prìqeirh èkdosh tou 2ou kef apì to biblÐo g. bougiatz c, g...

1

Prìqeirh èkdosh tou 2ou Kef apì to biblÐo

G. Bougiatz c, G. Mpìzhc, D.Papadìpouloc Dia-

forikèc Exis¸seic kai Efarmogèc , Ekdìseic Kleid-

rijmoc

S autì to keflaio parousizoume merikèc gnwstèc mejìdouc gia thn epÐlushdiaforik¸n exis¸sewn (D.E.) pr¸thc txhc. Ja xekin soume me exis¸seic se lumènhmorf

y′ = f(x, y) (1)

kai ja anazht soume th genik touc lÔsh, eÐte se lumènh morf

y = f(x, c), x ∈ I ⊆ R, c ∈ E ⊆ R,

eÐte se peplegmènh morf

Φ(x, y; c) = 0, (x, y) ∈ T ⊆ R2, c ∈ E ⊆ R.

Oi mèjodoi pou ja parousisoume sundèontai me sugkekrimènec morfèc kath-gorÐec thc diaforik c exÐswshc (1). EÐnai shmantikì, loipìn, ìtan mac dÐnetai miaexÐswsh na mporoÔme na anagnwrÐsoume se poia eidik kathgorÐa an kei ¸ste na e-farmìsoume thn katllhlh mèjodo gia thn epÐlus thc1. 'Omwc, ìpwc ja gÐnei safècstic epìmenec enìthtec, h anagwg miac D.E. se mia sugkekrimènh kathgorÐa mporeÐna apaiteÐ thn allag metablht¸n.

Prèpei na shmei¸soume ìti mìno elqistec eidikèc kathgorÐec exis¸sewn èqounlÔseic pou mporoÔn na brejoÔn kai na apodojoÔn me stoiqei¸deic majhmatikèc su-nart seic. Gia pardeigma h exÐswsh

y′ = x sin(y)

an kei sthn kathgorÐa twn exis¸sewn thc Enìthtac 0.1 kai lÔnetai antÐstoiqa. 'Omwch exÐswsh

y′ = x+ sin(y)

den an kei se kpoia kathgorÐa gia thn opoÐa gnwrÐzoume ènan antÐstoiqo algìrijmoepÐlushc.

1Μια Δ.Ε. βέβαια μπορεί να ανήκει σε περισσότερες από μία κατηγορίες και να λύνεται με περισ-

σότερους από έναν τρόπους

2

0.1 Diaforikèc exis¸seic qwrizìmenwn meta-

blht¸n

Prìkeitai gia exis¸seic thc morf c:

y′ ≡ dy

dx= f1(x)f2(y). (2)

SumbaÐnei dhlad o ìroc f(x, y) sto dexiì skèloc thc diaforik c exÐswshc y′ =f(x, y) na grfetai wc ginìmeno miac sunrthshc tou x epÐ mia sunrthsh tou y. Jaupojèsoume ìti oi sunart seic f1(x) kai f2(y) eÐnai suneqeÐc se anoiqt diast matatim¸n twn x kai y, antÐstoiqa. IdiaÐtera th sunrthsh f2(y) ja th jewroÔme mhmhdenik 2 se anoiqt diast mata tim¸n thc y

H exÐswsh (2) grfetai tìte:

dy

f2(y)= f1(x)dx,

kai oloklhr¸nontac èqoume:∫dy

f2(y)=

∫f1(x)dx ⇒ σ2(y) = σ1(x) + c, (3)

ìpou oi σ1 kai σ2 eÐnai sunart seic mìno tou x kai tou y, antÐstoiqa, kai h c eÐnaiaujaÐreth stajer3. Akrib¸c aut h sqèsh apoteleÐ th genik lÔsh thc D.E. (2).An h (3) mporeÐ na lujeÐ wc proc y, tìte paÐrnoume th lÔsh se lumènh morf .

Pardeigma :Gia th D.E. y′ = −x(y+ 1) na brejeÐ h genik lÔsh kai h merik lÔsh pou pernei apì to shmeÐo (x, y) = (0, 0).

H dedomènh D.E. eÐnai thc morf c (2), dhlad qwrizìmenwn metablht¸n, kai giay 6= −1 grfetai

dyy+1

= −xdx ⇒∫

dyy+1

= −∫xdx

′h

ln |y + 1| = −x2

2+ c ⇒ y + 1 = ±ece−x2

2 .

H c eÐnai h aujaÐreth stajer thc olokl rwshc, opìte kai h posìthta ±ec ek-frzei mia pragmatik aujaÐreth stajer; mporoÔme na th sumbolÐsoume wc c kai nagryoume telik th genik lÔsh sth lumènh morf

y = ce−x2

2 − 1 (4)

2Αν για κάποια διακριτή τιμή y0 είναι f(y0) = 0, τότε η y = y0 θα αποτελεί λύση της Δ.Ε. όπως

εύκολα μπορεί να αποδειχθεί.3Είναι προφανές ότι δεν χρειάζεται να γράψουμε δύο αυθαίρετες σταθερές, δηλαδή μία για την

ολοκλήρωση του αριστερού και μία για την ολοκλήρωση του δεξιού μέλους. Η αυθαίρετη σταθερά cμπορεί να τοποθετηθεί είτε στο αριστερό είτε στο δεξί μέλος κατά βούληση.

0.1. DIAFORIKES EXISWSEIS QWRIZOMENWN METABLHTWN 3

Sq ma 1: H oikogèneia lÔsewn (4). H merik lÔsh pou pernei apì to (0, 0) anti-stoiqeÐ sthn èntonh kampÔlh.

An sth (4) jèsoume x = 0 kai y = 0, paÐrnoume 0 = c− 1, dhlad h zhtoÔmenh merik lÔsh antistoiqeÐ se c = 1. Ftsame sth lÔsh (4) jewr¸ntac arqik ìti y 6= −1.ParathroÔme ìmwc ìti gia c = 0 h (4) dÐnei y = −1, h opoÐa faÐnetai amèswc apì thnÐdia th D.E. ìti apoteleÐ lÔsh thc. J

Oi lÔseic (4) parousizoun akrìtato sto x = 0 kai teÐnoun asumptwtik sthntim y = −1 kaj¸c x → ±∞. Ja mporoÔsate na sumpernete autèc tic idiìthtectwn lÔsewn apì th dedomènh diaforik exÐswsh?

Mia diaforik exÐswsh qwrizìmenwn metablht¸n mporeÐ na emfanisteÐ kai wc

g1(x)dx+ g2(y)dy = 0, g2(y) 6= 0,

, gia g1(x) = 1, kai wcg(y)dy = dx.

0.1.1 Eidikèc peript¸seic

Se morf D.E. qwrizìmenwn metablht¸n angontai kai exis¸seic thc morf c:

y′ = f(αx+ βy + γ), (5)

me α, β kai γ stajerèc. Prgmati, gia β = 0 autì eÐnai profanèc. Gia β 6= 0, anjèsoume

z = αx+ βy + γ,

paÐrnoumez′ = α + βy′

4

kai h (5) grfetai:z′ − αβ

= f(z),

dhlad z′ = α + βf(z). (6)

H (6) eÐnai qwrizìmenwn metablht¸n kai oloklhr¸netai.Pardeigma :JewroÔme thn exÐswsh

y′ =√x+ y + 1− 1. (7)

EÐnai thc morf c (5). Jètoume x + y + 1 = z, opìte paragwgÐzontac wc proc xpaÐrnoume 1 + y′ = z′, kai h (7) gÐnetai

z′ =√z. (8)

Oi metablhtèc sthn exÐswsh (8) qwrÐzontai kai, me olokl rwsh twn dÔo mer¸n,prokÔptei ìti

x+ c = 2√z,

me ton periorismìx+ c ≥ 0 x ≥ −c. (9)

Antikajist¸ntac to z paÐrnoume telik

y =(x+ c)2

4− (x+ 1), x ≥ −c, (10)

h opoÐa eÐnai genik lÔsh thc (7). JShmeÐwsh : Ac jewr soume tic kampÔles-lÔseic (10) pou antistoiqoÔn stic

timèc c = 0 kai c = 1 thc aujaÐrethc stajerc. Gia tic timèc autèc paÐrnoume

y =x2

4− (x+ 1), x ≥ 0,

kai

y =(x+ 1)2

4− (x+ 1), x ≥ −1.

