CHAPTER. Differential Equations. x( ) ( ) ( ) ( ) ( ) 4.1 The Harmonic Oscillator. The Undamped Oscillator = 1, ( ) , x ( 0)

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1 CHAPTER Higher-Order Second-Order Linear Differenial Equaions The Harmonic Oscillaor The Undamed Oscillaor x+ x= x =, x =, The general soluion of he harmonic oscillaor equaion x+ x= is given by () x = ccos+ csin x = csin + c cos Subsiuing he iniial condiions x ( ) =, x x so c =, x+ x= x =, gives = c = = c = c = Hence, he IVP has he soluion x( ) cos, x ( ) =, x = = The general soluion of he harmonic oscillaor equaion x+ x= is given by () x = ccos+ csin x = csin+ c cos Subsiuing he iniial condiions x ( ) =, x x or c = c = Hence, he IVP has he soluion x =, gives = c = = c = x( ) = cos+ sin 9

2 SECTION The Harmonic Oscillaor 95 In olar form, his would be, x ( ) =, x+ 9x= x = π x () = cos The general soluion of he harmonic oscillaor equaion x+ 9x= is given by () x = c cos+ c sin x = c sin+ c cos Subsiuing he iniial condiions x ( ) =, x x x =, gives = c = = c = so c =, c = Hence, he IVP has he soluion In olar form, his would be x() = cos+ sin x cos δ () = ( ) where δ = an This would be in he firs quadran, x ( ) =, x x+ x= = The general soluion of he harmonic oscillaor equaion x+ x= is given by () x = ccos + csin x = c sin+ c cos Subsiuing he iniial condiions x ( ) =, x x c x =, gives = = = c = so c =, c = Hence, he IVP has he soluion In olar form, his would be x( ) = cos sin π x () = cos +

3 96 CHAPTER Higher-Order Linear Differenial Equaions, x ( ) =, 5 x+ 6x= x = The general soluion of he harmonic oscillaor equaion x+ 6x= is given by () x = ccos + csin x = c sin+ c cos Subsiuing he iniial condiions x ( ) =, x c x x =, gives = = = c = so c =, c = Hence, he IVP has he soluion, x ( ) =, 6 x+ 6x= x = cos x = The general soluion of he harmonic oscillaor equaion x+ 6x= is given by () x = ccos + csin x = c sin+ c cos Subsiuing he iniial condiions x ( ) =, x x so c =, c = The IVP has he soluion x =, we ge = c = = c = 7 x+ 6π x=, x () =, x ( ) 6π ω = = π = π x = c cos π + c sin π x = π c sin π + π c cos π sin x = x() = = c x () = π = π c c = x = sin π

4 SECTION The Harmonic Oscillaor 97 8 x+ π x=, x () =, x ( ) π π ω = = = π x = c cos π + c sin π π π π π x = c sin + c cos x() = = c x () = π = c π, c = π π x = cos + sin Grahing by Calculaor 9 y = cos+ sin The equaion ells us T = π and because π T =, ω = We hen measure he delay ω δ 8 which we can comue as he hase ω δ 8 = 8 The amliude A can be angle y 5 δ ω 8 T = π π A measured direcly giving A Hence, cos+ sin cos 8 Š5 Comare wih he algebraic form in Problem 5 y = cos+ sin y The equaion ells us T = π and because π T =, ω = We hen measure he delay ω δ 5 ω, which we can comue as he hase δ 5 = 5 The amliude A can be angle δ ω 5 Š 5 T = π A 8 measured direcly giving A Hence, cos+ sin cos 5 Š5

5 98 CHAPTER Higher-Order Linear Differenial Equaions y = 5cos+ sin π The equaion ells us ha eriod is T = and π because T =, ω = We hen measure he ω delay δ 5 ω, which we can comue as he hase angle δ 5 = 5 The amliude A can be measured direcly giving A 5 Hence, 5cos+ sin 5cos y δ ω 5 T = π/ A 5 π y = cos+ 5sin π The equaion ells us he eriod is T = and π because T =, ω = We hen measure he ω delay δ 5 ω, which we can comue as he hase angle δ 5 = 5 The amliude A can be measured direcly giving A 5 Hence, cos+ 5sin 5cos 5 5 Š5 y δ ω 5 T = π / A 5 y = cos5+ sin5 π equaion ells us ha eriod is T = and 5 π because T =, ω = 5 We hen measure he ω delay δ π or ω, which we can comue as 8 he hase angle δ 5 = The amliude A can be measured direcly giving A Hence, cos5+ sin cos 5 y δ ω T = π /5 A

6 SECTION The Harmonic Oscillaor 99 Alernae Forms for Sinusoidal Oscillaions We have ( ) = ( + ) = + Acos ω δ A cosω cosδ cosω cosδ Acosδ cosω Asinδ sinω = c cosω + c sinω where c = Acosδ, c = Asinδ Single-Wave Forms of Simle Harmonic Moion 5 cos+ sin By Equaion () c =, c =, and ω = By Equaion (5) yielding (Comare wih soluion o Problem 9) 6 cos sin A =, π δ = π cos+ sin = cos By Equaion () c =, c =, and ω = By Equaion (5) yielding A =, π δ = π cos sin = cos + Because c is osiive and c is negaive he hase angle is in he h quadran 7 cos+ sin By Equaion () c =, c =, and ω = By Equaion (5) yielding A =, δ = π π cos+ sin = cos Because c is negaive and c is osiive he hase angle is in he nd quadran

7 CHAPTER Higher-Order Linear Differenial Equaions 8 cos sin By Equaion (5) c =, c =, and ω = By Equaion (6) yielding A =, δ = 5π 5π cos sin = cos Because c and c are negaive, he hase angle is in he rd quadran Comonen Form of Harmonic Moion Using cos A+ B = cos Acos B sin Asin B, we wrie: { } 9 ( π) ( π) ( π) cos = cos cos sin sin = cos π π π cos + = coscos sin sin = cos sin π π π cos = coscos sin sin = cos+ sin = cos+ sin { } π π π cos = cos cos sin sin = cos+ sin Inerreing Oscillaor Soluions, x ( ) =, x+ x= x = Because ω =, we know he naural frequency is π he iniial condiions, we find he soluion (see Problem ) cos x =, which ells us he amliude is and he hase angle δ = radians, x ( ) =, x+ x= x = Hz and he eriod is π seconds Using Because ω = radians er second, we know he naural frequency is Hz (cycles er π second), and he eriod is π Using he iniial condiions, we find he soluion (see Problem ) which ells us he amliude is π x () = cos, and he hase angle is π δ = radians

8 , x ( ) =, 5 x+ 9x= x = SECTION The Harmonic Oscillaor Because ω = radians er second, we know he naural frequency is Hz (cycles er π second), and he eriod is π Using he iniial condiions, we find he soluion (see Problem ) x cos δ () = ( ) where δ = an, which ells us he amliude is and he hase angle is, x ( ) =, x 6 x+ x= = δ = an 8 radians Because ω = radians er second, we know he naural frequency is Hz (cycles er second), π and he eriod is π Using he iniial condiions, we find he soluion (see Problem ) which ells us he amliude is, x ( ) =, 7 x+ 6x= x = π x () = cos +, and he hase angle is π δ = radians Because ω = radians er second, we know he naural frequency is Hz (cycles er second), π and he eriod is π Using he iniial condiions, we find he soluion (see Problem 5) cos( ) x = π, which ells us he amliude is and he hase angle is δ = π radians

9 CHAPTER Higher-Order Linear Differenial Equaions, x ( ) =, 8 x+ 6x= x = Because ω = radians er second, we know he naural frequency is Hz (cycles er second), π and he eriod is π Using he iniial condiions, we find he soluion (see Problem 6) 9 π x () = cos, which ells us he amliude is and he hase angle is x+ 6π x=, x () =, x ( ) From Problem 7, x = sin π = π π δ = radians Amliude =, x = π cos π, hase angle δ = π π, and eriod T = ω = π = 8 x+ π x=, x () =, x ( ) = π r + π = r = π ± i π π x = ccos + csin = c π π π π π c x = csin + ccos π = π π x = cos + sin c = Amliude: A = + = 5 Relaing Grahs π x = 5 cos (a) See grah, nex age k (b) x+ 5x= ω = = 5 m From () x = c cos 5 + c sin 5 x() = c = so x() = c sin and c x () = cos

10 SECTION The Harmonic Oscillaor Alernaively, you could use (5) x = A cos (5 δ) π π x() = δ = so x() = A cos = A sin, A and x () = cos (c) See grah (d) Grah for b) and d) Grah for a) A A A 5A Amliudes are aroximaely,,,, and A 6 Phase Porrais For comarison of hase orrais, he main observaion is ha he elliical shae deends on ω, which is k in all of hese roblems because x+ kx= If ω =, rajecories are circular As ω increases above, ellises become aller and hinner As ω decreases from o, ellises become shorer and wider xmax The asec raio of = x ω max Oher observaions include: All hese hase orrais show closed elliical rajecories ha circulae clockwise The rajecory of Problem has a greaer radius han ha of Problem because he iniial condiion is furher from he origin The rajecories in Problems 6 and 7 are on he same ellise wih differen saring oins ha give differen soluion equaions

11 CHAPTER Higher-Order Linear Differenial Equaions x+ x= x = From Problem x() = cos(), so x ( ) = sin() x+ x= x = From Problem x() = cos() + sin(), so x ( ) = sin() + cos() x+ 9x= x = From Problem, x() = cos + sin, so x ( ) = sin + cos

12 SECTION The Harmonic Oscillaor 5 5 x+ x= x = From Problem, x() = cos sin, so x ( ) = sin cos 6 x+ 6x= x = From Problem 5, x() = cos, so x ( ) = sin 7 x+ 6x= x = From Problem 6, x() = sin, so x ( ) = cos

13 6 CHAPTER Higher-Order Linear Differenial Equaions 8 x+ 6π x= x = π From Problem 7, x() = sin π, so x ( ) = cos π 9 x+ π x= x = π π π From Problem 8, x() = cos + sin, so x ( ) = π sin π + π cos π Maching Problems B A D C

14 SECTION The Harmonic Oscillaor 7 Changing Frequencies (a) ω = 5 gives x curve wih lowes frequency (fewes hums); ω = gives he x ω = 5 ω = highes frequency (mos hums) (b) ω = 5 gives he innermos haselane rajecory; as ω increases, he amliude of x increases In Figure 8 he rajecory ha is no oally visible is he one for ω = π ω = π Deecive Work 5 (a) 8π The curve y = cos 5 8π and hase angle δ 5 is a sinusoidal curve wih eriod π, amliude A, π (b) From his grah we esimae ω =, A, and δ Thus, we have Pulling a Weigh 6 (a) The mass is m = kg ha π π π x () = Acos( ω δ) = cos cos cos sin sin = = cos+ sin 6( cos+ sin ) Because a force of 8 n sreches he sring 5 meers, we find 8 k = = 6 n m If we hen release he weigh, he IVP describing he moion of 5 he weigh is x+ 6x= or The soluion of he differenial equaion is x+ 8x=, x ( ) = 5, () cos( 8 ) x = A δ Using he iniial condiions, we ge he simle oscillaion () 5cos( 8) x = x =

15 8 CHAPTER Higher-Order Linear Differenial Equaions π π (b) Amliude = m; T = = sec, ω 8 (c) Seing 8 f = π cycles er second cos 8 =, we find ha he weigh will ass hrough equilibrium a of he eriod or afer π = 56 seconds 8 A ha ime velociy is moving away from original dislacemen Finding he Differenial Equaion π x ( 56) = sin m sec 7 (a) The mass is m = 5 gm, which means he force acing on he sring is 5 98 dynes This sreches he sring 5 cm, so he sring consan is 5 98 k = = 98 dynes cm 5 The mass is hen ulled down cm from is iniial dislacemen, giving x ( ) = (as long as we measure downward o be he osiive direcion, which is yical in hese x = Thus, he roblems) The iniial velociy of he mass is assumed o be zero, so IVP for he mass is 5 x+ 98x= or, x ( ) =, 5x+ 98x= x = (b) (c) The soluion of he differenial equaion found in ar (a) is x() = Acos δ = cos 5 5 In ar (b) he amliude is cm, hase angle is, he eriod is m 5 T = π = π sec, k 98 and he naural frequency is given by he recirocal f = = 7 oscillaions er second

16 Iniial-Value Problems SECTION The Harmonic Oscillaor 9 8 (a) The weigh is 6 lbs, so he mass is roughly 6 = slugs (See Table in ex) This mass sreches he sring foo, hence 6 k = = lb f This yields he equaion x + x=, or x+ 6x= The iniial condiions are ha he mass is ulled down inches ( foo) from equilibrium and hen given an uward velociy of f sec This gives he iniial condiions of x ( ) = f, x osiive = f/sec, using he engineering convenion ha for x, down is (b) We have he same equaion x+ 6x= x = f/sec One More Weigh 9 The mass is, bu he iniial condiions are ( ) m = = slugs The sring is sreched 8 x = f, 6 foo, so he sring consan is k = = lb f The iniial osiion of he mass is inches ( f) uward so x = The iniial moion is f sec uward, and hus x mass is x+ 6x=, x ( ) =, x = Hence, he equaion for he moion of he =, which has he soluion x() = cos8 sin 8

