1. y 5y + 6y = 2e t Solution: Characteristic equation is r 2 5r +6 = 0, therefore r 1 = 2, r 2 = 3, and y 1 (t) = e 2t,

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1 Homework6 Soluions.7 In Problem hrough 4 use he mehod of variaion of parameers o find a paricular soluion of he given differenial equaion. Then check your answer by using he mehod of undeermined coeffiens.. y 5y + 6y = 2e Soluion: Characerisic equaion is r 2 5r +6 = 0, herefore r = 2, r 2 =, and y () = e 2, y 2 () = e. Wronskian W (y, y 2 )() = 2e 2 = e5 e2 e e Using he formula given in he ex, we have e Y () = e 2 2e e + e 2 2e e 5 e 5 = 2e 2 e + 2e e 2 = e. 2. y y 2y = 2e Soluion: Characerisic equaion is r 2 r 2 = 0, herefore r =, r 2 = 2, and y () = e, y 2 () = e 2. Wronskian W (y, y 2 )() = e e = e Using he formula given in he ex, we have e Y () = e 2 2e e + e 2 2e 2e e = 2 e 2 9 e We can choose Y () = 2 e. In problem 5 2 find he general soluion of he given equaion. 5. y + y = an, 0 < < π/2. Soluion: Characerisic equaion is r 2 + = 0, herefore r = i, r 2 = i, and y () = cos, y 2 () = sin. Wronskian W (y, y 2 )() = cos sin sin cos =. e 2 2e 2

2 Using he formula given in he ex, we have Y () = cos sin an + sin cos an sin 2 = cos + sin sin cos sin 2 = cos sin 2 d(sin ) sin cos = cos ( sin 2 ) d(sin ) sin cos = cos ( sin + + sin ln ) sin cos 2 sin = 2 cos ln + sin sin = cos ln + sin cos or Y () = cos ln(sec + an ). Noe he range for guaranees he funcion inside ln is posiive. 0. y 2y + y = e /( + 2 ) Soluion: Characerisic equaion is r 2 2r + = 0, herefore r = r 2 =, and y () = e, y 2 () = e. Wronskian W (y, y 2 )() = e e e e + e = e2 Using he formula given in he ex, we have Y () = e + e = 2 e ln( + 2 ) + e arcan., In problems 20 verify ha he given funcions y and y 2 are soluions o he corresponding homogeneous equaion; hen find a paricular soluion of he given nonhomogeneous equaion.. 2 y 2y = 2, > 0, y = 2, y 2 = Soluion: I s easy o verify ha he given funcions are soluions o he homogeneous equaion. We will concenrae on finding a paricular soluion. Firs rewrie he equaion so ha he leading coefficien is. y 2 2 y = 2. The Wronskian is W (y, y 2 )() = 2 2 / 2 = 2

3 and Y () = 2 ( / 2 ) = 2 ( ln ) ( ) = 2 ln ( / 2 ) 5. y ( + )y + y = 2 e 2, > 0, y = +, y 2 = e Soluion: Rewrie he equaion so ha he leading coefficien is. The Wronskian is and y + y + y = e2. W (y, y 2 )() = + e e = e e (e 2 ) ( + )e Y () = ( + ) + e 2 e e = ( + ) e 2 + e ( + )e = + 2 e2 + e (( + )e e ) = 2 e2 + 2 e2 = 2 e2 ( )..8 In problems 4 deermine ω 0, R, and δ so as o wrie he given expression in he form u = R cos(ω 0 δ).. u = cos sin 2 Soluion: u = 5( 5 cos sin 2) = 5 cos(2 arccos 5 ) Therefore R = 5, ω 0 = 2, δ = arccos(/5). 2. u = cos + sin Soluion: u = 2( 2 cos + 2 sin ) = 2 cos( 2π ). Therefore R = 2, ω 0 =, δ = 2π/.

4 6. A mass of 00 g sreches a spring 5 cm. If he mass is se in moion from is equilibrium posiion wih a downward velociy of 0 cm/sec, and if here is no damping, deermine he posiion u of he mass a any ime. When does he mass firs reurn o is equilibrium posiion? Soluion: m = 00 = 0. kg, L = 5 = 0.05 m, u(0) = 0 m, u (0) = 0 = 0. m/sec. Therefore k = /0.05 = 98/5 N/m. The equaion for he sysem is 0 u u = 0 Solving he characerisic equaion we ge r = ±4i. Therefore he general soluion is u() = C cos 4 + C 2 sin 4. Using he iniail condiion we ge C = 0, C 2 = /40. Thus u() = sin Seing u() = 0 we ge = π/4 sec as he ime when he mass firs reurn o is equilibrium posiion.. A cerain vibraing sysem saisfies he equaion u + γu + u = 0. Find he value of he damping coefficien γ for which he quasi period of he damped moion is 50% geaer han he period of he corresponding undamped moion. Soluion: We can solve he equaion o find he quasi period direcly. Bu since here is a formula in he ex for he raio of he quasi period o he period of he undamped sysem we simply use ha one. By his formula (28) in he ex, and noice ha m =, k =, we have T d ( T = γ2 4km solving for γ we ge γ = 2 5. ) /2 ) /2 = ( γ2 = 50% = 4 2, 7. A mass weighing 8 lb sreches a sping.5 in. The mass is also aached o a damper wih coefficien γ. Deermine he value of γ for which he sysem is criically damped; be sure o give he unis for γ. Soluion: m = 8/2 = /4 lb sec 2 /f, L =.5/2 = /8 f. Thus k = 8/(/8) = 64 lb/f. Thus he equaion is 4 u + γu + 64u = 0. So in order for he sysem o be criically damped we need or γ = 8 lb sec/f. γ 2 = 4mk = 64 4

