MATH 215/255 Fall 2014 Assignment 4 §2.2, §2.4, §2.5 Solutions to selected exercises can be found in [Lebl], starting from page 303. • 2.2.9: Solve y + 9y = 0 for y(0) = 1, y (0) = 1. Answer. The characteristic equation of y + 9y = 0 is: r2 + 9r = 0. Solve r2 + 9r = 0, we have r1 = 0, and r2 = −9. So the general solution to y + 9y = 0 is: y(x) = C1 + C2 e−9x . Then y (x) = −9C2 e−9x . Since y(0) = 1 and y (0) = 1, then 1 = C1 + C2 , So C1 = 10 , 9 and − 9C2 = 1. 1 and C2 = − . 9 So the solution to y + 9y = 0 for y(0) = 1, y (0) = 1 is: y(x) = 10 e−9x − . 9 9 • 2.2.102: Find the general solution to y − 6y + 9y = 0. Answer. The characteristic equation of y − 6y + 9y = 0 is: r2 − 6r + 9 = 0. Solve r2 − 6r + 9 = 0, we have r1 = r2 = 3. So the general solution to y − 6y + 9y = 0 is: y(x) = C1 e3x + C2 xe3x . • 2.2.105: Find the solution to z (t) = −2z (t) − 2z(t), z(0) = 2, z (0) = −2. Answer. Rewrite the equation: z + 2z + 2z = 0. The characteristic equation of z + 2z + 2z = 0 is: r2 + 2r + 2 = 0. Solve r2 + 2r + 2 = 0, we have r1 = −1 + i, and r2 = −1 = i. So the general solution to z + 2z + 2z = 0 is: z(t) = C1 e−t cos(t) + C2 e−t sin(t). Then z (t) = −C1 e−t cos(t) − C1 e−t sin(t) − C2 e−t sin(t) + C2 e−t cos(t). Since z(0) = 2 and z (0) = −2, then 2 = C1 + 0, and − 2 = −C1 + C2 . So C1 = 2, and C2 = 0. Therefore, the solution to z (t) = −2z (t) − 2z(t), z(0) = 2, z (0) = −2 is: z(t) = 2e−t cos(t) . • 2.4.2: Consider a mass and spring system with a mass m = 2, spring constant k = 3, and damping constant c = 1. a) Set up and find the general solution of the system. b) Is the system underdamped, overdamped or critically damped? c) If the system is not critically damped, find a c that makes the system critically damped. Answer. a). Let x(t) be the displacement of the mass at time t. By the assumption, our differential equation is: mx + cx + kx = 2x + x + 3x = 0. The characteristic equation of 2x + x + 3x = 0 is: 2r2 + r + 3 = 0. Solving 2r2 + r + 3 = 0, we have √ √ −1 + 1 − 4 · 2 · 3 −1 + i 23 r1 = = , 4 4 2 √ −1 − i 23 and r2 = . 4 So the general solution to 2x + x + 3x = 0 is: x(t) = C1 e − 4t cos √ 23 t 4 √ + C2 e − 4t sin 23 t 4 . b). Since 1 − 4 · 2 · 3 = −23 < 0, the system is underdamped . c). In order to make the system to be critically damped, that is, c2 − 4 · 2 · 3 = 0. Then c= √ √ 24 = 2 6 . • 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to find c). You have a spring with spring constant k = 5 N/m. You take the spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency 1 Hz. What is the friction? Answer. It’s easy to see that our differential equation is: 0.1x + cx + 5x = 0. Rewrite the equation as x + 10cx + 50x = 0. By the assumption, we know that our motion is underdamped with frequency 1 Hz, which implies that (10c)2 − 4 · 50 < 0 and √ 4 · 50 − (10c)2 200 − 100c2 1 · 2π = = = 50 − 25C 2 . 2 2 So we get 25c2 = 50 − 4π 2 , that is, c = 2− 4π 2 ≈ 0.648. Hence the friction is 25 0.648 . • 2.4.101: A mass of 2 kilograms is on a spring with spring constant k newtons per meter with no damping. Suppose the system is at rest and at time t = 0 the mass is kicked and starts traveling at 2 meters per second. How large does k have to be to so that the mass does not go further than 3 meters from the rest position? Answer. The only force acting on the spring is the spring compression force so the differential equation is mx = −kx where m = 2 kg is the mass and k is the spring constant. The spring is initially uncompressed so x(0) = 0, and it was kicked at a velocity of 2 m/s so x (0) = 2. Therefore the initial value problem we have to solve is x (t) + k x(t) = 0, m x(0) = 0, 3 x (0) = 2. The general solution is x(t) = A cos k x + B sin m k x . m The initial conditions give 0 = x(0) = A, 2 = x (0) = k B, m so A = 0 and B = 2 m/k. Therefore the solution is m sin k x(t) = 2 k x . m A maximum displacement of 3 m means that 4m 8 m ≤ 3 =⇒ k ≥ = . k 9 9 |x(t)| ≤ 3 ⇐⇒ 2 Therefore to ensure the mass doesn’t go farther than 3 m from the rest position we must have 8 k ≥ N/m . 