17 CHAPTER Higher-Order Linear Differenial Equaions Wriing his in olar form, we have x 5 A= c + c = + = c δ = an = an c 78 radians angle in rd quadran 6 8 Hence, we have he soluion in olar form 5 = cos 8 78 () ( ) x See figure Comaring Harmonic Moions Š 5 x ()= cos ( 8 78 ) Sring oscillaion 5 The eriod of simle harmonic moion is given by π T =, where ω = ω k m Noice ha his does no deend a all on our iniial condiions Period is he same so is he frequency, bu he amliude will be wice ha in he firs case Tesing Your Inuiion 5 x+ x+ x = Here we have a vibraing sring wih no fricion, bu a nonlinear resoring force F = x x ha is sronger han a urely linear force x For small dislacemen x he nonlinear F will no be much differen (for small x, x is very small), bu for larger x, he force F will be much sronger han in a linear sring; as F increases, he frequency of he vibraion increases This equaion is called Duffing s (srong) equaion, and he associaed srings are called srong srings 5 x+ x x = Here we have a vibraing sring wih no fricion, and a nonlinear resoring force F = x+ x For small dislacemen x he nonlinear erm x has lile effec, bu as x increases oward, he resoring force F diminishes (ie, he sring weakens when i is sreched a lo, and he resoring force becomes zero when x = ) The decreasing F causes decreasing frequency (and increasing eriod) This equaion is called Duffing s (weak) equaion, and he associaed srings are called weak srings

18 SECTION The Harmonic Oscillaor 5 x x= 5 This equaion describes a sring wih no fricion and a negaive resoring force You may wonder if here are such hysical sysems In he nex wo secions we will see ha his equaion describes he moion of an invered endulum ( Problems 58, 59), and i has soluions sinh and cosh ( Examle ), in conras o x+ x=, which has soluions sin and cos The resoring force for he equaion under discussion is always direced away from he equilibrium osiion; hence he soluion always moves away from he equilibrium, which us unsable x+ x + x= This equaion can be inerreed as describing he moion of a vibraing mass ha has infinie fricion x a =, bu fricion immediaely begins o diminish and aroaches zero as becomes very large You may simulae in your mind he moion of such a sysem Do you hink for large ha he oscillaion migh behave much like simle harmonic moion? (See Problem 68) x+ x x + x= 55 This is called van der Pol s equaion and describes oscillaions (mosly elecrical) where inernal fricion deends on he value of he deenden variable x Noe ha when x <, we acually have negaive fricion, so for a small dislacemen x we would exec he sysem o move away from he zero soluion (an unsable equilibrium) in he direcion of x = Bu when x >, we will have osiive fricion causing daming We will see in Problem 7 and in Chaer 7 ha here is a eriodic soluion beween small x and large x ha aracs all hese oher soluions 56 x+ x= Here we have a vibraing sring wih no fricion, bu he resoring force x ges sronger as ime asses Hence we exec o see no daming, bu faser vibraions as increases LR-Circui 57 (a) Wihou having a caacior o sore energy, we do no exec he curren in he circui o oscillae If here had been a consan volage V on in he as, we would exec he V curren o be (by Ohm s law) I = If we hen shu off he volage, we would exec R he curren o die off in he resence of a resisance (b) If a curren I asses hrough a resisor wih resisance R, hen he volage dro is RI; he volage dro across an inducor of inducance L is LI We obain he IVP:, LI + RI = I V = R

19 CHAPTER Higher-Order Linear Differenial Equaions (c) The soluion of he IVP is () I V RL = e R 8 (d) If R = ohms, L = 5 henries, V = vols, hen I() = e ohms LC-Circui 58 (a) Wih a nonzero iniial curren and no resisance, we do no exec he curren o dam o zero We would exec an oscillaory curren due o he caacior Thus he charge on he caacior would oscillae indefiniely The exac behavior deends on he iniial condiions and he values of he inducance and caaciance (b) Kirchoff s volage law saes ha he sum of he volage dros around he circui is equal o he imressed volage source Hence, we have LI + I = C or, in erms of he charge across he caacior, we have he IVP LQ + Q = C Q = 5, Q ( ) =, (c) The soluion of he IVP is () ( LC ) sin Q = 5 LC This agrees wih he oscillaory behavior rediced in ar (a) (d) Wih values L = henries, A Pendulum Exerimen C = farads, he charge on he caacior is () Q ( ) sin = 5 = sin 59 The endulum equaion is g θ + sinθ = L For small θ, we can aroximae sinθ θ, giving he differenial equaion g θ + θ = L

20 SECTION The Harmonic Oscillaor This is he equaion of simle harmonic moion wih circular frequency ω = g L frequency f = Hence, he eriod of moion is T = = π π L f g g L, and naural T T earh sun g g sun = =, = 6 earh Changing ino Sysems 6 x x + x= 7 cos x = y 6 6 y = ( x+ y+ 7 cos ) Lq + Rq + q = V () c q = I I = q RI + V() L c 5q + 5q + q= 5cos q = I I = q I + cos 5 6 x+ x + x= sin > sin x+ x + x= x = y x sin y = y+ 6 x+ 6x= sin x + x= sin x = y y = x+ sin Circular Moion 65 Wriing he moion in erms of olar coordinaes r and θ and using he fac ha he angular velociy is consan, we have θ = ω (a consan) We also know he aricle moves along a circle of consan radius, which makes r a consan We hen have he relaion x= rcosθ, and hence

21 CHAPTER Higher-Order Linear Differenial Equaions ( rsinθ) θ ( θ ) θ ( θ) θ x = x= rsin rcos Because θ =, θ = ω, we arrive a he differenial equaion Anoher Harmonic Moion x+ ω = x 66 For simle harmonic moion he circular frequency ω is kr ω = mr + I, so he naural frequency f is kr f = π mr + I Moion of a Buoy 67 The buoy moves in simle harmonic moion, so he eriod is π m T = 7 = = π ω k We have one equaion in wo unknowns, bu he buoyancy equaion yields he second equaion If we ush he buoy down foo, he force uwards will be F = V ρ, where V is he submerged volume and ρ is he densiy of waer In his case, V = π r h, r = 9 inches = 75 f, h = f, and 9 ρ = 65 f sec, so he force required o ush he buoy down foo is π ()( 65 ) lbs 6 Bu k is he force divided by disance, so k = = lbs f Finally, solving for m in he kt equaion for T, we ge m =, and subsiuing in all of our numbers, we arrive a m π mg = = 657 lbs slugs (see Table in he ex) The buoy weighs Los Angeles o Tokyo 68 (a) Along he unnel, mx = kr cosθ = kx x() = d if x is measured osiive o he lef of he cener of he unnel x () = means ha he rain sars from res (as soon as a brake is released)

22 SECTION The Harmonic Oscillaor 5 (b) The soluion o he IVP in ar (a) is x() = c cos ω + c sin ω, where ω = k m A he surface of he earh mg = kr where r = R, so ω k q = = m R Leing x() = d yields c = d, while leing x () = yields c = Hence we have q x() = dcos R For he rain o go from LA o Tokyo, x() goes from d o d and q R goes from o π Hence, f = π R q = π mi 58 f/mi f/sec = 55 sec 5 minues (c) The soluion f = R π from ar (b) does no deend on he locaion of he oins on he q earh s surface; π, R, and q are all consan Facoring Ou Fricion ( b m) 69 (a) Leing x() = e X (), we have b ( b m) ( b m) x () = e X() + e X () m b ( b m) b ( b m) ( b m) x() = e X () + e X () + e X () m m

23 6 CHAPTER Higher-Order Linear Differenial Equaions Subsiuing his ino he original equaion () and dividing hrough by a e ( b m), we arrive b b b [ ] = m X X X b X X k X m m m Rearranging erms gives b b mx + [ b + b] X + + k X = m m or b mx + k X = m (b) If we assume b k >, hen divide by m and le m ω = mk b ) m we find he soluion of his DE in X is cosω sinω cos( ω δ) X = c + c = A Thus, we have () () ( b m) ( b m) x = e X = Ae cos ω δ Suggesed Journal Enry 7 Suden Projec

24 SECTION Real Characerisic Roos 7 Real Characerisic Roos Real Characerisic Roos y = The characerisic equaion is is y y = The characerisic equaion is y 9y= The characerisic equaion is y y = The characerisic equaion is r =, so here is double roo a r = Thus, he general soluion r y = ce + ce = c+ c r =, which has roos, Thus, he general soluion is y = c+ ce r 9=, which has roos, Thus, he general soluion is y = ce + ce r =, which has roos, Thus, he general soluion is 5 y y + y = The characerisic equaion is r roos, Thus, he general soluion is 6 y y y= The characerisic equaion is roos, Thus, he general soluion is 7 y + y + y= The characerisic equaion is r r y = ce+ ce r+ =, which facors ino ( r )( r ) =, and hence has y = ce+ ce r =, which facors ino ( r )( r+ ) =, and hence has y = ce + ce he double roo, Thus, he general soluion is + r+ =, which facors ino ( r+ )( r+ ) =, and hence has y = ce + ce

25 8 CHAPTER Higher-Order Linear Differenial Equaions 8 y y + y = The characerisic equaion is r r + =, which facors ino ( r )( r ) =, and hence has he double roo, Thus, he general soluion is 9 y y + y = The characerisic equaion is y = ce + ce r r + =, which facors ino ( r )( r ) =, and hence has roos, Thus, he general soluion is y 6y + 9y= The characerisic equaion is r he double roo, Thus, he general soluion is y 8y + 6y= The characerisic equaion is r 8r 6 y = ce + ce 6r+ 9=, which facors ino ( r )( r ) =, and hence has y = ce + ce has he double roo, Thus, he general soluion is y y 6y = The characerisic equaion is roos, Thus, he general soluion is y + y y = The characerisic equaion is r r + =, which facors ino ( r )( r ) y = ce + ce =, and hence r 6=, which facors ino ( r+ )( r ) =, and hence has y = ce + ce hence has roos +, Thus, he general soluion is 9y + 6y + y = The characerisic equaion is 9r 6r double roo, + r =, which facors ino ( r+ )( r+ + ) =, and () ( ) y = e ce + c e Thus, he general soluion is + + =, which facors ino y = ce + ce r + =, and hence has he

26 SECTION Real Characerisic Roos 9 Iniial Values Secified =, y ( ) =, 5 y 5y y = The characerisic equaion of he differenial equaion is r 5 =, which facors ino r 5 r+ 5 =, and hus has roos 5, 5 Hence, Subsiuing in he iniial condiions ( ) 5 5 y = ce + c e y = gives c c + = Subsiuing in 5c 5c = Solving for c, c gives c = c = Thus he general soluion is 6 y y y + =, y ( ) =, y = 5 5 y() = e + e y = gives The characerisic equaion of he differenial equaion is r + r =, which facors ino r+ r =, and hus has roos, Thus, he general soluion is Subsiuing ino y ( ) =, 7 y y y + + =, y ( ) =, The characerisic equaion is y = ce + c e y = yields c =, c = y = r y() = e + e he double roo, Thus, he general soluion is Subsiuing ino y ( ) =, =, y ( ) =, 8 y 9y, so + r+ =, which facors ino ( r+ )( r+ ) =, and hence has y = ce + c e y = yields c =, c =, so y = The characerisic equaion is 9 roos are, Thus, he general soluion is Subsiuing ino y ( ) =, e y = r =, which facors ino ( r )( r ) y = ce + c e y = yields c = c =, so y() = e e + =, and hence has

27 CHAPTER Higher-Order Linear Differenial Equaions 9 y 6y 9y + =, y ( ) =, The characerisic equaion is y = he double roo, Thus, he general soluion is Subsiuing ino y ( ) =, y y 6y + =, y ( ) =, The characerisic equaion is r 6r+ 9=, which facors ino ( r )( r ) =, and hence has y = ce + ce y = yields c =, c =, so y = roos, Thus, he general soluion is Subsiuing ino y ( ) =, r e y = + r 6=, which facors ino ( r+ )( r ) =, and hence has y = ce + c e y = yields c =, c = 5 y y = y() =, y() = r r = r(r ) = r =, (Characerisic equaion) 5 y() = e + e 5 5, so y = c + c e = c + c y = c e = c, c = y = e y y y = y() =, y () = r r = (Characerisic equaion) (r + )(r 6) = r =, 6 y = ce + c e 6 y = ce + 6c e 6 y() = c + c = 7 c =, c = y () = c + 6c = 8 8 y = e + e 6

28 SECTION Real Characerisic Roos Bases and Soluion Saces y y = r r = (Characerisic equaion) r(r ) = r =, Basis: {, e } Soluion Sace: {y y = c + c e ; c, c } y y + 5y= r r + 5 = (Characerisic equaion) (r 5) = r = 5, 5 Basis: {e 5, e 5 } Soluion Sace: {y y = c e 5 + c e 5 ; c, c } 5 5y y 5y= 5r r 5 = (Characerisic equaion) r r = (r )(r + ) = r =, Basis: {e, e } Soluion Sace: {y y = c e + c e ; c, c } 6 y + y + = r + r+ = (Characerisic equaion) ( r+ )( r+ ) = r =, Basis: { e, e } Soluion Sace: { y y= ce + ce ; c, c } Oher Bases 7 y y = r = (Characerisic equaion) r = ± {e, e } is a basis To show {cosh, sinh } is a basis, we need only show ha cosh and sinh are linearly indeenden soluions: cosh sinh W = = cosh sinh sinh cosh

