29.4 m/s . The position of the ball 3 seconds after dropping it is given by 1 2 x(3) = − (9.8 m/s )(3s)2 + 200 m = 155.9 m. 2 § 2.1 #8 (a) Use implicit differentiation to show that t2 + y 2 = C 2 implicitly defines solutions of the differential equation t = yy 0 = 0. Solution: Differentiating both sides of the equation t2 + y 2 = C 2 with respect to t gives 2t + 2yy 0 = 0. Dividing both sides by 2, we get t + yy 0 = 0. Hence, t2 + y 2 = C 2 implicitly defines solutions of the differential equation t = yy 0 = 0. (b) Solve t2 + y 2 = C 2 for y in terms of t to provide explicit solutions. Show that these functions are also solutions of t + yy 0 = 0. Solution: Note t2 + y 2 = C 2 ⇒ y=± p C 2 − t2 . Note y 0 (t) = ∓ √ t C2 − t2 . Thus, t + yy 0 = t ± p C 2 − t2 · √ ∓t = 0. C 2 − t2 (c) Discuss the interval of existence for each of the solutions in part (b). √ Solution: The solutions are y = ± C 2 − t2 . Since we cannot take the square root of a negative number, we need t2 ≤ C 2 , which implies that t ∈ [−C, C]. In order that y 0 = √C∓t 2 −t2 is defined, we also can’t have t = ±C. Thus, the interval of existence is (−C, C). (d) Sketch solutions for C = 1, 2, 3, 4. The sketch of solutions for C = 1, 2, 3, 4 should be circles of radius 1, 2, 3, and 4, respectively. § 2.2 #14: Find the exact solution of the initial value problem y 0 = −2t(1 + y 2 )/y, y(0) = 1. Indicate the interval of existence. Solution: Note y0 = − 2t(1 + y 2 ) y ⇒ Z y dy = −2t 1 + y 2 dt ⇒ y dy = − 1 + y2 Z 2tdt Integrating, we get 1 ln(1 + y 2 ) = −t2 + C 2 ⇒ ln(1 + y 2 ) = −2t2 + 2C ⇒ 1 + y 2 = e−2t ⇒ y 2 = Ae−2t − 1 p y(t) = ± Ae−2t2 − 1. ⇒ 2 +2C 2 Now we use the initial condition y(0) = 1, which gives √ 1 = y(0) = ± A − 1 ⇒ 2 = e2C e−2t = Ae−2t A = 2. 2 Also, we must keep only the solution which contains the initial condition y(0) = 1, so we need the + sign in front of the square root. So the exact solution is p y(t) = 2e−2t2 − 1. 2 The square root exists provided that 2e−2t − 1 ≥ 0. Also, we don’t want y = 0 since otherwise the original differential equation y 0 (t) = − 2t(1 + y 2 ) y is not defined. Thus, we need 2 2e−2t − 1 > 0 ⇒ ⇒ ⇒ ⇒ 2 e−2t > 1 2 1 −2t2 > ln( ) = − ln(2) 2 1 2 t < ln(2) 2 r 1 ln(2). |t| < 2 Thus, the interval of convergence is r − 1 ln(2), 2 r 1 ln(2) . 2