PANEPISTHMIO DUTIKHS ATTIKHS SQOLH MHQANIKWN TMHMA POLITIKWN MHQANIKWN ANWTERA MAJHMATIKA II DIAFORIKES EXISWSEIS 6h Seirˆ Ask sewn OmogeneÐc grammikèc diaforikèc exis seic me stajeroôc suntelestèc Jèma 1. Na lujoôn oi diaforikèc exis seic: i. y 3y + 2y = 0 ii. y 9y = 0 iii. y + 2y = 0 iv. y 6y + 9y = 0 v. y + 2y + y = 0 vi. y 12y + 36y = 0 vii. y 4y + 5y = 0 viii. y + 16y = 0 ix. y 6y + 10y = 0 Jèma 2. Na lujoôn oi diaforikèc exis seic: i. y + 4y = 0 ii. y 3y + 2y = 0 iii. y 5y + 8y 4y = 0 iv. y + 3y + 3y + y = 0 v. y 2y + y = 0 vi. y + y 2y = 0 Anaplhrwt c Kajhght c : Dr. Pappˆc G. Alèxandroc 1
Jèma 1. Na lujoôn oi diaforikèc exis seic: i. y 3y + 2y = 0 ii. y 9y = 0 iii. y + 2y = 0 iv. y 6y + 9y = 0 v. y + 2y + y = 0 vi. y 12y + 36y = 0 vii. y 4y + 5y = 0 viii. y + 16y = 0 ix. y 6y + 10y = 0 LUSH Sthn ˆskhsh aut, ja doôme na efarmìzontai gia exis seic (grammikèc - omogeneðc) opoiasd pote taxewc me stajeroôc suntelestèc, ìsa efarmìsame sthn prohgoômenh ˆskhsh gia antðstoiqec D.E. (grammikèc - omogeneðc me stajeroôc suntelestèc) deôterhc tˆxhc. i. y + 4y = 0. Me << antikatˆstash >> y 1, y λ prokôptei h qarakthristik algebrik : H genik lôsh thc D.E. eðnai: λ + 4 = 0 λ = 4. y = c 1 e 4x afoô h e λx = e 4x eðnai mða (grammikˆ anexˆrthta) lôsh thc D.E. ii. y 3y + 2y = 0. λ 3 3λ 2 + 2λ = 0 λ ( λ 2 3λ + 2 ) = 0 pou proèkuye apì th D.E. me << antikatˆstash >> : y 1, y λ, y λ 2, y λ 3. Oi rðzec thc eðnai λ 1 = 0 kai oi rðzec tou triwnômou λ 2 3λ + 2 pou eðnai λ 2 = 1, λ 3 = 2. Treic grammikˆ anexˆrthtec lôseic eðnai e λ 1x = e 0 x = 1, e λ 2x = e x, e λ 3x = e 2x 2
opìte h genik lôsh thc D.E. eðnai: ìpou c 1, c 2, c 3 eðnai aujaðretec stajerèc. y = c 1 1 + c 2 e x + c 3 e 2x y = c 1 + c 2 e x + c 3 e 2x iii. y 5y + 8y 4y = 0. λ 3 5λ 2 + 8λ 4 H eôresh twn riz n mðac exðswshc trðtou bajmoô, ìpwc eðnai h algebrik qarakthristik, eðnai genikˆ èna polôploko prìblhma. GnwrÐzoume ìmwc ìti an upˆrqoun akèraiec rðzec, prèpei na eðnai diairètec tou stajeroô ìrou. Ja anazht soume rðzec anˆmesa stouc diairètec (±1, ±2, ±3, ±4) rðzec eðnai mìno oi λ 1 = 1, λ 2 = 2, ìpwc polô eôkola mporoôme na sumperˆnoume me apl antikatˆstash. To polu numo grˆfetai: λ 3 5λ 2 + 8λ 4 λ 3 5λ 2 + 8λ 4 = (λ 1)(λ 2)(λ λ 3 ). Metˆ thn ektèlesh twn prˆxewn kai exðswsh twn suntelest n twn omobajmðwn ìrwn, twn poluwnômwn tou λ, prokôptei λ 3 = 2. 