29 CHAPTER Higher-Order Linear Differenial Equaions cosh = e + e e + + e = sinh e e e e = = so cosh sinh = and W = cosh, sinh are linearly indeenden Subsiue y = cosh, y = sinh, y = cosh Then y y = cosh cosh = y = cosh is a soluion In similar fashion, we can show ha y = sinh is also a soluion To show ha { e,cosh } Then: is a basis, we use he facs ha e and cosh are soluions W = e cosh e sinh = e + e e e e e = (e ) (e + ) = e and cosh are linearly indeenden 8 y = r = (Characerisic equaion) so ha r =, Basis: {, } To show { +, } is also a basis: Noe ha for boh y = + and y =, y =, so boh are soluions W = + = ( + ) ( ) = +, are linearly indeenden To show {, } is anoher basis: Noe ha for boh y = and y =, y =, so boh are soluions W = - 6 ( ) =, are linearly indeenden

30 SECTION Real Characerisic Roos The Wronskian Tes 9 W = = ( + ) = ( + ) ( ) = Yes, { +,, +, } is a basis for he soluion sace for y () = W = e e e e (5+ ) e 5e e = e (5+ ) e 5e 5e = e + ( e ) = 5e 5 Yes, {e 5, e 5, e 5 } is a basis for he soluion sace for y y + 5y = The given se has only hree soluions, so i canno be a basis A basis for he soluion sace of y () = mus have linearly indeenden soluions Soring Grahs

31 CHAPTER Higher-Order Linear Differenial Equaions Relaing Grahs For Problems 5, x+ 5x + 6x= has (from Examle ) soluions x () = ce + ce () x () = ce ce () (a), (b) x() x () c + c = c c = c =, c = (c) From () in box, x() = e + e For >, each erm diminishes as increases; he resul remains negaive, below he -axis For <, each exonenial increases as decreases; he negaive erm cancels he osiive erm when e = e or e = 5, ha is, when = ln 5 5 which looks abou righ on he x-grah (d) From (), x ( ) = 6e 6e = 6e ( e ) which is always osiive for >, decreasing as increases x () reaches a maximum when x = e + 8e = + e = e =, so = ln 6, which looks abou righ on he x -grah

32 SECTION Real Characerisic Roos 5 (a) (b) (b) x() 5 x () c + c = 5 c c = Because all roblems for finding c i are of ye Ac = b, we solve for c= A b We have A = so A =, and here 5 b = so c = = (c) From () in box on revious age, x()= 5e e As increases from zero, boh exonenials decrease wih heir sum remaining osiive, which agrees wih he x grah (d) From (), x ( ) = e e = e ( + e ) For >, his quaniy is always negaive, and as long as increases, each erm ges closer o zero, in agreemen wih x -grah

33 6 CHAPTER Higher-Order Linear Differenial Equaions 5 (a) x() and x () 8 (b) From he mehod of (b), c 8 c = A = = so from (), x() = 8e + 8e (c) For >, e > e so he sum is always negaive and aroaches zero as increases, in agreemen wih he x grah (d) From () x ( ) = +6e e =, so e e =, which yields e = or 6, which looks abou righ on he x -grah For > 6, x > and decreases oward zero as increases

34 SECTION Real Characerisic Roos 7 For Problems 6 9, x x 6x= has (from Examle ) soluions x () = ce + ce () x = ce + ce () 6 (a) (b) From () c + c = From () c + c = x() = e + e x () = e + e 5 5 c = ; c = 5 5 (c) 7 (a) For >, e < e, so x() is always osiive, and as increases, so does x() This resul agrees wih he x-grah For <, e < e so x() is always negaive, and as becomes more negaive, x() becomes more negaive (d) x ( ) is always osiive For >, e < e, so he second erm dominaes as increases, and x ( ) increases as well These facs are in agreemen wih he x -grah

35 8 CHAPTER Higher-Order Linear Differenial Equaions (b) From (), c + c = From (), c + c = 6 x() = e + e 5 5 x () = e + e 5 5 c = 6 ; c = 5 5 (c) x() is always osiive 8 (a) For >, as increases, he firs erm decreases oward and he second erm increases ever more raidly, in agreemen wih he x-grah (d) For >, e > e, so x ( ) is osiive, and x ( ) increases as increases, as shown on he x grah For <, he firs erm will dominae and x ( ) will be negaive, ever more so as becomes more negaive, in agreemen wih he x -grah (b) From (), c + c = From (), c + c = c = 9 ; c 6 = x() = e e x () =+ e e 5 5 (c) x() is always negaive, wih a maximum a = (See ar (d) and se x ( ) = ) These facs agree wih he x grah (d) For >, e > e so he negaive erm dominaes in x ( ) and x ( ) is negaive, ever more so as increases For <, e > e so he osiive erm dominaes in x ( ) and x ( ) is osiive, ever more so as becomes more negaive These facs agree wih he x grah

36 SECTION Real Characerisic Roos 9 9 (a) (b) From (), c + c = From (), c + c = c = ; c = 5 5 x() = e e 5 5 x () = e e 5 5 (c) For > he second erm dominaes, so x() is negaive, ever more so as increases For < he firs erm dominaes, so x() is osiive, ever more so as becomes more negaive These facs agree wih he x grah (d) x ( ) is always negaive The maximum value will occur when =, as shown on he x - grah Phase Porrais Careful insecion shows: (B) (D) (A) (C)

37 CHAPTER Higher-Order Linear Differenial Equaions Indeenden Soluions Leing ce r r + ce = for all, hen by seing = and = we have, resecively c + c = r r ce + c e = When r r hen hese equaions have he unique soluion c = c =, which shows he given r r funcions e, e are linearly indeenden for r r Second Soluion 5 Subsiuing y = v( ) e b a ino gives ay + by + cy = b y = ve ve a b a b a b a b b a b b a y = v e ve + ve a a Subsiuing vv,, v ino he differenial equaion gives he new equaion (afer dividing by b a e ) b b b = a v v v b v v cv a a a Simlifying gives Because we have assumed roven b b av c v = a = ac, we have he equaion v =, which was he condiion o be Indeendence Again 6 Seing b a b a + ce = ce for all, we se in aricular = and hen = These yield, resecively, he equaions c = + = b a b a ce ce which have he unique soluion c = c = Hence, he given funcions are linearly indeenden

38 SECTION Real Characerisic Roos Reeaed Roos, Long-Term Behavior 7 Because b a e aroaches as (for a, b > ), we know he firs erm ends oward zero For he second erm we need only verify ha e b a = e b a does as well To use l Hôial s rule, we comue he derivaives of boh he numeraor and denominaor of he revious exression, geing b e, b a a which clearly aroaches as Then l Hôial s rule assures us ha he given exression ( b a) e aroaches as well Negaive Roos 8 We have r = b± b mk, so in he overdamed case where b mk >, hese characerisic roos are real Because m and k are boh nonnegaive, b mk < b causing and r = b+ b mk o be a negaive sum of negaive and osiive erms r = b b mk o be a negaive sum of wo negaive erms Circuis and Srings 9 (a) The LRC equaion is hold: LQ + RQ + Q =, hence he following discriminan condiions C L Δ= R < C ( underdamed) L Δ= R = C criically damed L Δ= R > C ( overdamed )

39 CHAPTER Higher-Order Linear Differenial Equaions (b) The condiions in ar (a) can be wrien L R < underdamed C L R = criically damed C L R > ( overdamed ) C These corresond o he analogy ha m, b, and k corresond resecively o L, R, and C (see Table in he exbook) A Tes of Your Inuiion 5 Inuiively, a curve whose rae of increase is roorional o is heigh will increase very raidly as he heigh increases On he oher hand, uward curvaure doesn necessarily imly ha he funcion is increasing! (The curve e has uward curvaure, ye decreases o as ) In his case, he resricion ha y ( ) = will cause he second curve o increase, bu robably no nearly as raidly as he firs curve Solving he equaions, he IVP y = y, y ( ) = has he soluion y = e, whereas he second curve described by y y soluion y() = e + e The firs curve is indeed above he second curve An Overdamed Sring 5 (a) The soluion of an overdamed equaion has he form r r x = ce + c e =, y ( ) =, y = has he Suose ha r r r + ce = for some Because e is never zero, we can divide by e o ge Solving for gives ce = ln c r r c r ( ) ce r r c + = This unique number is he only value for which he curve may ass hrough If he argumen of he logarihm is negaive or if he value of is negaive, hen he soluion does no cross he equilibrium oin (b) By a similar argumen, we can show ha he derivaive x ( ) also has one zero

40 SECTION Real Characerisic Roos A Criically Damed Sring 5 (a) Suose r ( c c ) e + = r We can divide by he nonnegaive quaniy e geing he equaion c + c =, which c has he unique soluion = Hence, he soluion of a criically damed equaion can c ass hrough he equilibrium a mos once If he value of is negaive, hen he soluion does no cross he equilibrium oin (b) By a similar argumen, we can show ha he derivaive x ( ) has one zero Linking Grahs Afer insecion, we have labeled he y and y grahs as follows 5 y 5 y' 5 y' 5 5 y 5 5 = 5 5

41 CHAPTER Higher-Order Linear Differenial Equaions 55 Damed Vibraion 56 The IVP roblem is The soluion is x+ x + x=, x ( ) = in = f, x() = e + e x = f sec This is zero only for =, whereas he hysical sysem does no sar before = Surge Funcions 57 For mx + bx + kx =, le m =, find b, k and iniial condiions for he soluion x = Ae r x + bx + kx = r + br+ k = (characerisic equaion) r = b± b k b b b k = o obain reeaed roos, r =, x = c e r + c e r r ( r r x = rce + c r e + e ) c = = x() c = A = x () x () = c b = r, k = b and from above we know k = r so ha k = ±r, for k >

42 SECTION Real Characerisic Roos 5 Resuls: r and A are given, and LRC-Circui I 58 (a) b = r k = r x() = x () = A LQ + RQ + Q = Q + Q + 5Q = C, Q ( ) = 99, Q = 5 5 (b) Q = e + e (c) I( ) Q( ) 5e = = 5e (d) As, Q and I( ) LRC-Circui II 59 (a) LQ + RQ + Q = Q + 5Q + 5Q = C, Q ( ) = 5, Q = 5 (b) 5 Q = e 5e (c) I( ) Q( ) 5e = = + 5e (d) As, Q and I( ) The Euler-Cauchy Equaion a y + by + cy = r 6 Le y () =, so y = r r r y = r r Hence r r r a y + by + cy = ar r + br + c = Dividing by r yields he characerisic equaion ar r + br + c =, which can be wrien as ar + b a r + c = If r and r are wo disinc roos of his equaion, we have soluions () y = y r r =

43 6 CHAPTER Higher-Order Linear Differenial Equaions Because hese wo funcions are clearly linearly indeenden (one no a consan mulile of he oher) for r r, we have for > r r = + y c c The Euler-Cauchy Equaion wih Disinc Roos For Problems 6 65, see Problem 6 for he form of he characerisic equaion for he Euler-Cauchy DE 6 y + y y = In his case a =, b =, c = Hence, we have roos r =, r =, and hus, so he characerisic equaion is r r + r = r + r = r+ r = y = c + c 6 y + 8y y= In his case a =, b = 8, c =, so he characerisic equaion is Hence, we have roos r =, r = r r + 8r = r + r = r r+ =, and hus y = c + c 6 y + y + y= In his case a =, b =, c =, so he characerisic equaion is Hence, we have roos r =, r =, and hus r r + r+ = r + r+ = r+ r+ = y = c + c

44 SECTION Real Characerisic Roos 7 6 y + y y = In his case a =, b =, c =, so he characerisic equaion is Hence, we have roos r =, r =, and hus Reeaed Euler-Cauchy Roos 65 We are given ha he characerisic equaion of Euler s equaion r r + r = r + r = r r+ = y = c + c ar + b a r + c = a y + by + cy = has a double roo of r Hence, we have one soluion y = To verify ha ln is also a soluion, we differeniae r r = ln +, y r ( ) ln ( ) ( ) ln ( ) r r r r r = + + = + y r r r r r r r By direc subsiuion we have r r r r r a y + by + cy = a r r ln r b r ln c ln We know ha r r = ar r + br + c ln + a r + b ar r + br + c =, so his las exression becomes simly Thus he roo of he characerisic equaion is To verify ha r and roo of he characerisic equaion), we se ( ) r a y + by + cy = a r + b r r b a r =, which makes his exression zero a ln are linearly indeenden (where r c + c ln = for secific values = and, which give, resecively, he equaions r c c + ln= r r c = r b a r = is he double a and yields he unique soluion c = c = Hence, r and soluions r ln are linearly indeenden

45 8 CHAPTER Higher-Order Linear Differenial Equaions Soluions for Reeaed Euler-Cauchy Roos For Problems 66 and 67 use he resul of Problem 6, ln r y = c + c r 66 y + 5y + y = In his case, a =, b = 5, and c =, so our characerisic equaion for r is double roo a The general soluion is = + y c c ln r + r+ =, wih a for > 67 y y + y= In his case, a =, b =, and c =, so our characerisic equaion for r is a double roo a The general soluion is = + y c c ln r r+ =, wih for > 68 9 y + y + y = Euler-Cauchy mehod: y = m, > 9m(m ) + m + = 9m 6m + = (m ) = m = (characerisic equaion) / / y () = c + c ln 69 y + 8y + y = Euler-Cauchy mehod: y = m, > m(m ) + 8m + = m + m + = (m + ) = m = / / y () = c + c ln Comuer: Phase-Plane Trajecories 7 (a) y = e + e The roos of he characerisic equaion are and, so he characerisic equaion is y( ) saisfies he differenial equaion r+ r+ = r + r+ = y + y + y=