Oi dÔo autèc kampÔlec orÐzontai se ìlo to disthma −∞ < x < ∞ kai tèmnontaisto shmeÐo (x, y) = (−1

2,− 7

16), ìpwc faÐnetai kai sto Sq ma 2a. 'Omwc, wc lÔseic

tic diaforik c exÐswshc (7), orÐzontai mìno gia x > 0 kai x > −1, antÐstoiqa,sÔmfwna me ton periorismì (9). 'Ara, wc lÔseic t c D.E. den tèmnontai kai ètsiden parabizetai to je¸rhma monadikìthtac twn lÔsewn tou Cauchy (je¸rhma ??,sel. ??). ShmeÐwsh : Se sunèqeia thc parapnw parat rhshc shmei¸noume kai ta


Sq ma 2: (a) Oi kampÔlec (10) gia c = 0 kai c = 1. LÔseic thc (7) apoteloÔn mìno tasumpag tm mata (x > −c). (b) Endeiktik mèlh thc monoparametrik c oikogèneiaclÔsewn (10) kaj¸c kai h idizousa lÔsh y = −(x+ 1).

akìlouja. 'Estw ìti zhtoÔme, gia pardeigma, th merik lÔsh thc (7) pou perneiapì thn arq twn axìnwn (0, 0). Tìte grfoume ìti prèpei

0 =c2

4− 1 ⇒ c = ±2.

All akrib¸c o periorismìc c ≥ −x (dhlad c ≥ 0) mac odhgeÐ sto na aporrÐyoumethn tim c = −2.

ShmeÐwsh :MporeÐ na deiqjeÐ me mesh antikatstash ìti h (7) èqei idizousalÔsh thn eujeÐa

y = −(x+ 1).

MÐa toulqiston lÔsh thc (7) pernei apì kje shmeÐo (x, y) tou epipèdou efìson x+ y+ 1 ≥ 0. To monos manto ìmwc twn lÔsewn eÐnai exasfalismèno mìno giata shmeÐa me x+ y + 1 > 0 4. Sto Sq ma 2b faÐnetai ìti ìlec oi lÔseic (10) xekinoÔnapì thn idizousa lÔsh kai efptontai s aut .

0.1.2 Probl mata

Prìblhma O plhjusmìc enìc eÐdouc auxnetai anloga me ton uprqonta plhjusmìse kje qronik stigm 5. P¸c exelÐssetai o plhjusmìc tou eÐdouc?

'Estw x0 o plhjusmìc toÔ eÐdouc th qronik stigm t = 0 kai x = x(t) > 0 oplhjusmìc se mia tuqaÐa qronik stigm . O rujmìc h taqÔthta metabol c thc

4Παρατηρήστε ότι για τη Δ.Ε. (7), η ποσότητα ∂f/∂y δεν ορίζεται για x + y + 1 = 0, επομένως

το θεώρημα του Cauchy δεν εγγυάται τη μοναδικότητα των λύσεων στην ευθεία αυτή.5Πρόκειται για το απλούστερο μοντέλο πληθυσμιακής αύξησης ενός είδους.

6

posìthtac x eÐnai x = dxdt

kai, sÔmfwna me ton nìmo thc aÔxhshc, sundèetai me tonplhjusmì x mèsw thc sqèshc

dx

dt= k x, k > 0, (11)

ìpou k o suntelest c analogÐac, o opoÐoc prèpei na eÐnai jetikìc afoÔ èqoume aÔxhshtou plhjusmoÔ, dhlad dx/dt > 0.

H (11) eÐnai qwrizìmenwn metablht¸n kai mac dÐnei

dx

x= k dt ⇒

∫dx

x= k dt ⇒ lnx = kt+ c ⇒ x = c ekt,

ìpou c mia aujaÐreth stajer, profan¸c jetik . Mlista, lìgw thc arqik c sunj -khc x(0) = x0, prokÔptei ìti c = x0 kai ra h lÔsh thc exÐswshc (11) eÐnai h

x = x0 ek t. (12)

An epiplèon upojèsoume ìti o plhjusmìc diplasizetai se qrìno T , dhlad x(T ) = 2x0, tìte apì th (12) paÐrnoume k = ln 2

Tkai h lÔsh grfetai x = x02t/T . J

ShmeÐwsh : An eÐqame meÐwsh tou plhjusmoÔ me ton Ðdio nìmo, tìte ja Ðsquekai pli h exÐswsh (11) all me suntelest analogÐac arnhtikì. 'Etsi ja grfameth D.E. wc dx

dt= −kx me k > 0 kai h lÔsh thc ja tan x = x0 e

−k t (bl. skhsh 21)ShmeÐwsh : An o rujmìc metabol c thc posìthtac x eÐnai anlogoc miac posì-

thtac x1, me suntelest analogÐac k1, kai miac posìthtac x2, me suntelest analogÐack2, tìte autì shmaÐnei ìti dx

dt= k x1 x2, ìpou k = k1 k2 (bl. ask seic 29 kai 30).

Prìblhma S¸ma mzac m af netai na pèsei sto omogenèc pedÐo barÔthtac thcghc, me antÐstash aèra anlogh thc taqÔthtac. Zhtme thn exÐswsh thc taqÔthtacpt¸shc.

Sq ma 3: To Prìblhma ??: pt¸sh s¸matoc me antÐstash

Oi dunmeic pou askoÔntai sto s¸ma eÐnai oi B = mg kai Fα = −kυ, ìpou k > 0o suntelest c analogÐac t c antÐstashc tou aèra. (Parathr ste ìti sto Sq ma 3a


orÐsame wc jetik th for proc ta ktw, dhlad th for thc kÐnhshc.) Gia thnepitqunsh z = dυ

dtpou apokt to s¸ma ja isqÔei o jemeli¸dhc nìmoc thc mhqanik c,

dhlad

mdυ

dt= mg − kυ

dυ

g − (k/m)υ= dt. (13)

Me olokl rwsh h (13) mac dÐnei g− kmυ = ce−

kmt, ìpou c h aujaÐreth stajer h opoÐa

gia thn arqik sunj kh υ(0) = 0 gÐnetai c = g, opìte h lÔsh grfetai

υ =mg

k

(1− e−

kmt).

To grfhma thc parapnw lÔshc parousizetai sto Sq ma 3b. ParathroÔme ìti giat→∞, h taqÔthta teÐnei sthn tim υoρ = mg

kh opoÐa onomzetai oriak taqÔthta. J

Prìblhma Zhtme na broÔme fjÐnousa kampÔlh y = y(x), gia x > 0, pou napernei apì to shmeÐo (1, 1) kai na qarakthrÐzetai apì thn ex c idiìthta: h efaptomènhthc kampÔlhc se tuqìn shmeÐo thc P (x, y) na tèmnei ton xona Ox se shmeÐo A tètoio¸ste to trÐgwno OPA na eÐnai isoskelèc.

