# Homework #7 Solutions

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1 MAT 0 Spring 201 Problems Homework #7 Solutions Section.: 4, 18, 22, 24, 4, 40 Section.4: 4, abc, 16, 18, 22. Omit the graphing part on problems 16 and Find the general solution to the differential equation 2y 7y + y = 0. Solution: We determine that the characteristic equation for this linear polynomial is 2r 2 7r + = 0, which factors as (2r 1)(r ) = 0. Thus, the roots are r = 1/2 and r =, so the general solution is y = c 1 e x/2 + c 2 e x Find the general solution to the differential equation y (4) = 16y. Solution: Writing the DE as y (4) 16y = 0, we see that its characteristic equation is r 4 16 = 0. Since this is a difference of squares, it factors as (r 2 4)(r 2 + 4) = (r 2)(r + 2)(r 2 + 4) = 0. Therefore, it has the roots r = 2 and r = 2 from the linear factors, and the r factor has pure imaginary roots r = ± 4 = ±2i. From the real roots, we have the solutions e 2x and e 2x, while we get the trigonometric functions cos 2x and sin 2x from the pure imaginary roots. Thus, the general solution is y = c 1 e 2x + c 2 e 2x + c cos 2x + c 4 sin 2x Solve the IVP 9y + 6y + 4y = 0, y(0) =, y (0) = 4. Solution: We first find the general solution to the homogeneous linear DE. Since its characteristic equation is 9r 2 + 6r + 4 = 0, which has roots r = 6 ± 6 2 4(4)(9) 2 9 = 1 ± = 1 ± 1 i, the general solution is the linear combination y = c 1 e x/ cos x + c 2 e x/ sin x. 1

2 MAT 0 Spring 201 We use the product rule to compute its derivative; after factoring out e x/, this is ( y = c 1 e x/ 1 cos x 1 sin x ) ( + c 2 e x/ 1 sin x + 1 cos x ). Evaluating at x = 0 and applying the initial conditions, and noting that the sin terms vanish, y(0) = c 1 =, y (0) = c 1 + c 2 = 4. Then c 1 =, and c 2 = (4 + c 1 /) = 5, so the solution to the IVP is y = e x/ cos x + 5 e x/ sin x Solve the IVP 2y () y 2y = 0, y(0) = 1, y (0) = 1, y (0) =. Solution: The characteristic equation for this DE is 2r r 2 2r = 0, which factors as r(2r 2 r 2) = r(2r + 1)(r 2) = 0. Then r = 0, r = 1/2, and r = 2 are its (distinct real) roots, so the general solution is Its first and second derivatives are y = c 1 + c 2 e x/2 + c e 2x. y = 1 2 c 2e x/2 + 2c e 2x, y = 1 4 c 2e x/2 + 4c e 2x. Evaluating at x = 0 and applying the initial conditions, we have the linear system c 1 + c 2 + c = 1, 1 2 c 2 + 2c = 1, 1 4 c 2 + 4c =. From the third equation, c 2 = 16c. Substituting this into the second, 1 2 ( 16c ) + 2c = 1, so 10c = 5, and c = 1/2. Then c 2 = 8 = 4, so c 1 = 1 c 2 c = 7/2. Thus, the solution to the IVP is y = e x/ e2x...4. One solution to the DE y () 2y + y 8y = 0 is y = e 2x/. Find the general solution. Solution: The characteristic equation for the DE is r 2r 2 + r 8 = 0. Since r = 2/ is a root, we expect r 2/ to be a linear factor of this polynomial. In fact, (r 2/) = r 2 is seen to be a linear factor, so that this polynomial factors as (r 2)(r 2 + 4) = 0. Thus, the two other roots are r = ±2i, from the r factor, so the general solution is c 1 e 2x/ + c 2 cos 2x + c sin 2x. 2