46 (b) To find he IC for he rajecory of y( ) in yy sace we differeniae y(, ) geing y = e e SECTION Real Characerisic Roos 9 (c) The IC of he given rajecory of ( y( ), y ( ) ) (, ) (, 5) in yy sace is y y = We lo he rajecory saring a, 5 along wih a few oher rajecories in yy sace DE rajecories in yy sace 8 7 y( ) e = + e (a) The roos of he characerisic equaion are and 8, so he characerisic equaion is y( ) saisfies he differenial equaion r+ r+ 8 = r + 9r+ 8= y + 9y + 8y = (b) The derivaive is 8 8 = y e e The IC for he given rajecory in yy sace is ( (, ) ( ) ) (, 9) y y = (c) We lo his rajecory in yy sace DE rajecory in yy sace 7 y() = e + e (a) The roos of he characerisic equaion are and, so he characerisic equaion is y( ) saisfies he differenial equaion ( r )( r ) r + = = y y =

47 CHAPTER Higher-Order Linear Differenial Equaions (b) The derivaive is y y ( ) = e e The IC for he given rajecory in yy (c) sace is ( (, ) ( ) ) (, ) y y = We lo his and a few oher rajecories of his DE in yy sace (, ) y DE rajecories in yy sace 7 y( ) e = + e (a) The characerisic equaion has a double roo a, so he characerisic equaion is (b) y( ) saisfies he differenial equaion The derivaive is e y = r+ = r + r+ = y + y + y= y' 5 (c) The IC for he given rajecory in yy sace is y =, y = See he figure o he righ (, ) y DE rajecory in yy sace y = + e 7 (a) The roos of he characerisic equaion are and, so he characerisic equaion is y( ) saisfies he differenial equaion r r = r r = y y =

48 SECTION Real Characerisic Roos (b) The derivaive is e = y The IC for he given rajecory in yy sace is ( (, ) ( ) ) ( 5, ) y y = (c) See he figure o he righ DE rajecories in yy sace Reducion of Order 75 (a) Le y = vy and Then y = vy + vy y = v y + vy + vy y + x y + q x y = v y + vy + vy + vy + vy + qvy = Because y + y + qy =, cancel he erms involving v, and arrive a he new equaion ( ) yv + y + x y v = (b) Seing v = w and using he fac ha ydx = dy, we obain ( ) y + ( x) y yw + y + x y w= w + = By convenion, he osiive sign is chosen y w dw y x y = dx w y y ln w = ( x) dx y ln w = dy ( x) dx y ln w = ln y x dx ( x) dx e w=± = v y v =± e x dx y

49 CHAPTER Higher-Order Linear Differenial Equaions (c) If v is a consan funcion on I, hen v and w because v = w The condiion w conradics our work in ar (b) as ln w where w = is undefined Because v is no consan on I, {, } y y is a linearly indeenden se of I Reducion of Order: Second Soluion 76 y 6y + 9y=, y = e We idenify = 6, so () d = 6 Subsiuing in he formula develoed in Problem 75, we have d () 6 e e y = y d = e d = e y e () 77 y y + y =, y = e We won use he formula his ime We simly redo he ses in Problem 75 We seek a second soluion of he form y = vy = ve Differeniaing, we have Subsiuing ino he equaion we obain y = ve + ve y = v e + ve + ve y y + y = v e = Dividing by e gives v = or v = c + c Hence, we have found new soluions y = ve = ce + c e Because y = e, we le c =, c =, yielding a second indeenden soluion y = e

50 SECTION Real Characerisic Roos 78 y y + y =, y = We won use he formula his ime We simly redo he ses in Problem 75 We seek a second soluion of he form y = vy = v Differeniaing, we have y = v + v y = v + v Subsiuing ino he equaion we obain y y + y = v + v = Leing w= v and dividing by yields w + w= We can solve by inegraing he facor mehod, geing w= c Inegraing we find v= c ln + c, so y = v= cln + c Leing c =, c =, we ge a second linearly indeenden soluion 79 + y y + y=, y = y = ln We won use he formula his ime We simly redo he ses in Problem 75 We seek a second soluion of he form y = vy = v Differeniaing yields y = v + v y = v + v Subsiuing ino he equaion we ge Leing w v = and dividing by ( + ) + y y + y = + v + v =, we can solve he new equaion using he inegraing facor mehod, geing ( + ) d = ln + + ln = ln +

51 CHAPTER Higher-Order Linear Differenial Equaions We arrive a + w= c = c + c Inegraing his, we ge so v= c + c, ( ) y = v= c + c Leing c =, c = we ge a second linearly indeenden soluion Classical Equaions 8 y y y + =, Leing y vy v( ) y = y = (Hermie s Equaion) = =, we have y = v v y = v 8v v and erform he long division, yielding he equaion 8 v + + v = Leing w= v and solving he firs-order equaion in w, we ge w= ce To find y we simly le c = and inegrae o ge Mulilying by ( ) 8 y y y + =, Leing y = vy = v, we have ( ) v= e d yields a final answer of () = ( ) ( ) y e d y = (Chebyshev s Equaion) y = v + v, y = v + v,

52 hence we have he equaion Dividing by ( ) y y + y = v + v =, and leing w= v, Using arial fracions yields ( ) so our inegraing facor is and w + w= ( ) d = ln + ln + ln +, w= c Leing c = and mulilying by yields a final answer of y () = v= d SECTION Real Characerisic Roos 5 This is a erfec examle of a formula ha does no ell us much abou how he soluions behave Check ou he IDE ool Chebyshev s Equaion o see he value of grahical soluions 8 y + ( ) y + y =, Leing y vy v( ) y = (Laguerre s Equaion) = =, we have hence we have he equaion = ( ) +, y v v y = v + v, y + y + y = v + + v = Dividing by ( ) and leing w= v yields + w + w= ( ) Hence by use of arial fracions, our inegraing facor is e w= C ( ) e + + d u = so ha Leing c = and mulilying by yields a final answer of e y = v = d ( )

53 6 CHAPTER Higher-Order Linear Differenial Equaions Lagrange s Adjoin Equaion 8 (a) (b) Differeniaing he righ side of we obain d μ() y y y () y g() y d [ + + ] = [ μ + ] μ y + μy + μy = μ y+ μy + gy + gy Seing he coefficiens of y, y, y equal, we find for y : μ = μ (no informaion) for y : μ = μ + g for y : μ = g The las equaion yields g = μd and subsiuing his ino he second equaion, and differeniaing, gives a differenial equaion for he inegraing facor μ μ + μ = (c) We erform he differeniaion on he righ-hand-side of he given equaion, yielding [ ] μ() y + () y + q() y = μy + μ y + g() y + g () y Mulilying ou he lef-hand side and subracing yields μ () μ g () y + μq () g () y= [ ] [ ] Seing he firs se of coefficiens equal o yields μ = μ g, hence μ = μ + μ g, so ha g = μ + μ + μ The second se of coefficiens yields μ q g = so ha g = μq Seing hese wo equaions for g equal o each oher yields μ μ + ( q ) μ = which was o be shown Suggesed Journal Enry 8 Suden Projec

54 SECTION Comlex Characerisic Roos 7 Comlex Characerisic Roos Soluions in General y + 9y= The characerisic equaion is r + 9=, which has roos i, i The general soluion is y = c cos+ c sin y + y + y= The characerisic equaion is r + r+ =, which has roos () y = e ccos + csin ± i The general soluion is y y + 5y= The characerisic equaion is r r+ 5=, which has roos ± i The general soluion is ( cos sin ) y = e c + c y + y + 8y= The characerisic equaion is r + r+ 8=, which has roos ± i 7 The general soluion is () ( cos 7 sin 7 ) y = e c + c 5 y + y + y = The characerisic equaion is r + r+ =, which has roos ± i The general soluion is () ( cos sin ) y = e c + c 6 y y + 7y= The characerisic equaion is r r+ 7=, which has roos ± i The general soluion is () ( cos sin ) y = e c + c 7 y y + 6y = The characerisic equaion is r r+ 6 =, which has roos 5 + i The general soluion is 5 ( cos sin ) y = e c + c

55 8 CHAPTER Higher-Order Linear Differenial Equaions 8 y + y + 9y = The characerisic equaion is r + r+ 9=, which has roos is () y = e ccos + csin 9 y y + y= ± i The general soluion The characerisic equaion is y + y + y= The characerisic equaion is r r+ =, which has roos ± i The general soluion is y () e = ccos + csin r + r+ =, which has roos () 7 7 y = e ccos + csin 7 ± i The general soluion is Iniial-Value Problems + =, y ( ) =, y y y = The characerisic equaion is r + =, which has roos ± i The general soluion is y = ccos + csin Subsiuing his ino he iniial condiions gives y( ) = c =, y c soluion of he iniial-value roblem is y y y + =, y ( ) =, The characerisic equaion is y = r y() = cos sin = = Hence, he r+ =, which has roos ± i The general soluion is ( cos sin ) y = e c + c Subsiuing his ino he iniial condiions yields y( ) = c =, y ( ) = c+ c =, resuling in c =, c = Hence, he soluion of he iniial-value roblem is y e cos sin () =

56 y y y + + =, y ( ) =, y = SECTION Comlex Characerisic Roos 9 The characerisic equaion is is r + r+ =, which has roos ± i Hence, he general soluion ( cos sin ) y = e c + c Subsiuing his ino he iniial condiions yields y( ) = c =, y c c c =, c = Hence, he soluion of he iniial-value roblem is y y y + =, y ( ) =, From Problem 6, y = ( cos sin ) y = e + () y e ccos () c = + sin () Subsiuing his ino he iniial condiions yields y ( ) =, c = 5 y y 7y Hence, he soluion of he iniial-value roblem is + =, y ( ) =, From Problem 6, y = y() = e sin { } () cos ( ) sin ( ) y = e c + c Subsiuing his ino he iniial condiions yields y ( ) =, = =, resuling in y =, resuling in c =, y =, resuling in c =, c = Hence, he soluion of he iniial-value roblem is 6 y y 5y + + =, y ( ) =, y = () sin ( ) y = e The characerisic equaion is r + r+ 5=, which has roos ± i Hence, he general soluion is ( cos sin ) y = e c + c

57 5 CHAPTER Higher-Order Linear Differenial Equaions Subsiuing his ino he iniial condiions yields y ( ) =, Hence, he soluion of he iniial-value roblem is Working Backwards 7 ( r ) = r r + r y y + y y= cos y = e y =, resuling in c =, c = 8 ( r )( r ( i))( r ( + i)) = r 6r + r 8 = y 6y + y 8y= 9 ( r )( r ( + i))( r ( i)) = r 6r + r y 6y + y y= ( r )( r ( + i))( r ( i)) = r r + r + 6r = () y 6y + y + 6y y = Maching Problems y y = r =, y() = c + c e y + y = r =, y() = c + c e Grah D Grah B y + y + y= r =, y() = c e + c e Grah A y 5y + 6y = r =, y() = c e + c e Grah C 5 y + y + y= r = ± i y() = e ccos + csin Grah G 6 y + y = r = ±i y() = c cos + c sin Grah F 7 y + y + y = r =, y () = ( c+ ce ) Grah E

58 SECTION Comlex Characerisic Roos 5 8 ± i y y + y= r = y () = e ccos + csin Grah H Euler s Formula 9 (a) The Maclaurin series for x e is iθ (b) e iθ ( iθ) ( iθ) ( iθ) x n e = + x+ x + x + + x +!! n! n = !! n! (c) Using he given ideniies for i, we can wrie iθ n e = + iθ + ( iθ) + ( iθ) + + ( iθ) +!! n! 5 = θ + θ + + i θ θ + θ = cosθ + isinθ!!! 5! (d) Done in ar (c) (e) Done in ar (c) Long-Term Behavior of Soluions r <, r < When r r, he soluion is r = + y ce ce and goes o as When r = r = r <, he soluion has he form r r r y = ce + c e In his case using l Hôial s rule we rove he second erm r < r <, r = The soluion r y = ce + c aroaches he consan c as because r < r α βi α = ±, y e ( ccos β csin β ) r e goes o zero as when = + For β he soluion y oscillaes wih decreasing amliude when α < ; oscillaes wih increasing amliude when α > ; oscillaes wih consan amliude when α =

59 5 CHAPTER Higher-Order Linear Differenial Equaions r =, r = The soluion y( ) = c+ c aroaches as when c > and when c < r >, r < The soluion 5 r βi r = + y ce c e aroaches as when c > and when c < y = c β + c sin β is a eriodic funcion of eriod π, and amliude β =±, cos + c c Linear Indeendence 6 Suose α β α r ce cos + ce sin β= on an arbirary inerval Dividing boh sides by dividing by β, yields c cos β + c sin β = c cos β c sin β = e α, hen differeniaing he new equaion and Hence, c =, c = and we have roven linear indeendence of he given funcions Real Coefficiens 7 Soluion of he differenial equaion is α α = ( cos β + sin β ) + ( cos β sin β ) α α = e ( k + k ) cos β+ ie ( k k ) sin β y k e i k e i For he soluion o be real, here mus exis real numbers r and s such ha k+ k = r k k = si Solving for k and k, we ge k = r+ si k = r si