9 • 2.4.103: A 5,000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses by 0.1 m. Assume no damping. a) Find k. b) Find out how far does the spring compress when a 10,000 kg railcar hits the spring at the same speed. c) If the spring would break if it compresses further than 0.3 m, what is the maximum mass of a railcar that can hit it at 1 m/s? d) What is the maximum mass of a railcar that can hit the spring without breaking at 2 m/s? Answer. a) Let M = 5000 kg be the mass of the railcar. Since the only force acting on the railcar is the spring compression force, this means that the differential equation is M x = −kx or equivalently x +(k/M )x = 0. Now the spring is initially uncompressed so x(0) = 0 and since the railcar is hitting the spring at 1 m/s this means x (0) = −1. Therefore we are looking at the initial value problem x (t) + k x(t) = 0, M x(0) = 0, x (0) = −1. The general solution is x(t) = A cos k t + B sin M k t . M The initial conditions give A = 0 and B = −1 M/k so the solution is x(t) = − M sin k 4 k t . M Since the spring compresses to 0.1 m this means that the maximum of |x(t)| is 0.1. But note that the maximum of the sine term is 1 so this means that M = 0.1 =⇒ k = 5000/(0.1)2 = 500, 000. k Therefore the spring constant is k = 500, 000 N/m . b) If we instead have M = 10, 000 kg, then the maximum compression is M = k 10000 1 = √ . 500000 5 2 Therefore the maximum compression is 1 √ = 0.14142 m . 5 2 c) In order to have a maximum compression of 0.3 m we need to constrain M according to M ≤ 0.3 =⇒ M ≤ (0.3)2 (500000) = 45000. k Therefore to keep the bumper from breaking the mass of the railcar can be at most 45, 000 kg . d) If instead of going at 1 m/s the railcar is going at 2 m/s, the initial condition changes to x (0) = −2 so the solution becomes M sin k x(t) = −2 k t . M Therefore to keep the spring from breaking we have to constrain M according to 2 M ≤ 0.3 =⇒ M ≤ (0.3)2 (500000)/4 = 11250. k So to keep the bumper from breaking the mass of the railcar travelling at 2 m/s can be at most 11, 250 kg . • 2.5.2: Find a particular solution of y − y − 6y = e2x . Answer. The characteristic equation of y − y − 6y = 0 is: r2 − r − 6 = 0. 5 Solve r2 − r − 6 = 0, then r1 = 3, and r2 = −2. Let y(x) = Ae2x be a particular solution to y − y − 6y = e2x , then y (x) = 2Ae2x , and y (x) = 4Ae2x . So we have 4Ae2x − 2Ae2x − 6Ae2x = e2x . That is, −4Ae2x = e2x . So 1 A=− . 4 Therefore, a particular solution of y − y − 6y = e2x is: y(x) = − e2x . 4 • 2.5.3: Find a particular solution of y − 4y + 4y = e2x . Answer. The characteristic equation of y − 4y + 4y = 0 is: r2 − 4r + 4 = 0. Solve r2 − 4r + 4 = 0, then r1 = r2 = 2. Let y(x) = Ax2 e2x be a particular solution to y − 4y + 4y = e2x , then y (x) = 2Axe2x + 2Ax2 e2x , and y (x) = 2Ae2x + 4Axe2x + 4Axe2x + 4Ax2 e2x . So we have 2Ae2x + 4Axe2x + 4Axe2x + 4Ax2 e2x − 4 2Axe2x + 2Ax2 e2x + 4Ax2 e2x = e2x . That is, 2Ae2x = e2x . So 1 A= . 2 Therefore, a particular solution of y − 4y + 4 = e2x is: y(x) = 6 x2 e2x . 2 • 2.5.102: a) Find a particular solution to y + 2y = ex + x3 . b) Find the general solution. Answer. a). The characteristic equation to y + 2y = 0 is: r2 + 2 = 0. Solve r2 + 2 = 0, then r1 = √ 2i, √ and r2 = − 2i. Let y(x) = Aex + Bx3 + Cx2 + Dx + E be particular solution to y + 2y = ex + x3 , then y (x) = Aex + 3Bx2 + 2Cx + D, and y (x) = Aex + 6Bx + 2C. Then Aex + 6Bx + 2C + 2 Aex + Bx3 + Cx2 + Dx + E = ex + x3 . That is, we have (3A − 1)ex + (2B − 1)x3 + 2Cx2 + (6B + 2D)x + 2C + 2E = 0. So we get 3A − 1 = 0, Then 2B − 1 = 0, 1 A= , 3 2C = 0, 1 B= , 2 6B + 2D = 0, C = 0, 3 D=− , 2 2C + 2E = 0. E = 0. That is, a particular solution to y + 2y = ex + x3 is: y(x) = ex x3 3 + − x. 3 2 2 b). By the computations in part a), we know that the general solution to y + 2y = 0 is: √ √ y(x) = C1 cos( 2x) + C2 sin( 2x). So the general solution to y + 2y = ex + x3 is: √ √ ex x3 3 y(x) = C1 cos( 2x) + C2 sin( 2x) + + − x. 3 2 2 7
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