H exÐswsh miac eujeÐac se trèqousec suntetagmènec X kai Y kai me klÐsh agrfetai:

Y − Y0 = a(X −X0). (14)

AfoÔ to shmeÐo P (x, y) thc zhtoÔmenhc kampÔlhc y = y(x) an kei sthn efaptomènheujeÐa (14), mporoÔme na jèsoume X0 = x, Y0 = y kai a = y′(x), opìte h (14)grfetai

Y − y = y′(X − x). (15)

EpÐshc, to shmeÐo A = (xA, 0) an kei sthn efaptomènh kai ra plhroÐ th (15). Mpo-roÔme loipìn na jèsoume X = xA = 2x kai Y = 0, epeid to trÐgwno OPT eÐnai

Sq ma 4: To Prìblhma ??.

8

isoskelèc. 'Etsi, h (15) gÐnetai h D.E. qwrizìmenwn metablht¸n

y′ = −yx,

me lÔsh (pou ikanopoieÐ thn arqik sunj kh y(1) = 1) thn uperbol

y =1

x. J

Prìblhma Na brejeÐ h atmosfairik pÐesh p wc sunrthsh tou Ôyouc h apìthn epifneia thc Ghc.

'Estw ènac stoiqei¸dhc kulindrikìc ìgkoc aèra (S× dh) mzac dm (Sq ma 5). Hsunj kh isorropÐac tou ja eÐnai6

S p− S(p+ dp)− g dm = 0⇒ dp

dh= −gρ, (16)

dedomènou ìti dm = ρ S dh, ìpou ρ h puknìthta thc atmìsfairac. All h ρ exarttaiapì to Ôyoc kai thn pÐesh. Gi autì diakrÐnoume dÔo peript¸seic.

1. An deqjoÔme ìti èqoume stajer jermokrasÐa, tìte isqÔei

p dV =dm

µRT ρ =

pµ

RT, (17)

ìpou µ to moriakì broc toÔ aèra, R h pagkìsmia stajer aerÐwn kai T hapìluth jermokrasÐa.

H (16) grfetai, sÔmfwna me th (17),

dp

dh= −k p, k =

gµ

RT. (18)

H (18) eÐnai mia D.E. qwrizìmenwn metablht¸n apì thn opoÐa, gia thn arqik sunj kh p = p0 ìtan h = 0, brÐskoume th lÔsh

p = p0e−k h.

2. An deqjoÔme adiabatik metabol gia th mza dm, èqoume thn exÐswsh

p (dV )γ = Cγ = staj. ρ =dmp1/γ

C.

6Η συνισταμένη των δυνάμεων που ασκούνται στον στοιχειώδη όγκο αέρα, δηλαδή της F1 =

S(p + dp) που εφαρμόζεται στην επάνω διατομή, της F2 = S p που εφαρμόζεται στην κάτω διατομήκαι του βάρους B = dmg, θα πρέπει να είναι μηδέν.


Sq ma 5: Stoiqei¸dhc katakìrufoc kulindrikìc ìgkoc aèra, mzac dm.

H (16) t¸ra grfetai

dp

p1/γ= −λdh, λ =

gdm

C. (19)

H stajer λ eÐnai aprosdiìristh afoÔ perièqei thn posìthta dm. 'Omwc, ìpwcgnwrÐzoume, gia γ → 1 ja prèpei h (19) na sumpÐptei me th (18). 'Ara λ = k.Apì th D.E. (19), kai pli me p = p0 gia h = 0, prokÔptei

p =

p

(γ−1)/γ0 − k(γ − 1)

γh

γ/(γ−1)

. J

Prìblhma Bzoume se leitourgÐa to sÔsthma klimatismoÔ th stigm pou o aèracenìc kleistoÔ dwmatÐou, ìgkou V0 m

3, èqei α% periektikìthta se CO2. Kajarìcaèrac, periektikìthtac β% se CO2 (me β < α), arqÐzei na mpaÐnei sto dwmtio meparoq ωm3

sec. ZhtoÔme thn periektikìthta se CO2 tou dwmatÐou se qrìno t.

'Estw ìti se qrìno t uprqoun sto dwmtio x m3 CO2. Th stigm ekeÐnh hperiektikìthta se CO2 eÐnai x

V0.

Sto disthma apì t mèqri t+ dt mpaÐnoun sto dwmtio ωdt m3 aèra pou perièqounV1 = βωdt/100 m3 CO2. Ston Ðdio qrìno bgaÐnoun apì to dwmtio ωdt m3 aèra pouperièqoun V2 = x

V0ωdt m3 CO2. H diafor V2 − V1 eÐnai Ðsh me th metabol −dx thc

posìthtac CO2 sto dwmtio kat to qronikì disthma dt (to arnhtikì prìshmo thcmetabol c dhl¸nei ìti to x eÐnai fjÐnousa sunrthsh tou qrìnou t).

'Eqoume ètsi th D.E. tou probl matoc

dx

dt= −ω

( xV0

− β

100

). (20)

H (20) eÐnai exÐswsh me qwrizìmenec metablhtèc. H genik thc lÔsh eÐnai:

x = V0

( β

100+ c e−t

ωΩ

). (21)

10

Qreizetai na kajoristeÐ h stajer c. Apì to dedomèno ìti gia t = 0 eÐnaix = αV0/100, prokÔptei ìti c = (α− β)/100. H (21) grfetai t¸ra

x =V0

100

(β + (α− β)e

−t ωV0

).

Epomènwc, h posostiaÐa periektikìthta tou aèra tou dwmatÐou se CO2 eÐnai Ðsh meβ + (α− β)e−tω/V0 . J

0.2 OmogeneÐc diaforikèc exis¸seic

Mia diaforik exÐswsh y′ = f(x, y) kaleÐtai omogen c7 an h sunrthsh f(x, y) eÐnaiomogen c bajmoÔ mhdèn, an dhlad , gia kje λ ∈ R, isqÔei

f(λx, λy) = f(x, y).

EÐnai fanerì ìti mia omogen c sunrthsh bajmoÔ mhdèn eÐnai telik sunrthsh toulìgou8 y/x. MporoÔme epomènwc na poÔme ìti genik morf miac omogenoÔc D.E.1hc txhc eÐnai h

y′ = f(y

x), x 6= 0. (22)

Gia th lÔsh thc (22) jètoume z = yx, opìte, paragwgÐzontac wc proc x, èqoume

y′ = z + xz′, toi

z′ =f(z)− z

x, x 6= 0. (23)

H (23), sthn opoÐa katèlhxe h (22), eÐnai D.E. me qwrizìmenec metablhtèc kailÔnetai sÔmfwna me ìsa perigrfoume sthn Enìthta §0.1.

Pardeigma: Na lujeÐ h D.E.

y′ =x2 + y2

x y, x 6= 0, y 6= 0.

H D.E. grfetai wc omogen c sth morf

y′ =x

y+y

x. (24)

7Στη θεωρία των διαφορικών εξισώσεων ο όρος ομογενής χρησιμοποιείται και με άλλες έννοιες.

Δείτε για παράδειγμα την Ενότητα §0.3.8΄Η, ισοδύναμα, του λόγου x/y

0.2. OMOGENEIS D.E. 11

Jètoume z = y/x, opìte eÐnai y = x z kai y′ = x z′ + z. Antikajist¸ntac sth (24),paÐrnoume

xdz

dx=

1

z,

, qwrÐzontac tic metablhtèc,

z dz =dx

x.

Oloklhr¸nontac paÐrnoume

z2 = 2 ln |x|+ c z = ±√

2 ln |x|+ c, (25)

ìpou c aujaÐreth stajer upì ton periorismì 2 ln |x| + c ≥ 0. Antikajist¸ntac thz = y/x sth (25) paÐrnoume

y = ±x√

2 ln |x|+ c. (26)

Merikèc apì tic parapnw lÔseic parousizontai sto Sq ma 6. J

Sq ma 6: Merikèc apì tic lÔseic thc oikogèneiac (26).

0.2.1 Eidikèc peript¸seic

H D.E. thc morf c

y′ = f

(α1x+ β1y

α2x+ β2y

)

12

mporeÐ pnta na grafeÐ wc omogen c sth morf

y′ = f

(α1 + β1

yx

α2 + β2yx

), x 6= 0.