60 SECTION Comlex Characerisic Roos 5 d y Solving = n d n 8 (a) d y = d d y = k d d y = k + k d dy = k + k + k d y= k + k + k+ k! (b) y = The characerisic equaion is r =, which has a fourh-order roo a Hence, he soluion is y = c+ c+ c + c, which is he same as in ar (a) (c) In general we have y() = k + k + + k+ k!! n n n n ( n ) ( n ) n n n n = c + c + + c+ c Higher-Order DEs because all of he consans are arbirary 9 5 d y d y d y + = 5 d d d The characerisic equaion is which has roos,,,,, Hence, 5 r r + r = r r r+ = r r =, d y d y 7dy y d + d d = The characerisic equaion is y = c + c + c + c e + c e 5 r + r 7r =, which has roos,,, 5 Hence, 5 y = ce + c e + c e

61 5 CHAPTER Higher-Order Linear Differenial Equaions 5 d y dy = d d 5 The characerisic equaion is which has roos,, ±, ±i Hence y y + 5y y= 5 r r = r r = r r r + = r r r+ r + =, () cos 5 r r + 5r = (characerisic equaion) f () = + 5 = r = is a roo y + 6y + y + 8y= r + 6r + r+ 8= (characerisic equaion) f ( ) = = r = is a roo y () y= r = (characerisic equaion) ( r + )( r ) = r =± i, ± y () = ccos+ c sin+ ce + ce Linking Grahs y = c+ ce+ ce + c + csin By long division, we obain ( r )( r r+ ) = ( r )( r )( r ) = r =,, y () = ce + ce + ce By long division, we obain ( r+ )( r + r+ ) = ( r+ ) = r =,, y () = ce + ce + ce 5 5 y 5 y' 5 y' = 5 y

62 SECTION Comlex Characerisic Roos y 5 y' 5 y' 5 5 y = Changing he Daming 7 The curves below show he soluion of x+ bx + x=, x ( ) =, x = for daming b =, 5,,, The larger he daming he faser he curves aroach The b = curve ha oscillaes has no daming x () b = b = b = b = ẋ 8 6 b = 5 b = b = x b = 5 b = b = In Figure (b) in he ex he larger he daming b he more direcly he rajecory heads for he origin The rajecory ha forms a circle corresonds o zero daming Noe ha every ime a curve in (a) crosses he axis wice he corresonding rajecory in (b) circles he origin

63 56 CHAPTER Higher-Order Linear Differenial Equaions Changing he Sring 8 (a) The soluions of x+ x + kx=, x ( ) =, x = x () k = k = 5 are shown for k =,,,, For larger k we have more oscillaions k = k = k = (b) For larger k, since here are more oscillaions, he hase-lane rajecory sirals furher around he origin ẋ k = k = 5 x k = 5 k = k = Changing he Mass 9 (a) b = and ω o = (b) k m so ha ω o is inversely roorional o m If m is doubled, ω o is decreased by a facor of (c) If m is doubled, he daming required for criical daming is increased by a facor of Finding he Maximum 5 (a) x+ x + x=, x() =, x () = r + r + = (characerisic equaion) r = ± i x= e ( ccos + csin ) x = e ( csin + ccos ) e ( ccos + csin ) = c = c so ha c = x = e cos + sin

64 To find maximum dislacemen, se x = SECTION Comlex Characerisic Roos 57 x = e sin + cos e cos + sin = + sin = π when = π, so ha π = sec Subsiuing for : x max = e π π cos sin + π Max Amliude x max = e (b) m =, b =, k = x() =, x () = The DE is x + x + x= for which he characerisic equaion r + r + = gives r = ± i x () = e ( ccos+ c sin) π = e x () = e ( c sin+ c cos) e ( c cos+ c sin) x() = e sin is he soluion To find he maximum dislacemen x max, se x ()= and solve for : = (e cos e sin) so ha an = and = 6 radians which gives x max = 7 (c) m =, b =, k = x() =, x () = The DE is x + x + = for which he characerisic equaion r + r + = gives r =, x () = ce + ce x ce c e e () = + ( + ) c =, c = The soluion is x() = e To find he maximum dislacemen x max, we se x () = and solve for : = ( e + e ) so ha = / which yields x max = e

65 58 CHAPTER Higher-Order Linear Differenial Equaions Oscillaing Euler-Cauchy r 5 We used he subsiuion y = and obained for r = α + iβ and r = α iβ he soluion α+ β α β ( α+ iβ) ( α iβ) α + β α β y = k + k = ke + ke = ke + ke i i ln ln ln i ln ln i ln αln α ( ( β ) ( β )) ( β ) ( β ) = e c cos ln + c sin ln = c cos ln + c sin ln This is he same rocess as ha used a he sar of Case in he ex uilizing he Euler s Formula () y y y + + =, r( r ) + r+ =, r + r+ =, r = ± i, y() = ccos ln + csin ln y + y + 5y= Leing y = yields and gives r r r r r r { r( r ) r } rr + r + 5 = =, + r+ 5=, which has roos ± i Hence, he soluion is cos( ln ) sin ( ln ) ( ) y = c + c y + 7y + 6y= Euler-Cauchy: y = m, > m(m ) + 7m + 6 = (characerisic equaion) m + 6m + 6 = 8 m = 6 ± (6) (6) ( ) y () = ccos( ln ) + c sin( ln ) = 8 ±

66 SECTION Comlex Characerisic Roos 59 Third-Order Euler-Cauchy 55 The hird-order Euler-Cauchy equaion has he form of y r = ( ) > are a y + b y + cy + dy = The derivaives r y = r y = r r y = r r r ( r ) ( )( ) Subsiue hese equaions ino he hird-order Euler-Cauchy equaion above o obain: Dividing by r r r r r a r r r + b r r + cr + d = r r r r a r r r + b r r + c r + d = r, we obain he characerisic equaion: Third-Order Euler-Cauchy Problems ar r r + br r + cr + d = 56 y + y y + y = has Euler-Cauchy characerisic equaion: Noe: r r r + r r r+ = r r + r+ r r r+ = r r r+ = r = is a zero of he olynomial f r = r r r+ because f ( ) = + = Therefore r is a facor of r r r+, which enables us o find he oher facors r r r+ = r r+ r so r =,, Hence, he general soluion o his Euler-Cauchy DE is for > y = c+ c + c, 57 y + y + 5y= Le y = m, > mm ( )( m ) + mm ( ) + 5m= (characerisic equaion) m m + m+ m m+ 5= m + m= m=, ± i y() = c + c cos ( ln ) + c sin ( ln )

67 6 CHAPTER Higher-Order Linear Differenial Equaions Invered Pendulum 58 The differenial equaion x x= has he characerisic equaion r = wih roos ± Hence, he general soluion is (a) Wih iniial condiions x ( ) =, Hence, he soluion of he IVP is x = ce + c e x =, we find c = and c = x() = e e = sinh (b) As, x if c =, and hen x Pendulum and Invered Pendulum x also This will haen whenever ( ) x( ) = 59 (a) The invered endulum equaion has characerisic equaion r =, which has roos ± Hence, he soluion x = ce + c e = c cosh + sinh + c cosh sinh = C sinh+ C cosh, where C = c c, C = c+ c (b) The characerisic equaion of he endulum equaion is r + =, which has roos ± i Hence, he soluion cos x = c + c sin (c) The reader may hink somehing srange abou his because one form (a) aears real and (b) comlex, bu hey are really he same; he difference is aken u by how one chooses he coefficiens, i i e, e is he same as he san of { sin, cos } c c in each case The san of { }

68 SECTION Comlex Characerisic Roos 6 Finding he Damed Oscillaion 6 The iniial condiions x ( ) =, x = give he consans c =, c = Hence, we have () ( cos sin ) x = e + x 5 5 x ()= e ( cos + sin) 5 5 Exremes of Damed Oscillaions 6 The local maxima and minima of he curve have nohing o do wih he exonenial facor which can be rewrien as Acos( ω δ ) α = ( cosω + sinω ) x e c c e α ; hey deend only on c cosω + c sinω, π having eriod T = Hence, consecuive maxima and ω minima occur a equidisan values of, he disance beween hem being one-half he eriod, or π (You can noe in Problem ha he ime beween he firs local maxima and he firs local ω π minima is = π ) Underdamed Mass-Sring Sysem 6 We are given arameers and iniial condiions m = 5, b =, x = k =, x ( ) =, Hence, he IVP is, x ( ) =, 5x+ x+ x= x =, which has he soluion x() = e cos + sin

69 6 CHAPTER Higher-Order Linear Differenial Equaions Damed Mass-Sring Sysem 6 The IVP is x+ bx + 6x=, x ( ) =, x = (a) (b) 6 (c) LRC-Circui I 6 (a) The IVP is 5 5 b = : (underdamed), () = + x e cos 9 sin b = : (criically damed), x( ) = ( + 8) e 6 b = : (overdamed), () ( x e = e ) LQ + RQ + Q = Q + 8Q + 5Q = C, Q ( ) =, 5 (b) () cos sin Q = e + = e cos( δ ) (c) () () 5 sin( ) I = Q = e δ e cos( δ ) where δ = an Q = where δ = an (d) Charge on he caacior and curren in he circui aroach zero as + LRC-Circui II 65 (a) The IVP is LQ + RQ + Q = Q + Q + Q = C, Q ( ) =, (b) () cos sin Q = e + = e cos( δ ) (c) () () cos( ) I = Q = e δ e sin ( δ ) Q =, anδ =, anδ = (d) As, boh Q and I( ) Comuer Lab: Damed Free Vibraions 66 IDE Lab

70 SECTION Comlex Characerisic Roos 6 Effecs of Nonconsan Coefficiens 67 x+ x= (a) This ODE describes (among oher hings) an undamed vibraing sring in which he resoring force is iniially very large (when is near zero), bu evenually decays o zero, causing he frequency of vibraion o decrease and he soluion eriod o increase as increases (b) We loed he soluion wih IC x ( ) =, x = in he x and xx lanes x x 68 (c) As we execed, he x grah shows ha he eriod of he oscillaion increases wih We see also ha he amliude increases in he absence of fricion The xx hase orrai shows ha as ime and amliude increase, velociy decreases, which is consisen wih he revious observaions A good quesion for furher exloraion would be wheher amliude increases indefiniely or levels off x+ x + x= (a) This ODE describes a damed vibraing sring in which he daming sars very large when is near zero, bu decays o zero We susec ha iniially he amliude of a soluion will raidly decay, bu as ime increases he moion could become almos like simle harmonic oscillaion, as here will be almos no fricion (b) We loed he soluion wih IC x ( ) =, lanes x = in he x as well as he xx

71 6 CHAPTER Higher-Order Linear Differenial Equaions x 5 5 x 5 (c) As firs execed, he x grah shows ha he soluion is raidly decaying However he xx hase orrai, consruced wih a longer ime inerval, shows ha our second execaion is no confirmed As ime increases he oscillaions do no become harmonic he amliude of he oscillaions coninues o decrease, gradually and indefiniely 69 x + x = (a) 7 If you divide by, you will see ha his equaion is he same as he equaion in Problem 67 x+ x x + x= (a) This ODE shows negaive fricion for x < and osiive daming for x > For a small iniial condiion near x =, we migh exec he soluion o grow and hen oscillae around x = (b) We loed he soluions in he x and xx lanes a iniial velociy differen iniial dislacemens: x ( ) = 5, x ( ) =, x ( ) = x x = for hree x

72 SECTION Comlex Characerisic Roos 65 (c) As execed, he x grah shows ha iniially he soluion is growing for x ( ) = 5 and decaying for x ( ) = We also see ha all he soluions seem o become eriodic wih 7 he same amliude and eriod, bu we noe ha he moion is no exacly sinusoidal and ha he amliude is abou raher han as we suseced The xx hase orrai confirms ha he long erm rajecories are no circular as in simle harmonic moion, bu disored as we see in he x grah This equaion is called van der Pol s equaion and describes oscillaions (mosly elecrical) where inernal fricion deends on he value of he deenden variable x; furher deails will be exlored in Chaer 7 x+ sin x + x= (a) In his ODE daming changes eriodically from negaive o osiive, so we can redic oscillaion in amliude as well as eriodic vibraory moion (b) We loed he soluion wih IC x x =, = in he x and xx lanes 7 (c) The x grah looks like a suerosiion of wo eriodic oscillaions The xx hase orrai for a longer ime inerval shows ha coninued oscillaions almos reea, bu never exacly This is called quasi-eriodic moion x+ x + x= (a) For his ODE daming is iniially large, bu vanishes as ime increases; he resoring force on he oher hand is iniially small bu increases wih ime How will hese effecs combine? (b) We loed he soluion wih IC x x =, = in he x and xx lanes

73 66 CHAPTER Higher-Order Linear Differenial Equaions x x (c) 7 x (a) As we execed, he x grah shows iniially large daming, which raidly decreases he amliude of he soluion, and increasing frequency, due o he effec of he increasing sring consan, which shorens he eriod The cener of he xx grah will coninue o fill in, very slowly, if you give i a much longer ime inerval + sin x= In his ODE he resoring force changes eriodically from osiive o negaive wih a frequency ha is differen from he naural frequency of he sring We exec some comlicaed bu eriodic moion (b) We loed he soluion wih IC x ( ) =, x x = in he x and xx lanes 6 8 (c) The x grah o = indeed looks almos eriodic, wih eriod 5 However he xx hase orrai over a longer ime inerval shows ha coninued moion almos reeas, bu never exacly This is anoher examle of quasi-eriodic moion, as in Problem 7 Exending he x grah will be anoher good way o see ha he long erm moion is indeed no erfecly reeaing