3 MAT 0 Spring Find a linear homogeneous constant-coefficient equation with general solution y = Ae 2x + B cos 2x + C sin 2x. Solution: Since the general solution contains both e 2x and a cos 2x-sin 2x pair, the original DE should have roots r = 2 and r = ±2i. Furthermore, its general solution has three independent parameters, so it should be a third-order DE. Hence, its characteristic equation is (r 2)(r 2i)(r + 2i) = (r 2)(r 2 + 4) = r 2r 2 + 4r 8 = 0, which comes from the linear homogeneous DE y () 2y + 4y 8y = 0. (Scalar multiples of this DE have the same solutions.).4.4. A body with mass 250 g is attached to the end of a spring that is stretched 25 cm by a force of 9 N. At time t = 0 the body is pulled 1 m to the right, stretching the spring, and set in motion with an initial velocity of 5 m/s to the left. (a) Find x(t) in the form C cos(ω 0 t α). (b) Find the amplitude and period of motion of the body. Solution (a): We normalize the constants to mks SI units, so m = 0.25 kg and k = 9/0.25 = 6 N/m. Then the circular frequency is ω 0 = k/m = 6/0.25 = rad/s, so the general solution for the motion of the body is x(t) = A cos t + B sin t, x (t) = A sin t + B cos t. Evaluating at t = 0, x(0) = A = 1 and x (0) = B = 5 (since the initial velocity is rightward). Then A = 1 and B = 5/. Computing C, C 2 = A 2 + B 2 = = ( 1 2 )2, so C = 1. The phase angle α has tan α = B/A = 5, but must satisfy C cos α = A > 0 and C sin α = B < 0 from the above values of A and B. Fortunately, since its cosine must be positive, we may then pick α from the principal branch of arctan, so α = arctan( 5/) = arctan (Of course, we may also add an arbitrary multiple of 2π to this angle, so 2π arctan is also a correct answer.) Thus, x(t) = 1 ( cos t + arctan 5 ). Solution (b): The amplitude is the constant C = 1 2π, and the period is T = = π 6 s.

4 MAT 0 Spring abc. Assume that the earth is a solid sphere of uniform density, with mass M and radius R = 960 miles. For a particle of mass m within the earth at a distance r from the center of the earth, the gravitational force attracting m toward the center is F r = GM r m/r 2, where M r is the mass of the part of the earth within a sphere of radius r. (a) Show that F r = GMmr/R. (b) Now suppose that a small hole is drilled straight through the center of the earth, thus connecting two antipodal paints on its surface. Let a particle of mass m be dropped at time t = 0 into this hole with initial speed zero, and let r(t) be its distance from the center of the earth at time t. Conclude from Newton s second law and part (a) that r (t) = k 2 r(t), where k 2 = GM/R = g/r. (c) Take g = 2.2 ft/s 2, and conclude from part (b) that the particle undergoes simple harmonic motion back and forth between the ends of the hole, with a period of about 84 min. Solution (a): The volume of the Earth (assumed to be a perfect sphere) is V = 4π R, so the density of the Earth is ρ = M V = M 4π. The mass of the radius-r portion of the Earth is R then M r = ρ 4π r = M r R. Hence, the force of gravity on the mass m at its surface is F r = Gm r 2 M r = Gm r 2 M r R = GMmr R. Solution (b): From Newton s second law, F = ma. Since the position of the particle is given by r(t), the acceleration is its second derivative, r (t). The only force on the body is gravity, so from part (a), we have the DE. GMmr R = mr (t). Dividing out the m and collecting the r terms, this is r + GM R r = 0, which gives simple harmonic motion with circular frequency ω 0 = that when r = R, the gravitational acceleration g is given by GMR R is also g R. Solution (c): Since ω 0 = units, GM. We also note R = GM, so this constant R 2 g 2π R, the period of oscillation is T = ω 0 = 2π Rg. Then, in fps T = 2π s 84.4 min. 4