Mia D.E. thc morf c

y′ = f

(α1x+ β1y + γ1

α2x+ β2y + γ2

)(27)

èrqetai sth morf (22) wc ex c: LÔnoume to sÔsthma

α1x+ β1y + γ1 = 0α2x+ β2y + γ2 = 0.

'Estw (x0, y0) h lÔsh toÔ sust matoc9. Jètoume10 x = x0 + z kai y = y0 + ω, opìte

y′ =dy

dx=dω

dz= f

(α1 + β1

ωz

α2 + β2ωz

), z 6= 0. (28)

H (28) eÐnai omogen c thc morf c (22) kai lÔnetai kat ta gnwst.Pardeigma :

y′ = −x− 2y + 5

2x− y + 4(29)

LÔsh tou sust matocx− 2y + 5 = 02x− y + 4 = 0

eÐnai h (x0 = −1, y0 = 2). Jètoume x = −1 + z kai y = 2 + ω. Epeid y′ = dydx

= dωdz,

h (29) gÐnetaidω

dz= −z − 2ω

2z − ω

dω

dz= −1− 2ω/z

2− ω/z, z 6= 0. (30)

Jètoume ω/z = φ, opìte h omogen c D.E. (30) èrqetai sth morf

(φ− 2)dφ

(1− φ2)=dz

z, φ 6= ±1.

9Δεν υπάρχει λύση αν α1β2 − α2β1 = 0. ΄Ομως, σε αυτή την περίπτωση η Δ.Ε. είναι της μορφής

(5)10Κάνουμε μια παράλληλη μετατόπιση στο σύστημα συντεταγμένων xy, έτσι ώστε οι ευθείες αix+

βiy + γi = 0, με i = 1, 2, να τέμνονται στο (0, 0) και, συνεπώς, οι σταθερές γ1 να απαλείφονται.

0.2. OMOGENEIS D.E. 13

H teleutaÐa aut exÐswsh eÐnai D.E. qwrizìmenwn metabsewn. H genik lÔshthc eÐnai h

|(φ− 1)/(φ+ 1)3| = cz2,

h opoÐa, an epanafèroume tic metabseic z kai ω, paÐrnei th morf |(ω−z)/(ω+z)3| =c. GurÐzontac akìmh pio pÐsw stic metablhtèc x kai y, brÐskoume wc genik lÔsh thc(29) thn

|x− y + 3| = c|x+ y − 1|3. J

0.2.2 Probl mata

OmogeneÐc diaforikèc exis¸seic prokÔptoun se probl mata eÔreshc troqi¸n katadÐw-xhc (pursuit curves). Prìkeitai gia troqièc pou akoloujeÐ èna s¸ma ìtan kateujÔneithn kÐnhs tou proc èna stajerì kinoÔmeno shmeÐo. 'Ena deÐgma aut c thc kathgo-rÐac problhmtwn apoteleÐ to paraktw prìblhma.

Prìblhma Kolumbht c kolumpei apì th mia ìqjh enìc potamoÔ (shmeÐo A)proc èna shmeÐo B akrib¸c apènanti, sthn llh ìqjh, se apìstash d. Kolumpei metaqÔthta stajeroÔ mètrou υκ kai me kateÔjunsh pnta proc to O. An to potmi rèeikai parasÔrei ton kolumbht me taqÔthta υπ, na brejeÐ h troqi pou akoloujeÐ okolumbht c.

JewroÔme to sÔsthma suntetagmènwn tou Sq matoc 7, ìpou èstw K(x, y), mex ≥ 0, y ≥ 0, h jèsh tou kolumbht thn tuqaÐa qronik stigm t. H taqÔthta me thnopoÐa prospajeÐ na kinhjeÐ o kolumbht c eÐnai

~υκ = υκ~KO

| ~KO|= −υκ cos θ~i− υκ sin θ~j,

ìpou ~i kai ~j eÐnai ta monadiaÐa dianÔsmata twn axìnwn Ox kai Oy antÐstoiqa.

Sq ma 7: Sqhmatik parstash thc troqic tou kolumbht kai to sÔsthma sunte-tagmènwn sto opoÐo perigrfetai.

14

H taqÔthta me thn opoÐa parasÔretai o kolumbht c apì to potmi eÐnai ~υπ = υπ~j.H sunolik taqÔthta ~υ tou kolumbht ja eÐnai h sunistamènh twn ~υκ kai ~υπ, dhlad

~υ = ~υκ + ~υπ = −υκ cos θ~i+ (υπ − υκ sin θ)~j.

Se kje stigm , h troqi y = y(x) ja efptetai sthn taqÔthta ~υ = υx~i+ υy~j, opìteja eÐnai

dy

dx=υyυx

= −υπ − υκ sin θ

υκ cos θ= −a− sin θ

cos θ, (31)

ìpou a = υπ/υκ.Gia to tuqaÐo shmeÐo K(x, y), kai sÔmfwna me to Sq ma 7b, isqÔoun oi sqèseic

sin θ =y√

x2 + y2kai cos θ =

x√x2 + y2

oi opoÐec, an antikatastajoÔn sth (31) kai met apì lÐgec prxeic, mac dÐnoun thsqèsh

dy

dx= −a

√1 +

(yx

)2

+y

x. (32)

H (32) apoteleÐ th D.E. thc troqic tou kolumbht kai eÐnai omogen c. An jèsoumez = y/x kai y′ = x z′+z, oi metablhtèc qwrÐzontai kai h D.E. oloklhr¸netai dÐnontac

ln(z +√

1 + z2)

= −a lnx+ c, x > 0. (33)

Gia x = d, eÐnai y = 0 z = 0 (arqikèc sunj kec) kai h aujaÐreth stajerprosdiorÐzetai sthn tim c = a ln d. Antikajist¸ntac th c kai th z = y/x sth (33),paÐrnoume met apì lÐgec prxeic

y

x+

√1 +

y2

x2=

(d

x

)a, x > 0,

kai h exÐswsh thc troqic tou kolumbht grfetai

y =x

2

(d

x

)a−(d

x

)−a. (34)

Sto Sq ma 8 parousizetai h troqi (34) gia diforec timèc tou lìgou taqut twna = υπ/υκ. ParathroÔme pwc an υκ > υπ, opìte a < 1, tìte o kolumbht c diagrfeikampÔlh troqi apì th mia ìqjh sthn llh, h opoÐa eÐnai tìso suntomìterh ìsomegalÔterh eÐnai h taqÔthta υκ me thn opoÐa mporeÐ na kolump sei. An υπ > υκ,opìte a > 1, o kolumbht c parasÔretai apì to reÔma tou potamoÔ kai den ftnei sthnapènanti ìqjh11. Sthn krÐsimh perÐptwsh ìpou υπ = υκ, opìte a = 1, o kolumbht cftnei sthn apènanti ìqjh all se apìstash d/2 apì ton arqikì tou stìqo (apodeÐxteto!). J

11Αν βέβαια συνεχίσει να κολυμπάει με κατεύθυνση τον αρχικό του στόχο (σημείο Β ή Ο) και δεν

αλλάξει στρατηγική.

0.3. GRAMMIKES DIAFORIKES EXISWSEIS 15

Sq ma 8: Peript¸seic troqi¸n (34) tou kolumbht gia diaforetikèc timèc tou lìgoua = υπ/υκ.

0.3 Grammikèc diaforikèc exis¸seic

Prìkeitai gia diaforikèc exis¸seic thc morf c

y′ + g(x)y + h(x) = 0. (35)

Gia thn anexrthth metablht ja jewr soume to disthma α < x < β sto opoÐo oisunart seic g(x) kai h(x) eÐnai suneqeÐc.