74 SECTION Comlex Characerisic Roos 67 Boundary-Value Problems 7 y + y =, y() =, y π = y() = c cos + c sin y() = = c y π = = c, so y = is he soluion 75 y + y =, y() =, y π = y() = c cos + c sin y() = = c y π = = c, so y = sin is he soluion 76 y + y =, y() =, y( π ) = y() = c cos + c sin y() = = c y( π ) = = c *No soluions 77 y + y =, y π =, y π = y() = c cos + c sin = c + c c + c = = c c = y () = cos+ sinis he soluion

75 68 CHAPTER Higher-Order Linear Differenial Equaions Exac Second-Order Differenial Equaions 78 y + y y= is he same as y + y = Inegraing we obain he linear equaion y + y = c, d ln for which μ = e = e = so we have y + y = c d Thus, ( y ) = c, so y = d c c + c and c y () = + c Subsiuing back ino he original equaion we find c =, so y () = is he general soluion 79 y + y = y + y = Inegraing and seing c =, we obain d y + y = μ = e = e = y + y= = c ln d ( y ) =, c y = c, y () = d 8 ( ) y + ( ) y + y = where Find ( gy ) : ( gy) = gy + yg ( gy) = ( gy + yg ) = gy + yg + yg + gy = gy + g y + yg Le g = Then g =, g = Then ( gy) = ( ) y + ( ) y + y ( gy) = ( gy) = c gy = c + c so y () = c+ c Suggesed Journal Enry 8 Suden Projec

76 SECTION Undeermined Coefficiens 69 Undeermined Coefficiens Insecion Firs y y = y ( ) = y + y = y = y = y ( ) = y + y = y ( ) = 5 y y + y= y ( ) = 6 cos y y = y = cos 7 y y + y = e y ( ) = e 8 Educaed Predicion y + y + y= + y = The homogeneous equaion y + y + 5y = has characerisic equaion r + r+ 5=, which has comlex roos ± i Hence, sin y = ce + c e cos, so for he righ-hand sides f ( ), we ry he following: 9 f = y = A + B + C+ D h f ( ) = e y ( ) = ( A+ B) e f e = sin y ( ) = e ( Acos+ Bsin) Guess Again f = sin y = Asin+ Bcos The homogeneous equaion y 6y + 9y= has characerisic equaion r 6r+ 9=, which has a double roo, Hence, We ry aricular soluions of he form: h y = ce + c e () = () = ( + ) + ( + ) f cos y A B sin C D cos = = ( + ) f e y A B e (We can have any erms here deenden on erms in he homogeneous soluion) 5 f = e + sin y = Ae + Bsin + Ccos f = + y = A + B + C + D+ E 6

77 7 CHAPTER Higher-Order Linear Differenial Equaions Deermining he Undeermined 7 y = The homogeneous soluion is yh ( ) = c, where c is any consan By simle insecion we observe ha y ( ) = is a soluion of he nonhomogeneous equaion Hence, he general soluion is 8 y y y( ) = + c + = The homogeneous soluion is y = ce where c is any consan By simle insecion we observe ha y soluion is 9 y y + = h = is a soluion of he nonhomogeneous equaion Hence, he general y = ce + yh = ce, y = A+ B, y = A Subsiuing ino he DE gives A+ ( A+ B) = Coefficien of : A = Coefficien of : A+ B= Hence, A =, B = y =, y = ce + y = The homogeneous soluion of he equaion is h y = c+ c, where c, c are arbirary consans By insecion, we noe ha soluion Hence, he soluion of he homogeneous equaion is y() = + c + c y = is a aricular If you could no find a aricular soluion by insecion, you could ry a soluion of he form y = A + B + C y + y = The characerisic equaion is r + r =, which has roos, Hence, he homogeneous soluion is h y = c + c e The consan on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y () = A, exce ha he homogeneous soluion has a consan soluion; hus we seek a soluion of he form y ( ) = A Subsiuing his exression ino he differenial equaion yields A =, or A = Hence, we have a aricular soluion y () =,

78 SECTION Undeermined Coefficiens 7 so he general soluion is y() = c+ ce + y + y = The characerisic equaion is r + =, which has roos ± i Hence, he homogeneous soluion is y = c cos + c sin h The consan on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y () = A Subsiuing his exression ino he differenial equaion yields A =, or A = We have a aricular soluion y () =, so he general soluion y + y = The characerisic equaion is r + r =, which has roos, Hence, he homogeneous soluion is y() = ccos + csin + h y = c + c e The erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y ( ) = A+ B However, he homogeneous soluion has a consan erm so we seek a soluion of he form y = A + B Subsiuing his exression ino he differenial equaion yields y + y = A+ 8A+ B= Seing he coefficien of, equal o each oher yields A =, 8 y() = c+ ce B = Thus, he soluion 6 y + y y = 6 The characerisic equaion is r + r =, which has roos and Hence, he homogeneous soluion h y = ce + c e The linear olynomial on he righ-hand side of he equaion indicaes we seek a aricular soluion of he form

79 7 CHAPTER Higher-Order Linear Differenial Equaions y ( ) = A+ B (Noe ha we don have any maches wih he homogeneous soluion) Subsiuing his exression ino he differenial equaion yields he equaion y + y y = A A B= 6 so A =, B = Hence, we have he general soluion y = ce + ce+ 5 y + y = e + The characerisic equaion is r + = homogeneous soluion is cos y = c + c sin h, which has roos ± i Hence, he The erms on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y = Ae + B Subsiuing his exression ino he differenial equaion yields y + y = Ae + B= e + Seing coefficiens of e, equal o each oher, we ge equaions for A, B, which yield B = Hence, we have he general soluion y() = ccos+ csin + e + 6 y y y= 6e The characerisic equaion is r he homogeneous soluion is h y = ce + c e A =, r =, which has roos and Hence, The exonenial erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y = Ae (Noe his is no linearly deenden on any of he exonenial erms in he homogeneous soluion) Subsiuing his exression ino he differenial equaion we ge y y y = Ae = 6e

80 SECTION Undeermined Coefficiens 7 Hence, A =, and we have a aricular soluion y = e, and hence y = ce + ce e 7 y + y = 6sin The characerisic equaion is r homogeneous soluion is h y = c + c e + r =, which has roos and Hence, he The sine erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form Subsiuing ino he differenial equaion yields cos sin y = A + B y + y = A+ B cos+ B A sin = 6sin Comaring coefficiens yields he equaions which has he soluion and he general soluion is A =, 5 A+ B= B A= 6, 6 B = Hence, we have 5 6 y () = cos sin, y() = c + ce cos sin y + y + 5y = e The characerisic equaion of he differenial equaion is r + r+ 5=, which has roos ± i Hence, he homogeneous soluion is h ( cos sin ) y = e c + c The exonenial on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y = Ae Subsiuing his exression ino he differenial equaion yields y + y + 5y = Ae = e,

81 7 CHAPTER Higher-Order Linear Differenial Equaions which yields is given by A = Hence, we have a aricular soluion y () = e, and he general soluion 5 5 y() = e ( ccos+ csin ) + e 5 9 y + y + y = e The characerisic equaion is given by r + r+ =, which has a double roo of, so he homogeneous soluion is h y = ce + c e The erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y = Ae + Be Subsiuing his exression ino he differenial equaion yields y + y + y = Ae + A+ B e = e Comaring coefficiens, yields equaions, which we solve, geing A =, B = Hence, he general soluion is y = ce + c e + e e y y= sin The characerisic equaion is r =, which has roos ± Hence, he homogeneous soluion is h y = ce + c e The erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion cos y = A+ B + C+ D sin Differeniaing his exression wo imes and subsiuing i ino he differenial equaion yields he algebraic equaion y y = Csin Acos+ A D sin+ C B cos = sin Comaring erms in sin, cos, sin Hence, A =,, cos, we ge equaions ha yield B =, C =, D = y() = ce + ce ( sin + cos )

82 SECTION Undeermined Coefficiens 75 + = cos The characerisic equaion is r + = y y homogeneous soluion is cos y = c + c sin h, which has roos ± i Hence, he Using he rigonomeric ideniy cos = + cos ( ) he erm on he righ-hand side of he differenial equaion yields Hence, we seek a aricular soluion of he form ( ) cos = 6 + cos cos sin y = A + B + C Subsiuing his ino he differenial equaion yields y + y= Acos Bsin+ C = 6+ 6cos Comaring coefficiens, we ge A =, B =, C = 6, so he general soluion is y = c cos+ c sin cos + 6 y y = 8e The characerisic equaion is r =, which has roos ± Hence, he homogeneous soluion is h y = ce + c e The erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion y = Ae + Be, bu one erm in he homogeneous soluion is linearly deenden on his erm, so we seek y = e A + B Subsiuing his exression ino he differenial equaion yields y y = Ae + A+ B e = 8e, which gives he wo equaions A = 8, A+ B=, which gives A =, B = Hence, he general soluion is y = ce + ce + e

83 76 CHAPTER Higher-Order Linear Differenial Equaions y y + y = e The characerisic equaion of he differenial equaion is r r+ =, which has a double roo of Hence, he homogeneous soluion is h y = ce + c e The erm on he righ-hand side of he differenial equaion indicaes we seek a aricular soluion of he form y = Ae + Be, bu boh erms are linearly deenden wih erms in he homogeneous soluion, so we choose y = A e + B e Differeniaing and subsiuing his exression ino he differenial equaion yields he algebraic equaion y y + y = e + + 6A+ B = e Comaring coefficiens, we ge A =, B = Hence, he general soluion is 6 y() = ce + ce + e 6 y y + y= cos The characerisic equaion of he differenial equaion is r which has roos, Hence, h y = ce + c e The erm on he righ-hand side of he differenial equaion indicaes we seek cos sin y = A + B Subsiuing his exression ino he differenial equaion yields y y + y= A+ B sin+ A B cos = cos Comaring coefficiens yields A =, B = Hence, () y = ce + ce + cos sin r+ =, 5 y y + y= e sin The characerisic equaion of he differenial equaion is r r+ =, which has roos, Hence, h y = ce + c e The erm on he righ-hand side of he equaion indicaes we seek a aricular soluion cos sin y = Ae + Be

84 Differeniaing and subsiuing his exression ino he equaion yields Comaring coefficiens, we find SECTION Undeermined Coefficiens 77 y y + y= A B e sin+ A B e cos = e sin A =, B =, yielding he general soluion y() = ce + ce + e ( cos sin ) 6 y + y = sin+ cos The characerisic equaion is r he homogeneous soluion is h y = c + c e + r =, which has roos, Hence, The sine and cosine erms on he righ-hand side of he equaion indicae we seek a aricular soluion of he form cos sin y = A + B Subsiuing his ino he equaion yields y + y = A+ B cos+ B A sin = sin+ cos Comaring erms, we arrive a ( A+ B) =, ( B A) his, ha he general soluion is =, yielding A =, B = From 7 y y = 6 y() = c+ ce + ( sin cos ) () Find y h : r r = r (r ) = r =,, y h = c + c + c e () Find y : y P = (A + B) = A + B y = A + B y = 6A + B y = 6A y y = 6A (6A+ B) = A+ 6A 8B= 6 coefficien of : A = 6, coefficien of : 6A 8B =

85 78 CHAPTER Higher-Order Linear Differenial Equaions 8 () so ha A = Hence y =, and 8B = 6 6 = so ha B = 6 y () = yh + y = c+ c+ ce 6 () y y + y y = e () Find y h : r r + r = (characerisic equaion) f(r) = r r + r = f() = + = so r = is a roo By long division, we obain r r + r = (r )(r r + ) = (r ) Trile roo r =,, y h = c e + c e + c e () Find y : y = (Ae ) = A e y = A e + A e y = A e + A e + A e + 6Ae = A e + 6A e + 6Ae () = y A e A e A e Ae A e Ae Ae Ae () y y y y A e A e Ae Ae () + = Thus y = e 6 A e 8A e 8Ae + e + A e + 9 A e + e + e A e + e + e + e y () = yh + y = ce + ce + ce + e 6 6 Ae = e so ha A = 6

86 SECTION Undeermined Coefficiens 79 9 y () y = () Find y h : r = ( r + )( r ) = ( r + )( r )( r+ ) r = ±i, ± () Find y : y h = c cos + c sin + c e + c e y = A, so ha y = y = y = y = () () y y = A= A= y = () h y () = y + y = ccos+ c sin+ ce + ce y = y y y = () Find y h : r r = r (r ) = r =,, () There is no y because he DE is homogeneous () y() = c + c + c e Iniial-Value Problems y + y y = 6, y() =, y () = () Find y h : r + r = (r )(r + ) = r =, y h = c e + c e () Find y : y = A + B, y = A, y = y + y y = A ( A+ B) = 6 coefficien of : A = 6, coefficien of : A B = A =, B = y = () y() = y h + y = c e + c e + ; = + y ce c e y() = c + c = ; y () = c c + = 5 c + c = c = c c = c = 5 y = e + e +

87 8 CHAPTER Higher-Order Linear Differenial Equaions y + y + y = e, y() =, y () = () Find y h : () Find y : r + r + = (r + ) = r =, y h = c e + c e = (c + c )e y = e (A + B) y = e ( A+ B) + Ae y = e ( A + B) Ae Ae = e ( A + B) Ae So y + y + y = e ( A+ B) Ae + ( e ( A+ B) + Ae ) + e ( A+ B) = e ( A+ A+ B) This gives A =, A + B = and so A = and B = Therefore y = e ( ) () y = y h + y = c e + c e + e ( ) y = ce + ce ce e ( ) + e y() = c = c = y () = c + c + + = c = y() = e + e ( ) y + y =, y() =, y () = () Find y h : r + = r = ± i y h = c cos + c sin () Find y : y = A + B, y = A, y = y + y = ( A+ b) = A+ B= coefficien of : A =, coefficien of : B = A =, B = y = () y = y h + y = c cos + c sin +, y = csin+ ccos+ y() = c = ; y () = c + = c = 5 y () = cos sin+ 8 5 c = 5 8

