MATH 3230 Test #2 Study Guide: Solutions for Diff. Equations & Motion under Forces, Exams of Differential Equations

Download MATH 3230 Test #2 Study Guide: Solutions for Diff. Equations & Motion under Forces and more Exams Differential Equations in PDF only on Docsity! Study Guide for Test # 2 Maymeskul MATH 3230 Sections 3.2, 3.4, 4.2−4.6 Show your work clearly to get credit. No graphing calculators! 1. (15 pts) A brine solution of salt flows at a constant rate of 4 L/min into a large tank that initially held 100 L of pure water. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 3 L/min. If the concentration of salt in the brine entering the tank is 0.2 kg/L, determine the concentration of salt in the tank after t min. What is the limiting concentration as t → ∞? Solution. Let x(t) denote the amount of salt in the tank at time t. We use rate of change = input rate – output rate. In this problem, the input rate is (4 L/min)(0.2 kg/L) = 0.8 kg/min. The volume of the solution in the tank at time t is 100+(4−3)t = 100+t (each minute 4 L go in, but only 3 L go out). Therefore, the output rate is (3 L/min)[x(t) kg/(100+t) L] = 3x(t)/(100+t) kg/min, and the equation becomes x′ = 0.8 − 3x 100 + t ⇒ x′ + 3x 100 + t = 0.8 . This is a linear equation in standard form. We find µ(t) = e ∫ [3/(100+t)]dt = e3 ln(100+t) = (100 + t)3 ⇒ x(t)(100 + t)3 = ∫ 0.8(100 + t)3dt = 0.2(100 + t)4 + C. For C, use the initial condition x(0) = 0 (pure water initially). 0 = 0.2(100)4 + C ⇒ C = −2(10)7 ⇒ x(t) = 0.2(100 + t) − 2(10)7(100 + t)−3. Dividing this amount of salt by the volume at time t, we get the concentration c(t). c(t) = 0.2 − 2(10)7(100 + t)−4. Finally, c(t) → 0.2 as t → ∞ (same as the concentration of the incoming flow, of course). 2. (15 pts) An object of mass 4 kg is given an upward initial velocity of 10 m/sec and then allowed to fall under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity, with the proportionality constant b = 8 N-sec/m. Determine the equation of motion of the object. If the object is initially 100 m above the ground, then (a) find its maximal height and (b) determine when it will strike the ground. Solution. The basic formula in this problem comes from the second Newton’s law: m dv dt = Fg − Fr , where Fg = mg (with positive downward direction) is the force due to gravity and Fr = bv is the resistance force. Thus we have m dv dt = mg − bv ⇒ dv dt = g − bv/m ⇒ dv dt = g − 2v. This is a separable (also linear) equation. Separating variables and integrating, we conclude that ∫ dv 2v − g = − ∫ dt ⇒ 1 2 ln |2v − g| = −t + C ⇒ v = g 2 + C1e −2t. From the initial condition, v(0) = −10, we find C (taking g ≈ 9.81). −10 = g 2 + C1 ⇒ C1 = g 2 − 10 = −14.905 ⇒ v = 4.905 − 14.905e−2t. Therefore, the position function x(t) is given by x(t) = ∫ v(t)dt = 4.905t + 7.4525e−2t + C. Since x(0) = 0 (with the origin at the initial position), we have 0 = 7.4525 + C ⇒ C = −7.4525 ⇒ x(t) = 4.905t + 7.4525e−2t − 7.4525 . For (a), at the highest point the velocity is zero. So, v(t) = 4.905 − 14.905e−2t = 0 ⇒ t = ln(14.905/4.905) 2 ≈ 0.5557 x(0.5557) = (4.905)(0.5557) + (4.905)(0.5) − 7.4525 ≈ −2.27 , and the maximal height is 100 − (−2.27) = 102.27 m. To answer (b), we solve the equation x(t) = 4.905t + 7.4525e−2t − 7.4525 = 100 ⇒ t ≈ 107.4525 4.905 ≈ 21.9 (sec). 2 Dividing these equations by e3t, we find that{ v′1 + tv ′ 2 = 0 3v′1 + (1 + 3t)v ′ 2 = t −1. We now subtract three times the first equation from the second one to get v′2 = t −1 ⇒ v2 = ∫ t−1dt = ln |t|. From the first equation, v′1 = −tv′2 = −1 ⇒ v1 = ∫ (−1)dt = −t. So, yp = v1y1 + v2y2 = −te3t + te3t ln |t| (or just te3t ln |t| since the first function equals −y2 and it is a solution to the corresponding homogeneous equation). EC. (10 pts) According to the Hooke’s law, the force Fs acting on an object due to a spring has magnitude proportional to the displacement of the spring from its natural length and has direction opposite to the displacement. A spring has one end attached to the ceiling and a mass m = 1 kg on its other end. The mass then displaced downward by 0.2 m and released. Assuming that the spring constant k = 20 N/m and the air resistance force Fr is proportional to the mass’ velocity with the proportionality constant b = 4 N-sec/m, find an equation of the motion. (Neglect the weight of the spring.) Solution. Let x(t) denote the position function of the mass with positive downward direction and the origin, x = 0, at the point corresponding to the other end of the spring in its natural length (i.e., without the mass attached). If Fg denotes the force due to gravity, then, according to the second Newton’s law, mx′′ = Fg − Fr − Fs = mg − bx′ − kx. With the given data, the equation becomes x′′ = g − 4x′ − 20x ⇒ x′′ + 4x′ + 20x = g. This equation is a second order linear equation with constant coefficients. Its characteristic equation, r2+4r+20 = 0 has roots r = −2±4i. A particular solution to this nonhomogeneous equation has the form xp = A. So, x ′ p = x ′′ p = 0. The substitution yields 20A = g ⇒ A = g 20 , 5 and a general solution to the equation is given by x(t) = g 20 + e−2t (C1 cos 4t + C2 sin 4t) . We note that the value of the position function, when the system is in its equilibrium position, is exactly g/20. This follows from the first Newton’s law (Fg = Fs) or from the equation of the motion since x′′ = x′ = 0 there. To find C1 and C2 we now solve the system{ e−2t (C1 cos 4t + C2 sin 4t) |t=0 = 0.2 [e−2t (C1 cos 4t + C2 sin 4t)] ′ |t=0 = 0 ⇒ { C1 = 0.2 −2C1 + 4C2 = 0 ⇒ C2 = C1/2 = 0.1 . Therefore, the equation of the motion is x(t) = g 20 + 0.1e−2t (2 cos 4t + sin 4t) . 6