An h(x) = 0, h exÐswsh (35) grfetai

y′ + g(x)y = 0. (36)

H (36) kaleÐtai omogen c grammik diaforik exÐswsh 1hc txewc, eÐnai exÐswsh meqwrizìmenec metablhtèc kai èqei profan merik lÔsh thn y = 0. MporoÔme eÔkolana doÔme ìti h genik lÔsh thc eÐnai h

y = c e−∫g(x)dx, (37)

ìpou h c eÐnai aujaÐreth stajer.ShmeÐwsh : EÐnai fanerì ìti an φ1(x) kai φ2(x) eÐnai dÔo merikèc lÔseic thc

(36), tìte kai kje èkfrash thc morf c

y = C1φ1 + C2φ2,

me C1 kai C2 stajerèc, eÐnai lÔsh thc D.E. Autì eÐnai apotèlesma thc grammikìthtacthc exÐswshc, oi lÔseic thc opoÐac apoteloÔn dianusmatikì q¸ro (blèpe kai Keflaio??).

16

Erqìmaste t¸ra na lÔsoume thn pl rh (ìpwc thn onomzoume) D.E. (35). Giaton skopì autìn ja qrhsimopoi soume th mèjodo pou eÐnai gnwst wc mèjodoc thcmetabol c twn aujairètwn stajer¸n mèjodoc tou Lagrange. Ja jewr soume lÔshthc morf c (37), all sth jèsh thc stajerc c ja èqoume mia sunrthsh z(x) kai japrosdiorÐsoume aut n th sunrthsh ètsi ¸ste lÔsh thc D.E. (35) na eÐnai h

y = z(x)e−∫g(x)dx. (38)

An antikatast soume th (38) sth (35) paÐrnoume

z′e−∫g(x)dx − g(x)z(x)e−

∫g(x)dx + g(x)z(x)e−

∫g(x)dx + h(x) = 0,

dhlad

z′ = −h(x)e∫g(x)dx

z = c−∫h(x)e

∫g(x)dxdx.

'Ara h genik lÔsh thc (35) eÐnai h

y = c−∫h(x)e

∫g(x)dxdxe−

∫g(x)dx. (39)

ShmeÐwsh : H lÔsh (37) prokÔptei wc eidik perÐptwsh thc lÔshc (39) an tejeÐh(x) = 0.

ShmeÐwsh : Oi stajerèc pou prokÔptoun apì tic oloklhr¸seic ston tÔpo (39)mporoÔn na enswmatwjoÔn sthn aujaÐreth stajer c kai, kat sunèpeia, mporoÔn naagnohjoÔn (ènac austhrìc anagn¸sthc ja prepe kai ja mporoÔse na to diapist¸seiautì o Ðdioc!).

EÐnai jèma merik¸n prxewn na deÐxei kaneÐc ìti h sunrthsh

φ0(x) = −∫

h(x)e∫g(x)dxdx

e−

∫g(x)dx

eÐnai mia merik lÔsh thc D.E. (35). Epeid ìmwc h (39) grfetai kai wc

y = c e−∫g(x)dx + φ0(x),

prokÔptei, sÔmfwna kai me thn exÐswsh (37), ìti h genik lÔsh miac pl rouc grammik cD.E. eÐnai jroisma dÔo ìrwn: thc genik c lÔshc thc antÐstoiqhc omogenoÔc D.E. kaimiac merik c lÔshc thc pl rouc D.E. H idiìthta aut qarakthrÐzei kai tic grammikècD.E. an¸terhc txhc (bl. Keflaio ??).

0.3. GRAMMIKES DIAFORIKES EXISWSEIS 17

Pardeigma : 'Estw h grammik D.E.

y′ +1− 2x

x2y − 1 = 0, x 6= 0.

'Eqoume apì ton tÔpo (39) ìti

y = c+

∫e∫

1−2x

x2 dxdxe−∫

1−2x

x2 dx.

EÐnai ìmwc ∫1− 2x

x2dx = −1

x− 2 ln | x | = −1

x− lnx2

kai ra

y =(c+

∫e−1x

x2 dx)x2e

1x

=(c+ e−

1x

)x2e

1x =

(1 + c e

1x

)x2, x 6= 0,

ìpwc llwste mporeÐ kai na epalhjeuteÐ. (SÔstas mac proc ìlouc touc anagn¸steceÐnai na epalhjeÔoun pnta tic lÔseic pou brÐskoun.) J

Pardeigma : JewroÔme th D.E.

y′ =1

xy2 + 1(40)

h opoÐa metatrèpetai se grammik an jewr soume wc gnwsth sunrthsh th x = x(y).Prgmati, h (40) grfetai wc

dx

dy− y2x− 1 = 0

kai eÐnai grammik me g(y) = y2 kai h(y) = 1. Efarmìzontac ton antÐstoiqo tÔpo(39), all gia th x = x(y), paÐrnoume12

x = (c+ γ(y))ey3/3, ìpou γ(y) =

∫e−y

3/3dy. J

0.3.1 Probl mata

Oi grammikèc D.E. brÐskoun efarmog se poll jèmata Mhqanik c kai HlektrismoÔ.Ja d¸soume ed¸ dÔo paradeÐgmata apì ton Hlektrismì. UpenjumÐzoume gia tonskopì autì merik basik peiramatik dedomèna sqetik me ta hlektrik kukl¸mata.

12Η γ(y) υπολογίζεται με τη βοήθεια της ειδικής συνάρτησης Γ(a, x) =

∫∞xta−1e−tdt.

18

(a) H pt¸sh thc tshc ER sta kra miac antÐstashc R eÐnai anlogh proc thnèntash I tou reÔmatoc pou diarrèei thn antÐstash. EÐnai dhlad

ER = RI.

(b) H pt¸sh tshc EL sta kra enìc phnÐou eÐnai anlogh proc thn taqÔthtametabol c thc èntashc tou reÔmatoc. EÐnai dhlad

EL = LdI

dt.

(g) H pt¸sh tshc EC sta kra enìc puknwt eÐnai anlogh proc to hlektrikìfortÐo Q tou puknwt . EÐnai dhlad

EC =1

CQ.

Me bsh ta parapnw ekfrzoume ton 2o Nìmo tou Kirchoff: H hlektregertik dÔnamh E = E(t) pou efarmìzetai sta kra enìc kleistoÔ kukl¸matoc RLC eÐnaiÐsh me to jroisma twn pt¸sewn tsewn kat m koc tou kukl¸matoc, dhlad

LdI

dt+RI +

1

CQ = E(t). (41)

Ja jewr soume kukl¸mata me mÐa antÐstash kai èna hlektrikì stoiqeÐo, phnÐo puknwt .

Sq ma 9: Ta hlektrik kukl¸mata (a) RL kai (b) RC.

SÔmfwna me ta parapnw, to reÔma sto kÔklwma me antÐstash kai phnÐo touSq matoc 9a ja dÐnetai apì th grammik D.E.

LdI

dt+RI = E(t). (42)

Gia suneqèc reÔma E = E0 =stajerì, h (42) èqei genik lÔsh thn

I(t) =E0

R+ ce−

RLt.

0.4. EIDIKES MORFES D.E. POU ANAGONTAI SE GRAMMIKES 19

Gia enallassìmeno reÔma thc morf c E(t) = E0 sin (ωt), h (42) èqei genik lÔshthn

I(t) = ce−RLt +

E0√R2 + ω2L2

sin (ωt− φ),

ìpou h c eÐnai mia stajer kai tan (φ) = ωLR

Gia to kÔklwma me antÐstash kai puknwt (Sq ma 9b), èqoume

RI +Q

C= E(t)

, epeid I = dQ/dt,

RdI

dt+I

C=dE

dt. (43)

'Eqoume dhlad kai pli mia grammik D.E. me gnwsth th sunrthsh I = I(t). GiaE = E0, apì th (43) prokÔptei I(t) = ce−

tRC .