'Etsi, h algebrik qarakthristik èqei telikˆ tic rðzec λ 1 = 1, λ 2 = λ 3 = 2 (dipl ). 'Ara, treðc grammikˆ anexˆrthtec lôseic thc D.E. eðnai: kai epomènwc h genik lôsh thc D.E. eðnai: e λ1x = e x, e λ2x = e 2x, xe λ3x = xe 2x y = c 1 e x + c 2 e 2x + c 3 xe 2x (c 1, c 2, c 3 aujaðreta). iv. y + 3y + 3y + y = 0. λ 3 + 3λ 2 + 3λ + 1 = 0 pou eðnai pˆli trðtou bajmoô. Oi diairètec tou stajeroô ìrou eðnai ±1. Me apl dokim brðskoume eôkola ìti h λ 1 = 1 eðnai mða rðza. An λ 2, λ 3 oi upìloipec rðzec, eðnai: λ 3 + 3λ 2 + 3λ + 1 = (λ + 1)(λ λ 2 )(λ λ 3 ). 3
Mèta thn ektèlesh twn prˆxewn kai exðswsh twn suntelest n twn omobajmðwn ìrwn twn dôo mel n, prokôptei λ 2 = λ 3 = 1. 'Etsi, h qarakthristik algebrik èqei rðza to 1 me bajmì pollaplìthtac 3 (tripl rðza): λ 1 = λ 2 = λ 3 = 1. TreÐc grammikˆ anexˆrthtec lôseic thc D.E. eðnai: H genik lôsh thc D.E. eðnai: ìpou c 1, c 2, c 3 aujaðretec stajerèc. e λ 1x = e x, xe λ 2x = xe x, x 2 e λ 3x = x 2 e x. y = c 1 e x + c 2 xe x + c 3 x 2 e x Genikˆ, an λ eðnai mða rðza thc qarakthristik c algebrik c, me bajmì pollaplìthtac ν, antistoiqoôn s> aut oi grammikˆ anexˆrthtec lôseic (ν to pl joc): e λx, xe λx, x 2 e λx,..., x ν 1 e λx. v. y 2y + y = 0. Me y 1, y λ 2, y λ 4 prokôptei h qarakthristik algebrik : λ 4 2λ 2 + 1 = 0. H exðswsh aut eðnai gnwst san ditetrˆgwnh. Me λ 2 = ω, grˆfetai: ω 2 2ω + 1 = 0. H teleutaða èqei rðzec ω 1 = ω 2 = 1 (dipl ). Oi rðzec thc algebrik c qarakthristik c eðnai: λ 2 = ω 1, λ 2 = ω 2 λ 1 = 1 (dipl ) λ 2 = 1 (dipl ). Sth dipl rðza λ 1 = 1 antistoiqoôn oi grammikˆ anexˆrthtec lôseic thc D.E. e x, xe x en sth dipl rðza λ 2 = 1 oi epðshc grammikˆ anexˆrthtec lôseic e x, xe x. 'Etsi h genik lôsh thc D.E. eðnai: y = c 1 e x + c 2 xe x + c 3 e x + c 4 xe x. 4
