Math 215 Spring 2010 Assignment 4 You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples. §3.6: 8, 11, 13, 15, §3.7: 1, 7, 9, 20 • (§3.6: 8) Find the general solution of y 00 + 4y = 3 csc 2t, 0 < t < π/2. Answer. The solutions of the homogeneous equation is yc (t) = c1 cos 2t + c2 sin 2t. The two functions y1 (t) = cos 2t and y2 (t) = sin 2t form a fundamental set of solutions, with Wronskian W (y1 , y2 ) = 2. A particular solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in which Z sin 2t(3 csc 2t) 3 u1 (t) = − dt = − t, W (t) 2 Z cos 2t(3 csc 2t) 3 u2 (t) = dt = ln | sin 2t|. W (t) 4 Hence the particular solution is Y (t) = − 32 t cos 2t + 34 (sin 2t) ln | sin 2t|. The general solution is 3 3 y(t) = c1 cos 2t + c2 sin 2t − t cos 2t + (sin 2t) ln | sin 2t|. 2 4 • (§3.6: 11) Find the general solution of y 00 − 5y 0 + 6y = g(t). Here g(t) is an arbitrary continuous function. Answer. Two linearly independent solutions of the homogeneous DE are y1 (t) = e3t and y2 (t) = e2t , with Wronskian W (y1 , y2 ) = −e5t . By Theorem 3.6.1, Z 2s Z 3s Z e g(s) e g(s) 3t 2t Y (t) = −e ds + e ds = [e3(t−s) − e2(t−s) ]g(s)ds. −e5s −e5s The general solution is y(t) = c1 e3t + c2 e2t + Y (t). • (§3.6: 13) Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. t2 y 00 − 2y = 3t2 − 1, t > 0; y1 (t) = t2 , y2 (t) = t−1 . Answer. t2 y100 − 2y1 = 2t2 − 2t2 = 0 and t2 y200 − 2y2 = 2t−1 − 2t−1 = 0. Their Wronskian W (y1 , y2 ) = −3. The equation can be rewritten as y 00 − 2t−2 y = g(t) = 3 − t−2 . Using the method of variation of parameters, a particular solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in which Z −1 t (3 − t−2 ) 1 u1 (t) = − dt = t−2 + ln t, W (t) 6 Z 2 t (3 − t−2 ) 1 u2 (t) = dt = − t3 + t/3. W (t) 3 Hence the particular solution is Y (t) = 1 6 + t2 ln t − t2 /3 + 1/3 = t2 ln t + 1/2 − t2 /3. (The general solution is y(t) = c1 t2 + c2 t−1 + t2 ln t + 1/2 and hence we can take Y (t) = t2 ln t + 1/2.) • (§3.6: 15) Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. ty 00 − (1 + t)y 0 + y = t2 e2t , t > 0; y1 (t) = 1 + t, y2 (t) = et . Answer. ty100 − (1 + t)y10 + y1 = 0 − (1 + t) + (1 + t) = 0 and ty200 − (1 + t)y20 + y2 = tet − (1 + t)et + et = 0. Their Wronskian W (y1 , y2 ) = tet . The equation can be rewritten as y 00 − (1 + t−1 )y 0 + t−1 y = g(t) = te2t . Using the method of variation of parameters, a particular solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in which Z t 2t e (te ) u1 (t) = − dt = −e2t /2, tet Z (1 + t)(te2t ) u2 (t) = dt = tet . tet Hence the particular solution is Y (t) = −(1 + t)e2t /2 + et tet = 12 e2t (t − 1). • (§3.7: 1) Determine ω0 , R and δ so as to write the expression u = 3 cos 2t + 4 sin 2t in the form u = R cos(ω0 t − δ). √ Answer. R cos δ = 3 and R sin δ = 4, so R = 32 + 42 = 5 and δ = arctan(4/3) ≈ 0.9273. Hence u = 5 cos(2t − 0.9273). 2 • (§3.7: 7) A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in., and then set in motion with a downward velocity of 2 ft/s, and if there is no damping, find the position u of the mass at any time t. Determine the frequency, period, amplitude, and phase of the motion. Answer. We use ft-lb-s for the units and note 1 ft = 12 in. The spring constant is k = 3/(1/4) = 12 lb/ft. Mass m = 3/32 lb-s2 /ft. Since there is no damping, the 3 00 equation of motion is 32 u + 12u = 0, that is, u00 + 128u = 0. The initial conditions are u(0) = −1/12, u0 (0) = 2. √ √ The general solution is u(t) = A cos 8 2t+B sin 8 2t. Invoking the initial conditions, we have √ √ √ 1 2 u(t) = − cos 8 2t + sin 8 2t. 12 8 p √ √ √ Thus R = 11/288 ft, δ = π−arctan(3/ 2) rad, ω0 = 8 2 rad/s, and T = π/(4 2)s. • (§3.7: 9) A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 400 dyn·s/cm. If the mass is pulled down an additional 2 cm and then released, find its position u at any time t. Plot u versus t. Determine the quasi frequency and the quasi period. Determine the ratio of the quasi period to the period of the corresponding undamped motion. Also find the time τ such that |u(t)| < 0.05 cm for all t > τ . Answer. We use cm-g-s as the units. The spring constant is k = (20)(980)/5 = 3920 dyne/cm. Mass m = 20 g. The equation of motion is 20u00 + 400u0 + 3920u = 0 or u00 + 20u0 + 196u = 0. The initial conditions are u0 (0) = 0. √ √ −10t sin 4 6t. Invoking the initial The general solution is u(t) = Ae−10t √ cos 4 6t + Be conditions, we have A = 2, B = 5/ 6, √ √ 5 u(t) = e−10t [2 cos 4 6t + √ sin 4 6t] (cm). 6 √ π The quasi-frequency is µ = 4 6. The quasi period is Td = 2π/µ = 2√ . The 6 u(0) = 2, undamped motion has period T = 2π/14 = π/7. The ratio Td /T = 7 √ . 2 6 To find an upper bound for τ , write u in the form p √ u(t) = 4 + 25/6 e−10t cos(4 6t − δ). p To guarantee u(t) ≤ .05, it suffices that 4 + 25/6 e−10t ≤ .05, which yields τ = .4046. 3 • (§3.7: 20) Assume that the system described by the equation mu00 + γu0 + ku = 0 is critically damped and that the initial conditions are u(0) = u0 and u0 (0) = v0 . If v0 = 0, show that u → 0 as t → ∞ but that u is never zero. If u0 is positive, determine a condition on v0 that will ensure that the mass passes through its equilibrium position after it is released. Answer. The general solution for the critical damping case is u(t) = (A+Bt)e−γt/2m . The I.C. u(0) = u0 implies A = u0 , and u0 (0) = v0 implies A(−γ/2m)+B = v0 . Hence u(t) = [u0 + (v0 + γu0 /2m)t]e−γt/2m . If v0 = 0, then u = u0 (1 + γt/2m)e−γt/2m , which is never zero since γ and m are positive. By L’Hospital’s Rule u → 0 as t → ∞. If u0 > 0 and u(t1 ) = 0 for some t1 > 0, we need (v0 + γu0 /2m) < 0, that is v0 < −γu0 /2m. 4

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