Gia E = E0 sin (ωt), prokÔptei apì th (43)

I(t) = ce−t

RC +ωE0C√

1 + ω2R2C2sin (ωt− φ),

ìpou tan (φ) = − 1ωRC

.

0.4 Eidikèc morfèc D.E. pou angontai se

grammikèc

0.4.1 H diaforik exÐswsh tou BernoulliPrìkeitai gia D.E. thc morf c

y′ + g(x)y + h(x)yρ = 0, ρ 6= 0, 1, (44)

me tic sunart seic g(x) kai h(x) suneqeÐc se kpoio disthma I = (α, β) ⊆ R. Gia ρ = 0 h(44) eÐnai pl rhc grammik , en¸ gia ρ = 1 eÐnai omogen c grammik diaforik exÐswsh.

Prosèqoume pnta na èqei ènnoia h dÔnamh yρ. 'Etsi, p.q., gia y > 0 h yρ èqei ènnoia giaìlec tic pragmatikèc timèc pou mporeÐ na prei to ρ. 'H, an to ρ eÐnai akèraioc, to yρ orÐzetaigia ìla ta y. An ìmwc, gia pardeigma, ρ = 1/2, h dÔnamh yρ orÐzetai mìno gia y ≥ 0.

H exÐswsh (44) metatrèpetai se grammik D.E. me ton metasqhmatismì

y = z1/(1−ρ). (45)

Prgmati, eÐnai tìte

y′ =1

1− ρzρ/(1−ρ)z′

20

kai h (44) grfetai

1

1− ρzρ/(1−ρ)z′ + g(x)z1/(1−ρ) + h(x)zρ/(1−ρ) = 0

, met apì pollaplasiasmì me z−ρ/(1−ρ),

1

1− ρz′ + g(x)z + h(x) = 0.

Katal goume dhlad se mia grammik D.E. me gnwsth th sunrthsh z = z(x). Me thbo jeia thc (45) brÐskoume sth sunèqeia thn y = y(x).

Pardeigma :

y′ + xy − x3y3 = 0 (46)

EÐnai exÐswsh Bernoulli me ρ = 3. 'Etsi, sÔmfwna me th (45), jètoume y = z−1/2 me z > 0kai katal goume sth grammik D.E.

z′ − 2xz + 2x3 = 0,

me genik lÔsh thn

z = c ex2

+ 1 + x2, z > 0.

Epomènwc, h genik lÔsh thc (46) eÐnai h

y =1√

c ex2 + 1 + x2, c ex

2+ 1 + x2 > 0. J

0.4.2 H diaforik exÐswsh tou RiccatiPrìkeitai gia D.E. pou mporoÔme na th fèroume sth morf

y′ = f(x)y2 + g(x)y + h(x). (47)

Upojètoume ìti oi sunart seic f(x), g(x) kai h(x) eÐnai suneqeÐc se kpoio disthma I =(α, β) ⊆ R. ParathroÔme ìti an h(x) ≡ 0, ∀x ∈ I, h (47) eÐnai thc morf c Bernoulli meρ = 2. EpÐshc parathroÔme ìti an f(x) ≡ 0, ∀x ∈ I, h (47) eÐnai grammik .

Sth genik thc morf h exÐswsh (47) den mporeÐ na lujeÐ par mìno an eÐnai gnwst miamerik lÔsh thc, p.q. h y = ψ(x). Th lÔsh aut n th brÐskoume ap eujeÐac apì thn exÐswsh(47), manteÔontac endeqomènwc th morf thc. An brejeÐ mia merik lÔsh ψ(x), efarmìzoumeton metasqhmatismì

y = ψ(x) +1

z, ìpou z = z(x) 6= 0, (48)

kai h exÐswsh (47) angetai se grammik me gnwsth th sunrthsh z = z(x). Prgmati,antikajist¸ntac th (48) sto dexÐ mèloc thc (47) paÐrnoume

y′ = f(x)

(ψ(x)2 +

1

z2+

2ψ(x)

z

)+ g(x)

(ψ(x) +

1

z

)+ h(x). (49)

0.5. PLHREIS DIAFORIKES EXISWSEIS KAI POLLAPLASIASTES OôLER21

EpÐshc, me parag¸gish thc (48) wc proc x,

y′ = ψ′(x)− z′

z2. (50)

Exis¸nontac ta dexi mèlh twn (49) kai (50) kai lambnontac upìyh ìti h ψ(x) eÐnai lÔshthc (47), dhlad eÐnai ψ′ = fψ2 + gψ + h, paÐrnoume telik th sqèsh

z′ + (2f(x)ψ(x) + g(x))z + f(x) = 0 (51)

pou apoteleÐ mia grammik D.E. gia th sunrthsh z = z(x). BrÐskoume th genik lÔshz = z(x, c) thc (51), opìte, sÔmfwna me th (48), h genik lÔsh thc D.E. tou Riccati (47)ja eÐnai h

y = ψ(x) +1

z(x, c).

Pardeigma : Na lujeÐ h D.E.

y′ = −y2 + (2x2 − 1)y − (x4 − x2 − 2x) (52)

ParathroÔme ìti y = ψ(x) = x2 eÐnai mia merik lÔsh thc (52). Jètoume

y = x2 +1

z, z 6= 0,

opìte, akolouj¸ntac th mejodologÐa pou perigryame parapnw, ap th (52) prokÔptei h

z′ − z − 1 = 0. (53)

H (53) eÐnai grammik kai èqei wc genik lÔsh thn

z = c ex − 1, c ex − 1 6= 0.

ProkÔptei, loipìn, ìti genik lÔsh thc exÐswshc (52) tou Riccati eÐnai h

y = x2 +1

c ex − 1. J

0.5 Pl reic diaforikèc exis¸seic kai polla-

plasiastèc 'Oôler

0.5.1 Pl rhc D.E.

'Estw h diaforik exÐswsh 1hc txhc

y′ = −P (x, y)

Q(x, y),

22

me P kai Q suneqeÐc sunart seic se ènan tìpo T , kai Q(x, y) 6= 0. Grfoume th D.E.sth morf

P (x, y)dx+Q(x, y)dy = 0. (54)

An uprqei mia sunrthsh Φ(x, y), pou na orÐzetai kai na èqei suneqeÐc merikècparag¸gouc ∂Φ

∂xkai ∂Φ

∂yston apl¸c sunektikì tìpo T , tètoia ¸ste to diaforikì thc

dΦ =∂Φ

∂xdx+

∂Φ

∂ydy

na isoÔtai me to aristerì mèloc thc (54), h D.E. onomzetai pl rhc13. 'Etsi miapl rhc D.E. grfetai wc dΦ(x, y) = 0 kai oloklhr¸netai mesa dÐnontac wc genik lÔsh th

Φ(x, y) = c. (55)

H (55) paristnei mia monoparametrik oikogèneia kampul¸n pou dÐnontai se pe-plegmènh morf . MporeÐ pntwc na elegqjeÐ ìti, an gia kpoia tim thc c = c0,h kampÔlh Φ(x, y) = c0 pernei apì to shmeÐo (ξ, η) ∈ T , tìte aut eÐnai kai hmình lÔsh thc (54) pou pernei apì to sugkekrimèno shmeÐo. Kai toÔto diìti h(∂Φ∂y

)(ξ,η) = Q(ξ, η) 6= 0 kai epomènwc h exÐswsh Φ(x, y) = c0 lÔnetai monìtima wcproc y kai dÐnei y = φ(x).

SÔmfwna me ton parapnw orismì thc plhrìthtac, ja isqÔoun oi sqèseic

∂Φ

∂x= P (x, y),

∂Φ

∂y= Q(x, y), (x, y) ∈ T . (56)

IsodÔnama, mporoÔme na orÐsoume th D.E. (54) wc pl rh an isqÔoun oi sqèseic(56) gia mia sunrthsh Φ = Φ(x, y).