88 SECTION Undeermined Coefficiens 8 y + y + y = 6cos, y() =, y () = () Find y h : r + r + = (r + ) = r =, y h = c e + c e () Find y : y = A cos + B sin, y = Asin + Bcos, y = Acos Bsin y + y + y = Acos Bsin + ( Bcos Asin ) + Acos+ Bsin B cos A sin = 6 cos coefficien of cos : B = 6, coefficien of sin : A = A =, B = y = sin () y = y h + y = ce + ce + sin, y = ce ce + ce + cos y() = c = ; y () = c + c + = c = y () = e e + sin 5 y + y = cos, y() =, y () = () Find y h : r + = r = r = ± i y h = ccos + csin () Find y : y = A cos + B sin, y = Asin + Bcos, y = Acos Bsin y + y = 6Acos 6Bsin + Acos + Bsin = 5Acos 5Bsin = cos coefficien of cos : 5A =, coefficien of sin : 5 B = A =, 5 B = y = cos 5

89 8 CHAPTER Higher-Order Linear Differenial Equaions () y = y h + y = c cos c sin cos + 5 y = csin ccos sin y() = c 5 = c = 6 5 y () = c = 6 y () cos = cos y + 9y= cos, y() =, y () = () Find y h : r + 9 = r = ± i y h = c cos + c sin () Find y : y = (A cos + B sin ) y = ( Asin+ Bcos ) + ( Acos+ Bsin ) y = ( 9Acos 9Bsin ) + ( Asin+ Bcos ) Asin+ Bcos y + 9 y = (9A 9 A) cos + (9B 9 B) sin + (6 B)cos + ( 6Asin ) = cos coefficien of cos : 6B =, coefficien of sin : 6A = A =, B = 6 so ha y = sin 6 () y = y h + y = c cos + c sin + sin 6, y = c sin + c cos + cos+ sin 6 y() = c = ; y () = c = c = y () = cos sin+ sin 6

90 SECTION Undeermined Coefficiens 8 7 y y + y= e, y() =, y () = () Find y h : r r + = (r )(r ) = r =, r =, so ha y h = c e + c e () Find y : y = Ae, y = Ae, y = Ae y y + y = Ae ( Ae ) + Ae = 6Ae = e so ha A= y = () y = y h + y = c e + c e +, = + e y ce c e e e y() = c + c + = c + c = y () = c+ c = c + c = Thus = + y () e e 8 y y + y = e +, y() =, y () = () Find y h : c = c = () Find y : r r + = (r )(r ) = so ha y h = c e + c e y = Ae + B + C, y = Ae + B, y = Ae Thus y y + y = Ae ( Ae + B) + ( Ae + B + C) = Ae ( Ae + B) + ( Ae + B + C) = 8Ae + B B + C = e + 8A =, B =, B + C = A =, B=, C = B= 8 9 Thus, y = e

91 8 CHAPTER Higher-Order Linear Differenial Equaions () Find y: y = y h + y = c e + c e + e y = ce + c e e + 8 y() = c + c + + = c + c = 8 9 y () = c + c + = c + c = 8 c = 7 5 c = 7 Therefore, y() = e + e + e y y y = cos, y() =, y () = () Find y h : r r = (r + )(r ) = so r =, y h = c e + c e () Find y : y = A cos + B sin, y = A sin + B cos y = A cos B sin y y y = A cos B sin + A sin B cos A cos B sin = ( 6A B) cos + (A 6B) sin = cos coefficien of cos : 6A B =, coefficien of sin : A 6B = so A = B and B = Thus y = A = 5 cos sin 5 5 5

92 SECTION Undeermined Coefficiens 85 () y = y h + y = c e + c e cos sin 5 5 y = c e + c e + 6 sin cos 5 5 y() = c + c 5 = c + c = 5 y () = c + c 5 = c + c = 5 c = 5 c = 5 y y () = e + e cos sin y + y =, y() =, y () =, y () = () Find y h : () Find y : r r + r = r(r r + ) = r(r )(r ) =, so r =,, y h = c + c e + c e y = (A + B + C) = A + B + C y = A + B + C, y = 6A + B, y = 6A y y + y = 6A A 8B + 9A + 6B + C = coefficien of : 9A =, coefficien of : A + 6 B =, coefficien of : 6A 8B + C = A =, B =, C = y = () y = y h + y = c + c e + c e Using his general soluion and he iniial condiions we obain: y () = e e

93 86 CHAPTER Higher-Order Linear Differenial Equaions 5 y () y = e, y() = y () = y () = y () = (a) Find y h : () Find y : r = (r + )(r ) r = ± i, ± y h = c cos + c sin + c e + c e y = Ae, y = Ae, y = Ae, y = 8Ae, () y = 6Ae Thus y () y = 6Ae Ae = e 5Ae = e A = 5 so ha y = () y = y h + y = c cos + c sin + c e + c e + e 5 y = c sin + c cos + c e c e + e 5 y = c cos c sin + c e + c e + e 5 y = c sin c cos + c e c e 8 + e 5 y() = c + c + c + 5 = y () = c + c c + = 5 y () = c+ c + c + = 5 8 y () = c + c c + = 5 e 5 From hese equaions in unknowns, we obain (by he mehods of Chaer ), c =, c =, 5 c = and c = y () = cos+ sin e + e + e y () = e, y() =, y () =, y () =, y () = () Find y h : r = r = (muliliciy ) y h = c + c + c + c

94 SECTION Undeermined Coefficiens 87 () Find y : y = Ae, y = Ae, y = Ae, y = Ae, () y = Ae = e A = so ha y = e () y = y h + y = c + c + c + c + e y = c + c + c + e y = c + 6c + e y = 6c + e y() = c + = c = y () = c + = c = y () = c + = c = y () = 6c + = c = 6 y () = + e 6 () y = Ae 5 y + y = cos Find y h : y h = c cos + c sin Find y : y = A + B y = ccos + Dsin y() = y + y cos sin = A+ B+ C D + 5 y y = + e Find y h : r r = r (r ) = r =,, y h = c + c + c e

95 88 CHAPTER Higher-Order Linear Differenial Equaions Find y : y = ( A + B + C), y = ( De ) y () = y + y = A + B + C + De 55 y 5y + 6y= cos e Find y h : r 5r + 6 = (r )(r ) = r =, y h = c e + c e Find y : y = Acos+ Bsin, y = e ( C+ D) y () = y + y = Acos+ Bsin + e ( C+ D) 56 y () y = e + sin r = (r + )(r ) = r = ± i, ± y h = c cos + c sin + c e + c e y = e ( A+ B), y = ( Ccos+ Dsin ) y () = e ( A + B) + Ccos+ Dsin Judicious Suerosiion 57 (a) The characerisic equaion is r r 6= has roos r =,, so he general soluion is (b) (i) Subsiuing which yields y = Ae yields (ii) Subsiuing h y = ce + c e Ae Ae 6Ae = e, A = Hence, y () = e 6 6 y = Ae yields or A = Hence, y () = e Ae + Ae 6Ae = e

96 L y = y y y we found in ar (b) ha (c) Calling 6 SECTION Undeermined Coefficiens 89 L e = e 6, and L e = e Mulilying each equaion by and using basic roeries of derivaives yields L e = e, and L e = e 8 and L e e = e + e = cosh 8 Hence, a soluion of y y 6y = cosh is Wholesale Suerosiion 58 We firs solve he equaion firs geing he homogeneous To find a aricular soluion, we ry Subsiuing his ino he equaion yields y () = e e 8 n y + y= n! h y = ce ( n ) n n () y = A + A + + A+ A n n n n n n nan + n An + + A + An + An + + A + A = n! Comaring coefficiens, we have ( ) ( ) n

97 9 CHAPTER Higher-Order Linear Differenial Equaions and so on Hence, we have An = n! An =! A A n n = = ( n )! ( n ),! ( n ) n n n ( n) y = + n! n! n! Furher, we have y ( ) ( ) = () y () = ( ) y () = +! ( ) y () = +!! ( ) y () = + +!!! n n n ( n ) y () = + n! n! n! By suerosiion, he sum of hese soluions is a soluion of y + y = e (We agree our discussion is formal in he sense ha we have roven suerosiion for finie sums) There is a sligh roblem in adding he receding funcions because he sum changes form deending on wheher we add an even or odd number of erms We have () () ( () ) n n Sn = y + y + y () = !! ( n)! () () ( () ) 5 n+ n+ Sn+ = y + y + y () = ! 5! n +!

98 SECTION Undeermined Coefficiens 9 However, because he sequence S n converges, i converges o he average of he nh and ( n + ) s erms Tha is, Hence, we found y () = e Disconinuous Forcing Funcions 59 y + y = ( Sn + Sn+ ) = = e!! < y() = y () = Par : y + y = y () = y () = Find (y ) h : r + r = r(r + ) = r =, (y ) h = c + c e Find (y) : (y ) = by insecion, so y = c + c e + = c + c y ce = + = c + c =, c = y = + e + Par : y + y = y = c + c e + y = c e + y () = y () = + e + 8 = 6 + e y () = y () = e + Thus, when =, 6+ e = c + c e + ce e + = + 8 = c + 5 c = e + = c e + e = c c = e y = + (e e )e + and y() = + e + < + + ( e ) e

99 9 CHAPTER Higher-Order Linear Differenial Equaions 6 cos π y() =, y () = y + 6y= > π Par : y + 6y = cos π y () =, y () = Find (y ) h : (y ) h = c cos + c sin Find (y ) : (y ) = A cos + B sin ( y ) = Acos Bsin B =, A = 5 Therefore, y () = y = c cos + c sin + cos 5 y = csin+ ccos sin 5 cos + cos 5 5 Par : y + 6y = > π y = c cos + c sin y = c sin+ c cos y () = = c + 5 c = y () = = c c = y (π) = + = = y( π ) = c c = y ( π ) = = y ( π ) = c c = Thus y () = cos + sin 5 cos + cos π 5 5 and y() = cos + sin > π 5 5 5

100 SECTION Undeermined Coefficiens 9 Soluions of Differenial Equaions Using Comlex Funcions 6 y y + y = sin The homogeneous soluion is y h = c e + c e For he aricular soluion we use y y + y = e i and seek he imaginary ar of he aricular soluion We le y = Ae i Then y i = iae and y = Ae i By subsiuion, we obain Ae i iae i + Ae i = e i ia = A = = i i y = ie i = i(cos + i sin ) = i cos sin Im(y ) = cos y() = y h + Im y = c e + c e + cos 6 y + 5y = 6sin We will use y + 5y = 6e i The homogeneous soluion is y h = c cos 5 + c sin 5 For he aricular soluion we wan Im(y ) where y = Ae i ; i y = iae and y = Ae i By subsiuion, we obain Ae i + 5Ae i = 6e i A = 6 so ha A = Im(y ) = Im i e = sin y() = c cos 5 + c sin 5 + sin 6 y + 5y= sin5 We will use y + y = e 5 5 i The homogeneous soluion is y h = c cos 5 + c sin 5 For he aricular soluion we noe ha e 5i is included in y h, so we mus use an exra facor of in y We wan Im(y ) where y = Ae 5i 5i 5i, so y = A(5 ie + e ), and 5i 5i 5i = 5(5 + ) + 5 = y A i ie e Ai e 5i 5i A(ie 5 e )

101 9 CHAPTER Higher-Order Linear Differenial Equaions By subsiuion, we obain 5i 5i 5i 5i A(ie 5 e ) + 5Ae = e so ha Ai = Thus A = i and y = ie 5i Im(y ) = Im( i(cos5 isin5 ) ) + = cos 5 y = c cos 5 + c sin 5 cos 5 Comlex Exonens 6 y y + y= e i The homogeneous soluion is y h = c e + c e We seek a aricular soluion of he form y = Ae i Then y i = iae and y i = Ae By subsiuion, we obain i i i i Ae ( iae ) + ( Ae ) = e i i i Ae 6Aie = e A( + 6 i) = 6i 9 A= = + i + 6i 6i 9 i 9 y = + i e = + i (cos + isin 9 9 = cos sin i cos sin + I can be verified direcly by subsiuion ha 9 Re(y ) = cos sin saisfies y y + y= cos and ha Im(y ) = 9 cos sin saisfies y y + y = sin Suggesed Journal Enry 65 Suden Projec

102 SECTION 5 Variaion of Parameers 95 5 Variaion of Parameers Sraigh Suff y + y= The homogeneous soluions o he equaion are y( ) = and y ( ) = e To find a aricular soluion of he form v + e v = y v v e e v = = +, we solve he equaions for v, v This gives v = and v = e Inegraing yields v ( ) = and v ( ) e ( ) Hence, we have a aricular soluion ( ) = y = yv + y v = + e e = + Combining he consan erm wih he homogeneous soluion, we wrie he general soluion as y ()= c + ce + y y = e The homogeneous soluions o he equaion are y( ) = and y ( ) = e y v y v y v v e = + = +, we solve To find a aricular soluion of he form v + = = e v ev e This gives v = e and v = e Inegraing yields v ( ) e = and v () = e Hence, we have a aricular soluion y () = yv + yv= ( e ) + e e e e e = = The general soluion is y() = c+ ce + e