vi. y + y 2y = 0 Me y 1, y λ 2, y λ 4 brðskoume th qarakthristik algebrik : λ 4 + λ 2 2 = 0 pou eðnai ditetrˆgwnh exðswsh. Me λ 2 = ω grˆfetai: ω 2 + ω 2 = 0 ω 1 = 1, ω 2 = 2 opìte apì tic exis seic λ 2 = ω 1, λ 2 = ω 2 prokôptoun oi rðzec thc qarakthristik c: λ 1 = 1, λ 2 = 1, λ 3 = i 2, λ 4 = i 2. Stic λ 1 = 1, λ 2 = 1 antistoiqoôn oi grammikˆ anexˆrthtec lôseic e x, e x, en sto zeugˆri twn (suzug n) migadik n riz n λ 3, λ 4 me α ± βi = 0 ± i 2 oi epðshc grammikˆ anexˆrthtec lôseic: ( ) ( ) ( ) e 0x sin 2 x = sin 2 x, e 0x cos 2 x ( ) = cos 2 x. H genik lôsh eðnai: y = c 1 e x + c 2 e x + c 3 sin ( ) ( ) 2 x + c 4 cos 2 x. 5
Jèma 2. Na lujoôn oi diaforikèc exis seic: i. y + 4y = 0 ii. y 3y + 2y = 0 iii. y 5y + 8y 4y = 0 iv. y + 3y + 3y + y = 0 v. y 2y + y = 0 vi. y + y 2y = 0 LUSH Sthn ˆskhsh aut, ja doôme na efarmìzontai gia exis seic (grammikèc - omogeneðc) opoiasd pote taxewc me stajeroôc suntelestèc, ìsa efarmìsame sthn prohgoômenh ˆskhsh gia antðstoiqec D.E. (grammikèc - omogeneðc me stajeroôc suntelestèc) deôterhc tˆxhc. i. Me << antikatˆstash >> prokôptei h qarakthristik algebrik : y + 4y = 0. y 1, y λ λ + 4 = 0 λ = 4. H genik lôsh thc D.E. eðnai: y = c 1 e 4x afoô h e λx = e 4x eðnai mða (grammikˆ anexˆrthta) lôsh thc D.E. ii. y 3y + 2y = 0. λ 3 3λ 2 + 2λ = 0 λ ( λ 2 3λ + 2 ) = 0 pou proèkuye apì th D.E. me << antikatˆstash >> : y 1, y λ, y λ 2, y λ 3. Oi rðzec thc eðnai λ 1 = 0 kai oi rðzec tou triwnômou λ 2 3λ + 2 pou eðnai λ 2 = 1, λ 3 = 2. Treic grammikˆ anexˆrthtec lôseic eðnai opìte h genik lôsh thc D.E. eðnai: ìpou c 1, c 2, c 3 eðnai aujaðretec stajerèc. e λ1x = e 0 x = 1, e λ2x = e x, e λ3x = e 2x y = c 1 1 + c 2 e x + c 3 e 2x y = c 1 + c 2 e x + c 3 e 2x 6
iii. y 5y + 8y 4y = 0. λ 3 5λ 2 + 8λ 4 H eôresh twn riz n mðac exðswshc trðtou bajmoô, ìpwc eðnai h algebrik qarakthristik, eðnai genikˆ èna polôploko prìblhma. GnwrÐzoume ìmwc ìti an upˆrqoun akèraiec rðzec, prèpei na eðnai diairètec tou stajeroô ìrou. Ja anazht soume rðzec anˆmesa stouc diairètec (±1, ±2, ±3, ±4) rðzec eðnai mìno oi λ 1 = 1, λ 2 = 2, ìpwc polô eôkola mporoôme na sumperˆnoume me apl antikatˆstash. To polu numo grˆfetai: λ 3 5λ 2 + 8λ 4 λ 3 5λ 2 + 8λ 4 = (λ 1)(λ 2)(λ λ 3 ). Metˆ thn ektèlesh twn prˆxewn kai exðswsh twn suntelest n twn omobajmðwn ìrwn, twn poluwnômwn tou λ, prokôptei λ 3 = 2. 'Etsi, h algebrik qarakthristik èqei telikˆ tic rðzec λ 1 = 1, λ 2 = λ 3 = 2 (dipl ). 