SÔmfwna me to je¸rhma thc plhrìthtac, ikan kai anagkaÐa sunj kh gia na eÐnaih (54) pl rhc eÐnai h

∂P (x, y)

∂y=∂Q(x, y)

∂x, ∀(x, y) ∈ T . (57)

'Ara, ìtan mac dÐnetai mia exÐswsh sth morf thc (54), elègqoume an isqÔei hsunj kh (57). An autì sumbaÐnei, tìte h olokl rwsh thc dΦ(x, y) kai, sunep¸c, heÔresh thc Φ(x, y), dÐnetai apì ton tÔpo

Φ(x, y) =

∫ x

x0

P (χ, y)dχ+

∫ y

y0

Q(x0, ψ)dψ, (58)

ìpou (x0, y0) èna stajerì shmeÐo tou T . Proôpìjesh gia na isqÔei h (58) eÐnaito parallhlìgrammo pou orÐzetai apì ta shmeÐa (x0, y0), (x, y0), (x0, y) kai (x, y) na

13Θυμηθείτε ότι ο όρος πλήρης χρησιμοποιήθηκε με άλλη έννοια στην Ενότητα 0.3.


Sq ma 10: H diadrom olokl rwshc thc pl rouc diaforik c exÐswshc (54).

an kei exolokl rou ston apl¸c sunektikì tìpo T . EpÐshc, h Φ(x, y) mporeÐ na brejeÐme olokl rwsh twn sqèsewn (??). Sth sunèqeia parajètoume dÔo paradeÐgmata aut¸ntwn dÔo mejìdwn olokl rwshc, h diadrom twn opoÐwn apeikonÐzetai sto Sq ma 10.

Pardeigma :

(1

y+ x)dx− x

y2dy = 0 (59)

H (59) eÐnai pl rhc diìti

∂

∂y

(1

y+ x

)=

∂

∂x

(− x

y2

)= − 1

y2.

Epomènwc uprqei Φ(x, y) thn opoÐa mporoÔme na broÔme mèsw thc sqèshc (58).'Eqoume ∫ x

x0

P (χ, y)dχ =

∫ x

x0

(1

y+ χ)dχ =

χ

y+χ2

2

xx0

=x

y+x2

2− x0

y+x2

0

2

kai ∫ y

y0

Q(x0, ψ)dψ =

∫ y

y0

−x0

ψ2dψ =

x0

ψ

∣∣yy0

=x0

y− x0

y0

.

Sunep¸c

Φ(x, y) =x

y+x2

2+x2

0

2− x0

y0︸︷︷︸staj.

.

24

'Ara, sÔmfwna me th (55), h genik lÔsh thc (59) ja eÐnai h

x

y+x2

2= c y =

2x

2c− x2, x2 6= 2c. J

Pardeigma :

x(1− y2)dx+ y(1− x2)dy = 0, y 6= 0, x 6= ±1. (60)

H (60) eÐnai pl rhc diìti

∂

∂y

(x(1− y2)

)=

∂

∂x

(y(1− x2)

)= −2xy.

'Ara uprqei sunrthsh Φ = Φ(x, y) gia thn opoÐa na isqÔoun oi sqèseic (??), dhlad

∂Φ

∂x= x(1− y2), (61)

∂Φ

∂y= y(1− x2). (62)

Oloklhr¸noume th (61) wc proc x jewr¸ntac to y stajerì:

Φ =

∫(y=σταθ)

x(1− y2)dx =x2

2(1− y2) + C(y). (63)

ParagwgÐzontac th (63) wc proc y, paÐrnoume

∂Φ

∂y= −x2y +

dC(y)

dy. (64)

Exis¸nontac ta dexi mèlh twn (62) kai (64), èqoume

y − x2y = −x2y +dC(y)

dy⇒ dC(y)

dy= y ⇒ C(y) =

y2

2+ c0,

ìpou c0 mia stajer. 'Ara apì th (63) èqoume ìti Φ = x2

2(1 − y2) + y2

2+ c0 kai h

genik lÔsh thc (60) ja eÐnai h

x2

2(1− y2) +

y2

2= c. J


0.5.2 Pollaplasiast c 'Oôler

Ac upojèsoume t¸ra ìti mac dìjhke h exÐswsh (54) kai ìti den isqÔei h sunj khplhrìthtac (57). MporoÔme wstìso na jèsoume to er¸thma: M pwc uprqei diafo-rÐsimh sunrthsh R(x, y) 6= 0 tètoia ¸ste h D.E.

R(x, y)P (x, y)dx+R(x, y)Q(x, y)dy = 0 (65)

na eÐnai pl rhc?Mia tètoia sunrthsh (an mporeÐ na brejeÐ) onomzetai pollaplasiast c tou Euler

alli¸c oloklhrwtikìc pargontac.Pollaplasizoume ta dÔo mèlh thc (54) me ton pargonta R(x, y) ¸ste na odhgh-

joÔme se mia isodÔnamh D.E. (th 65) h opoÐa na eÐnai pl rhc. H lÔsh tìte ja mporeÐna brejeÐ sÔmfwna me ìsa ekjèsame prohgoumènwc.

Pìte ìmwc h exÐswsh (65) eÐnai pl rhc? Gia na sumbaÐnei autì prèpei na eÐnai

∂(RP )

∂y=∂(RQ)

∂x⇒ PyR + PRy = QxR +QRx,

ìpou oi deÐktec stic sunart seic dhl¸noun tic antÐstoiqec merikèc parag¸gouc. Hparapnw exÐswsh grfetai telik:

RyP −QRx = R(Qx − Py). (66)

Prèpei loipìn h sunrthsh R(x, y) na epilegeÐ ètsi ¸ste na ikanopoieÐ th (66). HteleutaÐa aut sqèsh eÐnai D.E. me merikèc parag¸gouc kai h lÔsh thc eÐnai, en gènei,prìblhma polÔ pio dÔskolo apì th lÔsh thc arqik c exÐswshc (54). Wstìso, denzhtme th genik lÔsh thc (66) all mia apì tic lÔseic thc. SumbaÐnei loipìn kpoiecforèc na nai eÔkolh h eÔresh miac tètoiac lÔshc, dhlad enìc pollaplasiast Euler.Merikèc tètoiec peript¸seic parousizoume sth sunèqeia.

a) pollaplasiast c thc morf c R = R(y)

'Estw ìti anazhtme ènan pollaplasiast pou na exarttai mìno apì th metablht y, dhlad R = R(y). H exÐswsh (66) grfetai tìte

Ry

R=Qx − Py

P. (67)

Apì th (67) sumperaÐnoume ìti gia na uprqei pollaplasiast c Euler thc morf cR = R(y), prèpei h parstash (Qx − Py)/P na eÐnai anexrthth tou x, dhlad

Qx − PyP

= σ(y). (68)

26

'Ara, ìtan mac dojeÐ h (54) elègqoume m pwc prgmati h èkfrash (68) eÐnai anexrththtou x (exarttai dhlad mìno apì to y). An autì sumbaÐnei, tìte h (67) eÐnai sun jhcD.E. pr¸thc txhc, qwrizìmenwn metablht¸n, me mia lÔsh thn

R(y) = e∫σ(y)dy.

Pardeigma : H D.E.

(y + x y2)dx− xdy = 0 (69)

den eÐnai pl rhc diìti∂(y + xy2)

∂y6= ∂(−x)

∂x.

SumbaÐnei ìmwc na isqÔei gi aut thn exÐswsh h sqèsh (68), eÐnai dhlad

Qx − PyP

=−1− (1 + 2xy)

y(1 + xy)= −2

y= σ(y).