103 96 CHAPTER Higher-Order Linear Differenial Equaions y y + y =, ( > ) e The wo linear indeenden soluions y and y of he homogeneous equaion are y ()= e and y ( ) = e Using he mehod of variaion of arameers, we seek he aricular soluion y v e v e = + In order for y () o saisfy he differenial equaion, υ and υ mus saisfy yv + yv = evev = yv + yv = eve ( + ) v = e Solving algebraically for v and v we obain v ( ) = and v = = and v = ln Inegraing gives he values v ( ) Subsiuing hese values ino y yields he aricular soluion y ( ) = e + eln Hence, he general soluion is y ()= ce ce e + + ln y + y = csc The wo linearly indeenden soluions y and y of he homogeneous equaion are y ()= cos and y ( ) = sin Using he mehod of variaion of arameers, we seek he aricular soluion y = v cos+ v sin In order for y () o saisfy he differenial equaion, v and v mus saisfy yv + yv = (cos v ) + (sin v ) = yv + yv = ( sin v ) + (cos v ) = csc Solving algebraically for v and v we obain v ( ) Inegraing gives he values v ( ) = and v = co Subsiuing hese values ino y yields he aricular soluion cos sin ln ( sin ) y = + = and v = ln ( sin ) Hence, he general soluion is y ccos csin cos sin ln ( sin ) = + +

104 SECTION 5 Variaion of Parameers 97 5 y + y = sec an The homogeneous soluions are y ( ) = cos and y ( ) = sin We seek he soluion ( cos ) ( sin ) y y = v + v We form he sysem (cos v ) (sin v ) = (sin v ) + (cos v ) = secan Solving algebraically for v and v yields v = an and v = an Inegraing gives he values v = an and v = ln sec The aricular soluion is y ( an ) cos + sinln sec = sin cos + sinln sec = Thus, he general soluion is y = ccos + c sin cos + sinln sec 6 y y + y = e sin The homogeneous soluions are y e = cos and y e = sin y = v e cos+ v e sin, To find a aricular soluion of he form we solve he equaions for v and v This yields e cosv + e sin v = e cos e sin v e sin e cos v e sin + + = = and sin cos v sin v = Inegraing yields he funcions v () ( cossin ) Hence, a aricular soluion = + and v () = sin y yv yv e e e () = + = cos ( + cos sin ) + sin ( sin ) = ( sin cos ) and he general soluion is y ce ce ()= + e cos sin cos

105 98 CHAPTER Higher-Order Linear Differenial Equaions 7 y y + y = + e The homogeneous soluions are y ()= e and y ( ) = e Hence we seek he aricular soluion y y = e v + e v o form he sysem ev + = e v ev + e v = + e Solving algebraically for v and v yields e v = + e and The firs inegral is rivial; v = ln( + e ) e v = + e The second one is more difficul However, if we erform some algebra, we can wrie ( e e + e ) e e v = = = e, which inegraes o give v = e + ln ( + e ) + e + e + e Wih υ and υ we have he aricular soluion y = e ln + e e + e ln + e and he general soluion is b g b g () ln ( ) y = ce + c e + e + e + e (The erm e in y was absorbed in he homogeneous soluion, giving a beer form for he soluion) 8 y + y + y = e ln, ( > ) The homogeneous soluions are y ()= e and y ( ) = e y y = e v + e v o form he sysem We seek a aricular soluion e v+ e v = e v + ( e e ) v = e ln Solving algebraically for v and v, yields v = ln and v = ln Inegraing yields v v = = ln + and ln Hence, we have a aricular soluion y = e ln + e + e ln e = e ln Thus he general soluion is y ()= ce + ce + e ( ln )

106 SECTION 5 Variaion of Parameers 99 9 y + y = an y h = ccos + csin y y ( an )(sin ) sin cos v = d = sec cos d = d d cos = cos = ( ln sec + an sin ) an cos v = d = cos So y = y v + y v = cos ( ln sec+ an sin ) sin cos General soluion: y() = c cos + c sin + y y + 5y + 6y = cos( e ) y h = ce + ce y y ( cos( e )) e v = d = e cos( e ) d = e sin( e ) + cos( e ) 5 e e 5 cos( e ) v = d = e cos( e ) d = sin( e ) e sin( e ) = e cos( e ) e So y = y v + y v = e ( e sin( e ) + cos( e )) + e ( sin( e ) e sin( e ) e cos( e )) = e sin( e ) e cos( e ) General soluion: y() = c e + c e + y y + y = sec y h = ccos + csin y y v = ( sec )sin d = ( sec an ) d = sec v = sec cos d = sec d = ln sec+ an So, y = y v + y v = cossec+ sin ln sec+ an = + sinln sec+ an General soluion: y = coscos+ csin + sin ln sec+ an

107 CHAPTER Higher-Order Linear Differenial Equaions e y y= y h = ce + ce y y e v = ( e ) d d ln = = s e e e v = e d = d ds = s So y = v y + v y = General soluion: s e e ln e ds s y() = c e + c e + y Variable Coefficiens y y + y = sin, y ()=, y ( ) = We begin by dividing he equaion by, o ge he roer form for using variaion of arameers y + = y y sin Subsiuion verifies ha y and y for a fundamenal se of soluion o he associaed homogeneous equaion, so y = c+ c, h y y = v+ v, we seek a aricular soluion where v and v saisfy he condiions v + v = v + v = sin Solving algebraically for v and v, yields v = sin and v = sin Inegraing yields v = cos sin and v = cos Thus, y ()= cos sin cos = sin Hence, he general soluion of his equaion is y = c + c sin y + y y = + b g, y()=, y( ) = We begin by dividing he equaion by, o ge he roer form for using variaion of arameers: y + = + y y

108 SECTION 5 Variaion of Parameers Subsiuion verifies ha y and y form a fundamenal se of soluions o he associaed homogeneous equaion, so y = c + c h y y = v + v, We seek a aricular soluion where υ and υ saisfy he condiions v v + = Solving algebraically for v and v yields Inegraing yields Thus, y ()= + ln 8 6 v v = + + v = and v = ln + and 8 ( ) + v = v 6 6 = Hence, he general soluion of his equaion is y bg= c + c + ln + (Noice ha he erm in y can be absorbed in he homogeneous soluion) 5 ( y ) + y y= ( ) e, y ( ) =, y e bg= We begin by dividing he equaion by ( ), o ge he roer form for variaion of arameers y + = y y e Susbiuion verifies ha y and y form a fundamenal se of soluions o he associaed homogeneous equaion, so y = c + ce We seek a aricular soluion h y y v v e = +, where v and v saisfy he condiions v + ev = b v + e v = ( ) e Solving algebraically for v and v yields v = e and yields v = e and g v = e + Thus, y () = e + e + e = e Hence, he general soluion of his equaion is y ()= c+ ce + e H F v = e Inegraing I K

109 CHAPTER Higher-Order Linear Differenial Equaions 6 y + + y H F I y = K / /, y ( ) = sin, = y cos Subsiuion verifies ha y and y form a fundamenal se of soluions o he associaed / / homogeneous equaion, so y c = sin+ c cos h / / We seek a aricular soluion sin y y = v + v cos, where v and v saisfy he condiions sinv + cosv = / / + v + v = / / / / / sin cos cos sin Mulilying hrough by / hen solving for v and v : v = cos and v = sin Thus, () ( sin cos ) Third-Order Theory / / v = sin and v = cos y = + =, and he general soluion of his equaion is 7 Ly = y + y () + qy () + ry () = f () / / / sin cos y = y + y = c + c + Given h yh () = c y + c y + c y, we seek y () = vy + vy + vy Differeniaing yields y = vy + v y+ vy + v y+ vy + v y = vy + vy + vy (if we se yv + yv + yv = ) = Differeniaing, again y = v y + vy + v y + vy + v y + vy = vy + vy + vy (if now we se yv + yv + yv = ) Differeniaing ye again: y = v y + vy + v y + vy + v y + vy Subsiuing y, y, y and y ino he Ly = f, hen regrouing all erms in v and v, we see ha he coefficien of each is because each y i is a soluion of L(y i ) = Thus we are lef wih y = yv + y v + y v = f This las equaion, ogeher wih he wo assumions (in arenheses) ha we made while differeniaing, gives a sysem o solve for v, v, v : yv + yv + yv = yv + yv + yv = yv + yv + yv = f

110 SECTION 5 Variaion of Parameers We use Cramer s Rule o solve he sysem, hen inegrae o find v, v, v and hence, obain a aricular soluion y = v y + v y + v y Third-Order DEs 8 y y y + y = e The characerisic equaion ( λ )( λ )( λ ) se is y = e, y = e, and y = e Hence, By variaion of arameers, we seek + = and has roos,, and The fundamenal y = ce + c e + c e h y () = ve + v e + v e, as in Problem 7 Hence he sysem o solve is ev + e v + e v = ev e v + e v = ev + e v + e v = e Using Cramer s rule and comuing he deerminans yields: e e e e e e e e e e e W = e e e = 6 e ; v = = = W 6e e e e e e e e e e e e v = = = e W 6e 6 e e e e e e e e v = = = e W 6e Hence we obain v = v = e v = e 6 Hence, v = v = e v = e We ge a aricular soluion of y e ()= + e e and he general soluion is y ()= ce + ce + ce e e

111 CHAPTER Higher-Order Linear Differenial Equaions 9 y + y = sec Find y h : y h = r + r = r(r + ) = r =, ±i y h = c + c cos + c sin W = y = v + v cos + v sin cos sin sin cos = cos sin cos sin sin cos sec cos sin cos sin v = = sec = sec sin cos v = ln sec+ an sin cos sec sin cos cos v = = = = sec sin sec v = cos sin cos sec sin sin v = = = sinsec = cos sec cos v = ln cos y() = c+ ccos+ csin + ln sec+ an cos+ sin ln cos y + 9y = an Find y h : r + 9r = r(r + 9) = r = ± i y h = c + c cos + c sin y = v + v cos + v sin cos sin sin cos W = sin cos = 9cos 9sin 9cos 9sin = 7

112 SECTION 5 Variaion of Parameers 5 cos sin sin cos an 9cos 9sin an cos sin an an v = = = (cos + sin ) = 7 7 sin cos 7 9 v = ln cos 7 sin cos an 9sin cosan v = = = sin v = cos 7 cos sin 9cos an sinan sin v = = = cos cos d ln sec an sin 9 cos 7 v = = ( + ) sin y = c + c cos+ c sin ln cos + cos ( ln sec+ an sin ) Mehod Choice y y = f() We firs find he homogeneous soluion The characerisic equaion λλ ( ) = has roos, ±, so he homogeneous soluion is y = c+ ce + ce h (a) y y = e Because e is in y h, we mus ry y = a e The mehod of undeermined coefficiens is sraighforward and gives a =, so y = e and he general soluion can be wrien y () = c+ ce + ce + e (b) y y = sin We canno use underermined coefficiens on sin, so we use variaion of arameers o seek a aricular soluion of he form y () = v + v e + v e, wih he derivaives of v, v, and v deermined from he equaions v + e v + e v = v + e v e v = v + e v + e v = sin Discussion coninues on nex age

113 6 CHAPTER Higher-Order Linear Differenial Equaions Using Cramer s rule (as oulined in Problem 8), we obain v = sin v = e sin v = e sin The aniderivaive of υ is easy o find; he oher wo mus be lef as inegrals v = sin cos v = e sin d v = e sin d (c) Hence, he general soluion is y c ce ce e e d e e d () = ( sin cos ) + sin + sin y y = an As in Par (b), we mus use variaion of arameers o find y, wih v + e v + e v = v + e v e v = v + e v + e v = an Using Cramer s rule (as oulined in Problem 8), o solve hese equaions we find v = an v = e an v = e an The aniderivaive of υ is easy o find; he oher wo mus be lef as inegrals v = ln cos v = e an d v = e an d Hence, he general soluion is y() = c+ ce + ce + ln cos + e e an d e e an d + Pars (b) and (c) demonsrae he ower of grahical mehods because he algebraic exressions for y() are rey meaningless I is easier and more informaive o use DE sofware o aroximae soluions of his equaion in y sace han i is o ursue he analyical formula for he soluion The figures show curves for several iniial condiions o show he variey ha can occur For any IVP here would be only one soluion Noe: We used a D grahc DE solver wih he following equaions for y y = f(): y = x x = y y = x = z relised as y = x y = z = x+ f() z = x+ f()

114 SECTION 5 Variaion of Parameers 7 (b) f() = sin The exression for y() on he revious age can be furher evaluaed using he ideniy sin = ( cos ), bu soluion behavior is more easily seen on a grah of y() (c) f() = an The exression for y() on he revious age is even more comlicaed han ha for ar (b); again, soluion behavior is more readily undersood wih a grah of y() Green s Funcion Reresenaion y + y= f() We know ha y = cos and y = sin are he soluions of he corresonding homogeneous equaion Their Wronksian is Wy a, y f()= cos sin sin cos = which is makes i easy o use he suggesed variaion of arameers formulas y () f () y ( ) f ( ) v = = sin( ) f( ), v = = cos( ) f( ) W y, y W y, y Inegraing yields = ( s) f ( s) ds v v sin y = yv + y v Hence, Green Variaion = cos s f s ds = cos() sin() s f() s ds+ sin() cos() s f() s ds = cos( )sin( s) + sin( )cos( s) f( s) ds = sin( s) f( s) ds [ ] The homogeneous soluions are y = e and y = e We seek a aricular soluion of he form y = vy+ vy, where v and v saisfy ev + e v = ev e v = f ()