'Ara, treðc grammikˆ anexˆrthtec lôseic thc D.E. eðnai: kai epomènwc h genik lôsh thc D.E. eðnai: e λ1x = e x, e λ2x = e 2x, xe λ3x = xe 2x y = c 1 e x + c 2 e 2x + c 3 xe 2x (c 1, c 2, c 3 aujaðreta). iv. y + 3y + 3y + y = 0. λ 3 + 3λ 2 + 3λ + 1 = 0 pou eðnai pˆli trðtou bajmoô. Oi diairètec tou stajeroô ìrou eðnai ±1. Me apl dokim brðskoume eôkola ìti h λ 1 = 1 eðnai mða rðza. An λ 2, λ 3 oi upìloipec rðzec, eðnai: λ 3 + 3λ 2 + 3λ + 1 = (λ + 1)(λ λ 2 )(λ λ 3 ). Mèta thn ektèlesh twn prˆxewn kai exðswsh twn suntelest n twn omobajmðwn ìrwn twn dôo mel n, prokôptei λ 2 = λ 3 = 1. 7
'Etsi, h qarakthristik algebrik èqei rðza to 1 me bajmì pollaplìthtac 3 (tripl rðza): λ 1 = λ 2 = λ 3 = 1. TreÐc grammikˆ anexˆrthtec lôseic thc D.E. eðnai: H genik lôsh thc D.E. eðnai: ìpou c 1, c 2, c 3 aujaðretec stajerèc. e λ 1x = e x, xe λ 2x = xe x, x 2 e λ 3x = x 2 e x. y = c 1 e x + c 2 xe x + c 3 x 2 e x Genikˆ, an λ eðnai mða rðza thc qarakthristik c algebrik c, me bajmì pollaplìthtac ν, antistoiqoôn s> aut oi grammikˆ anexˆrthtec lôseic (ν to pl joc): e λx, xe λx, x 2 e λx,..., x ν 1 e λx. v. y 2y + y = 0. Me y 1, y λ 2, y λ 4 prokôptei h qarakthristik algebrik : λ 4 2λ 2 + 1 = 0. H exðswsh aut eðnai gnwst san ditetrˆgwnh. Me λ 2 = ω, grˆfetai: ω 2 2ω + 1 = 0. H teleutaða èqei rðzec ω 1 = ω 2 = 1 (dipl ). Oi rðzec thc algebrik c qarakthristik c eðnai: λ 2 = ω 1, λ 2 = ω 2 λ 1 = 1 (dipl ) λ 2 = 1 (dipl ). Sth dipl rðza λ 1 = 1 antistoiqoôn oi grammikˆ anexˆrthtec lôseic thc D.E. e x, xe x en sth dipl rðza λ 2 = 1 oi epðshc grammikˆ anexˆrthtec lôseic e x, xe x. 'Etsi h genik lôsh thc D.E. eðnai: y = c 1 e x + c 2 xe x + c 3 e x + c 4 xe x. 8
vi. y + y 2y = 0 Me y 1, y λ 2, y λ 4 brðskoume th qarakthristik algebrik : λ 4 + λ 2 2 = 0 pou eðnai ditetrˆgwnh exðswsh. Me λ 2 = ω grˆfetai: ω 2 + ω 2 = 0 ω 1 = 1, ω 2 = 2 opìte apì tic exis seic λ 2 = ω 1, λ 2 = ω 2 prokôptoun oi rðzec thc qarakthristik c: λ 1 = 1, λ 2 = 1, λ 3 = i 2, λ 4 = i 2. Stic λ 1 = 1, λ 2 = 1 antistoiqoôn oi grammikˆ anexˆrthtec lôseic e x, e x, en sto zeugˆri twn (suzug n) migadik n riz n λ 3, λ 4 me α ± βi = 0 ± i 2 oi epðshc grammikˆ anexˆrthtec lôseic: ( ) ( ) ( ) e 0x sin 2 x = sin 2 x, e 0x cos 2 x ( ) = cos 2 x. H genik lôsh eðnai: y = c 1 e x + c 2 e x + c 3 sin ( ) ( ) 2 x + c 4 cos 2 x. 9