5 MAT 0 Spring A mass m = is attached to both a spring with spring constant k = 6 and a dashpot with damping constant c = 0. The mass is set in motion with initial position x 0 = 2 and initial velocity v 0 = 2. Find the position function x(t) and determine whether the motion is overdamped, critically damped, or underdamped. If it is underdamped, write the position function in the form x(t) = Ce pt cos(ω 1 t α 1 ). Also, find the undamped position function u(t) = C 0 cos(ω 0 t α 0 ) that would result if the mass on the spring were set in motion with the same initial position and velocity but wth the dashpot disconnected (c = 0). Solution: We first find that ω0 2 = m k c = 21, and p = 2m = 2 0 = 5. Then p2 = 25, which is larger than ω0 2, so the motion is overdamped. The roots of the DE are r = p ± p 2 ω0 2 = 5 ± 2, or r = 7 and r =. Hence, the general solution is x(t) = c 1 e 7t + c 2 e t, v(t) = x (t) = 7c 1 e 7t c 2 e t. Setting t = 0 and matching the initial conditions, x(0) = c 1 + c 2 = 2 and v(0) = 7c 1 c 2 = 2. Solving for c 1 and c 2, c 2 = 2 c 1, so 7c 1 (2 c 1 ) = 2, and 4c 1 = 8, Then c 1 = 2, so c 2 = 4, and the solution is x(t) = 4e t e 7t. Removing the damping, we get simple harmonic motion with the frequency ω 0 = 21, so the general solution is u(t) = A cos ω 0 t + B sin ω 0 t, u (t) = Aω 0 sin ω 0 t + Bω 0 cos ω 0 t. Then u(0) = A = 2 and u (0) = Bω 0 = 2, so A = 2 and B = 2. Then u(t) = 21 C cos(ω 0 t α) C = A 2 + B 2 = = , and α = arctan A B = arctan A mass m = 2 is attached to both a spring with spring constant k = 50 and a dashpot with damping constant c =. The mass is set in motion with initial position x 0 = 0 and initial velocity v 0 = 8. Find the position function x(t) and determine whether the motion is overdamped, critically damped, or underdamped. If it is underdamped, write the position function in the form x(t) = Ce pt cos(ω 1 t α 1 ). Also, find the undamped position function u(t) = C 0 cos(ω 0 t α 0 ) that would result if the mass on the spring were set in motion with the same initial position and velocity but wth the dashpot disconnected (c = 0). Solution: We have that ω0 2 = m k = 50 2 = 25, so ω 0 = 5, and p = 2m c = 2 2 =, so the motion is underdamped. The pseudofrequency is ω 1 = ω0 2 p2 = 25 9 = 16 = 4, so the general solution is x(t) = c 1 e t cos 4t + c 2 e t sin 4t, v(t) = c 1 e t ( cos 4t 4 sin 4t) + c 2 e t ( sin 4t + 4 cos 4t) 5

6 MAT 0 Spring 201 At t = 0, x(0) = c 1 = 0. Then, using c 1 = 0, v(0) = 4c 2 = 8, so c 2 = 2. Hence, x(t) = 2e t sin 4t, which we convert to Ce pt cos(ω 1 t α 1 ). We have that C = 2, and A = 2 = 2 sin α 1, so α 1 = π 2 (taking an angle between 0 and 2π), so x(t) = 2e t cos (4t 2 ) π. With c = 0, the solution is u(t) = A cos 5t + B sin 5t, with u (t) = 5A sin 5t + 5B cos 5t; applying the initial conditions, A = 0 and 5B = 8, so u(t) = 5 8 sin 5t = 8 5 cos(5t π) A -lb weight (mass m = 0.75 slugs in fps units) is attached both to a vertically suspended spring that it stretches 6 inches and to a dashpot that provides lb of resistance for every foot-per-second of velocity. (a) If the weight is pulled down 1 foot below its static equilibrium position and then released from rest at time t = 0, find its position function x(t). (b) Find the frequency, time-varying amplitude, and phase angle of the motion. lb 0.5 ft = Solution (a): The mass in fps units is m = 0.75, and the spring constant is k = 24 lb/ft. The circular frequency is given by ω0 2 = m k = 64, so ω 0 = 8 rad/s. The damping constant is c = lb-s/ft, so p = 2m c = 4, and the system is therefore underdamped. Then ω 1 = ω0 2 p2 = 48 = 4, so the general solution is x(t) = Ae 4t cos 4 t + Be 4t sin 4 t, v(t) = Ae 4t ( 4 cos 4 t 4 sin 4 t) + Be 4t ( 4 sin 4 t + 4 cos 4 t). We measure the displacement vertically, considering a displacement downwards as being positive, since it corresponds to stretching the string further. Thus, at time t = 0 we have that x(0) = 1 and v(0) = 0. Then A = 1 and 4A + 4 B = 0, so B = 4A 4 = 1. Thus, the solution is x(t) = e 4t cos 4 t + 1 e 4t sin 4. Solution (b): We reformulate our answer to part (a) in the form Ce pt cos(ω 1 t α). Then C 2 = A 2 + B 2 = = 4, so C = 2, and the time-varying amplitude is 2 e 4t. From above, the pseudofrequency is ω 1 = 4 rad/s. Finally, tan α = B A = 1, with α in Quadrant I so that both A and B are positive. Thus, the phase angle is α = π 6. 6

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