MATH 215/255 Fall 2014 Assignment 4 В§2.2, В§2.4, В§2.5 Solutions to selected exercises can be found in [Lebl], starting from page 303. вЂў 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 = в€’9. So the general solution to y + 9y = 0 is: y(x) = C1 + C2 eв€’9x . Then y (x) = в€’9C2 eв€’9x . Since y(0) = 1 and y (0) = 1, then 1 = C1 + C2 , So C1 = 10 , 9 and в€’ 9C2 = 1. 1 and C2 = в€’ . 9 So the solution to y + 9y = 0 for y(0) = 1, y (0) = 1 is: y(x) = 10 eв€’9x в€’ . 9 9 вЂў 2.2.102: Find the general solution to y в€’ 6y + 9y = 0. Answer. The characteristic equation of y в€’ 6y + 9y = 0 is: r2 в€’ 6r + 9 = 0. Solve r2 в€’ 6r + 9 = 0, we have r1 = r2 = 3. So the general solution to y в€’ 6y + 9y = 0 is: y(x) = C1 e3x + C2 xe3x . вЂў 2.2.105: Find the solution to z (t) = в€’2z (t) в€’ 2z(t), z(0) = 2, z (0) = в€’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 = в€’1 + i, and r2 = в€’1 = i. So the general solution to z + 2z + 2z = 0 is: z(t) = C1 eв€’t cos(t) + C2 eв€’t sin(t). Then z (t) = в€’C1 eв€’t cos(t) в€’ C1 eв€’t sin(t) в€’ C2 eв€’t sin(t) + C2 eв€’t cos(t). Since z(0) = 2 and z (0) = в€’2, then 2 = C1 + 0, and в€’ 2 = в€’C1 + C2 . So C1 = 2, and C2 = 0. Therefore, the solution to z (t) = в€’2z (t) в€’ 2z(t), z(0) = 2, z (0) = в€’2 is: z(t) = 2eв€’t cos(t) . вЂў 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 в€љ в€љ в€’1 + 1 в€’ 4 В· 2 В· 3 в€’1 + i 23 r1 = = , 4 4 2 в€љ в€’1 в€’ i 23 and r2 = . 4 So the general solution to 2x + x + 3x = 0 is: x(t) = C1 e в€’ 4t cos в€љ 23 t 4 в€љ + C2 e в€’ 4t sin 23 t 4 . b). Since 1 в€’ 4 В· 2 В· 3 = в€’23 < 0, the system is underdamped . c). In order to make the system to be critically damped, that is, c2 в€’ 4 В· 2 В· 3 = 0. Then c= в€љ в€љ 24 = 2 6 . вЂў 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. ItвЂ™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 в€’ 4 В· 50 < 0 and в€љ 4 В· 50 в€’ (10c)2 200 в€’ 100c2 1 В· 2ПЂ = = = 50 в€’ 25C 2 . 2 2 So we get 25c2 = 50 в€’ 4ПЂ 2 , that is, c = 2в€’ 4ПЂ 2 в‰€ 0.648. Hence the friction is 25 0.648 . вЂў 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 = в€’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 в‰¤ 3 =в‡’ k в‰Ґ = . k 9 9 |x(t)| в‰¤ 3 в‡ђв‡’ 2 Therefore to ensure the mass doesnвЂ™t go farther than 3 m from the rest position we must have 8 k в‰Ґ N/m . 9 вЂў 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 = в€’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) = в€’1. Therefore we are looking at the initial value problem x (t) + k x(t) = 0, M x(0) = 0, x (0) = в€’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 = в€’1 M/k so the solution is x(t) = в€’ 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 =в‡’ 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 = в€љ . 500000 5 2 Therefore the maximum compression is 1 в€љ = 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 в‰¤ 0.3 =в‡’ M в‰¤ (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) = в€’2 so the solution becomes M sin k x(t) = в€’2 k t . M Therefore to keep the spring from breaking we have to constrain M according to 2 M в‰¤ 0.3 =в‡’ M в‰¤ (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 . вЂў 2.5.2: Find a particular solution of y в€’ y в€’ 6y = e2x . Answer. The characteristic equation of y в€’ y в€’ 6y = 0 is: r2 в€’ r в€’ 6 = 0. 5 Solve r2 в€’ r в€’ 6 = 0, then r1 = 3, and r2 = в€’2. Let y(x) = Ae2x be a particular solution to y в€’ y в€’ 6y = e2x , then y (x) = 2Ae2x , and y (x) = 4Ae2x . So we have 4Ae2x в€’ 2Ae2x в€’ 6Ae2x = e2x . That is, в€’4Ae2x = e2x . So 1 A=в€’ . 4 Therefore, a particular solution of y в€’ y в€’ 6y = e2x is: y(x) = в€’ e2x . 4 вЂў 2.5.3: Find a particular solution of y в€’ 4y + 4y = e2x . Answer. The characteristic equation of y в€’ 4y + 4y = 0 is: r2 в€’ 4r + 4 = 0. Solve r2 в€’ 4r + 4 = 0, then r1 = r2 = 2. Let y(x) = Ax2 e2x be a particular solution to y в€’ 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 в€’ 4 2Axe2x + 2Ax2 e2x + 4Ax2 e2x = e2x . That is, 2Ae2x = e2x . So 1 A= . 2 Therefore, a particular solution of y в€’ 4y + 4 = e2x is: y(x) = 6 x2 e2x . 2 вЂў 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 = в€љ 2i, в€љ and r2 = в€’ 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 в€’ 1)ex + (2B в€’ 1)x3 + 2Cx2 + (6B + 2D)x + 2C + 2E = 0. So we get 3A в€’ 1 = 0, Then 2B в€’ 1 = 0, 1 A= , 3 2C = 0, 1 B= , 2 6B + 2D = 0, C = 0, 3 D=в€’ , 2 2C + 2E = 0. E = 0. That is, a particular solution to y + 2y = ex + x3 is: y(x) = ex x3 3 + в€’ x. 3 2 2 b). By the computations in part a), we know that the general solution to y + 2y = 0 is: в€љ в€љ y(x) = C1 cos( 2x) + C2 sin( 2x). So the general solution to y + 2y = ex + x3 is: в€љ в€љ ex x3 3 y(x) = C1 cos( 2x) + C2 sin( 2x) + + в€’ x. 3 2 2 7

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