5 9. Assume he sysem decribed by he equaion mu + γu + ku = 0 is criically damped or overdamped. Show ha he mass can pass hrough he equilibrium posiion a mos once, regardless of he iniial condiion. Soluion: Firs assume he sysem is criically damped, hen he general soluion o he equaion is u = C e r + C 2 e r where r is he double roo o he characerisic equaion, and r is negaive. Se u = 0 and we ge e r (C + C 2 ) = 0. Clearly his equaion has no soluion if C 2 = 0 and exacly one soluion if C 2 0. For he overdamped case he general soluion is u = C e r + C 2 e r 2 and he proof is similar A mass of 6 kg sreches a spring 0 cm. The mass is aced on by an exernal force of 0 sin(/2) N and moves in a medium ha impars a viscous force of 2 N when he speed of he mass is 4 cm/sec. If he mass is se in mosion from is equilibrium posiion wih an iniial velociy of cm/sec, formulae he iniial value problem describing he moion of he mass. Soluion: m = 5 kg, L = 0.m, γ = 2/4 N sec/cm = 50 N sec/m, k = 5 9.8/0. = 490 N/m. Therefore he equaion is 5u + 50u + 490u = 0 sin /2. Iniial condiion is u(0) = 0, u (0) = /00 m/sec. 8. (a) Find he soluion of he iniial value problem in Problem 6. (b) Idenify he ransien and seady-sae pars of he soluion. (c) Plo he graph of he seady-sae soluion. (d) If he given exernal force is replaced by a force 2 cos ω of frequency ω, find he value of ω for which he ampliude of he forced response is maximum. Soluion (a) Solving he characerisic equaion we ge r = 5 ± 7i, so he general soluion o he homogeneous equaion is e 5 (C cos 7 + C 2 sin 7). One way o find a paricular soluion is using undeermined coefficiens, bu ha s a bi edious. Wha we will do here is o use equaion (8) and (9) in he ex. Le s firs rewrie he equaion as u + 0u + 98u = 2 sin /2. Thus m =, γ = 0, ω 0 = 7, ω = /2, F 0 = 2. Then according o equaion (9) = m 2 (ω0 2 ω 2 ) 2 + γ 2 ω 2 = 72.92, 5

6 and R = F 0 / = , δ = cos m(ω0 2 ω2 ) = cos = However we canno say u() = R cos(ω δ) is a paricular soluion o our equaion since ha s a soluion o equaion () in he ex in which he righ hand side is F 0 cos ω, while in our equaion he righ hand side is F 0 sin ω. To remedy he siuaion we consider he funcion u() = R sin(ω δ). I s easy o check ha his is a paricular soluion o our equaion. Therefore u() = e 5 (C cos 7 + C 2 sin 7) sin( ) is he general soluion. Using he iniial condiion we ge C = m = 0.87 cm, C 2 = m = 0.62 cm. (b) The ransien par is u() = e 5 (0.87 cos sin 7) cm, seady-sae par is u() = 2.74 sin( ) cm. (c) Use your calculaor (or compuer :). (d) The ampliude of he forced response R is given as F 0 /, herefore in order for i o reach maximum should reach he minimum. Noe = m 2 (ω 2 0 ω 2 ) 2 + γ 2 ω 2. To find he minimum we ake he derivaive of 2 wih respec o ω and se i o 0. Thus 2m 2 (ω0 2 ω 2 )( 2ω) + 2γ 2 ω = 0 ω = ω0 2 γ2 2m = If an undamped spring-mass sysem wih a mass ha weighs 6 lb and a spring consan lb/in. is suddenly se in moion a = 0 by an exernal force of 4 cos 7 lb, deermine he posiion of he mass a any ime and draw a graph of he displacemen versus. Soluion: m = 6/2 = /6 lb sec 2 /f, k = 2 lb/f, hus he equaion is 6 u +2u = 4 cos 7 wih iniial condiion u(0) = 0, u (0) = 0. Rewrie he equaion as u + 64u = 64 cos 7 we see ha ω 0 = 64 = 8. Using equaion (5) in he ex we have he soluion u = F 0 m(ω 2 0 ω 2 ) (cos ω cos ω 0) = 64 (cos 7 cos 8) Consider he forced bu undamped sysem described by he iniial value problem (a) Find he soluion u() for ω. u + u = cos ω, u(0) = 0, u (0) = 0. (b) Plo he soluion u() versus for ω = 0.7, ω = 0.8 and ω = 0.9. Describe how he response u() changes as ω varies in his inerval. Wha happens as ω akes on values closer and closer o? Noe ha he naural frequency of he unforced sysem is ω 0 =. Soluion: (a) Again using equaion (5) we ge he soluion u() = (cos ω cos ). ω2 6

7 (b) We will no show he graph here, bu we can describe wha happens as ω ges closer o. Rewrie he soluion as u() = 2( ω 2 ) sin ω sin + ω 2 2 We may view his as a vibraion wih a varying ampliude 2( ω 2 ) closer o he erm 2( ω 2 ) sin ω. As ω ges 2 becomes very large, which means he maximal ampliude is very big and also he period according o which he ampliude changes ges very big oo since i s equal o 4π/( ω). 7

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