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 seadysae pars of he soluion. (c) Plo he graph of he seadysae 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, seadysae 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 springmass 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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