'Ara uprqei pollaplasiast c Euler thc morf c R = R(y), gia ton opoÐo brÐskoumeapì thn exÐswsh (67) ìti eÐnai

Ry

R= −2

y⇒ R =

1

y2.

Prgmati, pollaplasizontac ta dÔo mèlh thc (69) me th sunrthsh R = 1y2 , paÐr-

noume thn isodÔnamh D.E.

(1

y+ x)dx− x

y2dy = 0 (70)

h opoÐa, ìpwc mporoÔme na diapist¸soume amèswc, eÐnai pl rhc. LÔnontac kat tagnwst, brÐskoume apì th (70) ìti

y = − 2x

x2 + c, x2 + c 6= 0. (71)

H (71) eÐnai lÔsh tìso thc (70) ìso kai thc (69).J

b) pollaplasiast c thc morf c R = R(x)

An to aristerì mèloc thc (68) den eÐnai anexrthto tou x (prgma pou mìno sum-ptwmatik ja mporoÔse na sumbaÐnei), anazhtoÔme pollaplasiast Euler thc morf cR = R(x). Tìte h exÐswsh (66) gÐnetai

Rx

R= −Qx − Py

Q. (72)


'Ara, gia na uprqei tètoioc pollaplasiast c, prèpei

Qx − PyQ

= φ(x), (73)

dhlad to dexÐ mèloc thc (72) na eÐnai anexrthto tou y. BrÐskoume tìte apì th (72)ìti h

R(x) = e−∫φ(x)dx

eÐnai ènac katllhloc pollaplasiast c.

g) Pollaplasiastèc R = R(x, y) eidik c morf c

Oi dÔo peript¸seic pou anafèrame gia thn eÔresh miac lÔshc thc D.E. (54) den eÐnaioi mìnec. Se arket biblÐa diaforik¸n exis¸sewn parousizontai diforec morfècpollaplasiast¸n Euler pou ja mporoÔsame na anazht soume, kai endeqomènwc naprosdiorÐsoume, ¸ste mia exÐswsh ìpwc h (54) na gÐnei pl rhc kai epilÔsimh. Denja exantl soume ed¸ to jèma, ja anafèroume mìno ìti suqn qreizetai kaneÐc naautenerg sei kai, èqontac exoikeiwjeÐ me thn exÐswsh pou èqei na lÔsei, na anazht seiènan pollaplasiast Euler me ìpoion trìpo jewreÐ pio prìsforo. DÐnoume ed¸ ènatètoio pardeigma.

Pardeigma : H D.E.

(4xy + 3y4)dx+ (2x2 + 5xy3)dy = 0 (74)

den eÐnai pl rhc. Kai ìqi mìno autì, all oÔte h (68) oÔte h (73) isqÔoun, ìpwceÔkola mporoÔme na diapist¸soume.

H poluwnumik morf thc (74) wstìso mac dÐnei thn idèa na anazht soume ènanpollaplasiast thc morf c

R(x, y) = xκyλ.

QwrÐc fusik na eÐnai bèbaio ìti tètoioc pollaplasiast c uprqei, epiqeiroÔme nakajorÐsoume ta κ kai λ ¸ste h exÐswsh

(4xκ+1yλ+1 + 3xκyλ+4)︸︷︷︸P (x,y)

dx+ (2xκ+2yλ + 5xκ+1yλ+3)︸︷︷︸Q(x,y)

dy = 0

na eÐnai pl rhc, dhlad ∂P/∂y = ∂Q/∂x

4(λ+ 1)xκ+1yλ + 3(4 + λ)xκy3+λ = 2(κ+ 2)xκ+1yλ + 5(κ+ 1)xκy3+λ.

'Ara prèpei4(λ+ 1) = 2(κ+ 2)3(4 + λ) = 5(κ+ 1)

⇒ κ = 2, λ = 1.

28

Br kame loipìn ènan pollaplasiast Euler, ton R(x, y) = x2y, kai h exÐswsh (74)gÐnetai

(4x3y2 + 3x2y5)dx+ (2x4y + 5x3y4)dy = 0.

Aut h teleutaÐa D.E. eÐnai pl rhc kai lÔnetai kat ta gnwst. JPrìblhma Na brejeÐ mia kampÔlh tou epipèdou xy pou na pernei apì to shmeÐo

(2, 1) kai na èqei thn ex c idiìthta: to trÐgwno pou sqhmatÐzetai apì thn epibatik aktÐna OM sto tuqaÐo shmeÐo M(x, y) thc kampÔlhc, thn efaptomènh sto shmeÐo Mkai ènan apì touc hmixonec Ox na èqei stajerì embadìn E = 1.

Ta gewmetrik probl mata, ìpwc autì, prèpei na diatup¸nontai kai na antimetw-pÐzontai me th megalÔterh dunat prosoq , epeid to sq ma pou ja qrhsimopoi soumegia thn epÐlus tou mporeÐ na mac odhg sei se paraplanhtik apotelèsmata. H exoi-keÐwsh me to sugkekrimèno prìblhma mac odhgeÐ se kpoiouc qr simouc periorismoÔc.Sto jèma mac p.q. mporoÔme na sumpernoume gia thn kampÔlh pou anazhtme ìti(a) den mporeÐ na tèmnei ton xona twn x kai (b) den mporeÐ h efaptomènh se kpoioshmeÐo thc na pernei apì thn arq twn axìnwn. Diìti, stic peript¸seic autèc, jamhdenizìtan to embadìn tou trig¸nou pou kajorÐzei h ekf¸nhsh.

H kampÔlh pou zhtoÔme loipìn ja qei eÐte y > 0 eÐte y < 0 s ìla ta shmeÐa thc.AfoÔ, sÔmfwna me thn ekf¸nhsh, jèloume h kampÔlh na pernei apì to shmeÐo (2, 1),ja qei y > 0 s ìla ta shmeÐa thc.

H exÐswsh thc efaptomènhc t c kampÔlhc sto tuqaÐo shmeÐo thc M(x, y) ja eÐnai

Y − y = (X − x)dY

dX|M . (75)

Apì th (75) brÐskoume ìti h tetmhmènh tou shmeÐou K sto opoÐoh efaptomènh MK tèmnei ton xona twn x ja eÐnai Ðsh me

x− y

dy/dx6= 0,

sÔmfwna me ton parapnw periorismì (b).Ja prèpei loipìn na èqoume gia to embadìn tou trig¸nou

1

2|x− y

dy/dx|y = 1

ε(x− y

dy/dx)y = 2, (76)

ìpou ε = ±1 gia to endeqìmeno na tèmnei h efaptomènh MK ton jetikì arnhtikìhmixona twn x antÐstoiqa.


Sq ma 11: Oi lÔseic (80) kai (81) (kampÔlec I kai II antÐstoiqa).

H (76) grfetaiεy2dx+ (2− εxy)dy = 0. (77)

H (77) den eÐnai pl rhc. Epidèqetai ìmwc, ìpwc eÔkola mporeÐ na apodeiqjeÐ, ènanpollaplasiast Euler thc morf c

R(y) =1

y3.

Pl rhc loipìn eÐnai h D.E.

ε

ydx+ (

2

y3− εx

y2)dy = 0. (78)

Apì aut n prokÔptei

Φ(x, y) =εx

y− 1

y2.

Genik lÔsh thc (78), isodÔnama thc (77), eÐnai loipìn h

Φ(x, y) = c ⇒ x =1

ε(c y +

1

y). (79)

Apì th (79) paÐrnoume wc merikèc lÔseic, pou pernoÔn apì to shmeÐo (2, 1), tic ex cdÔo sunart seic (Sq ma 11):

I. gia ε = 1 thn

x = y +1

y(80)

II. gia ε = −1 thn

x = 3y − 1

y(81)

30

0.6 Allag metablht¸n

Se pollèc apì tic kathgorÐec D.E. pou melet same wc ed¸, jewr same skìpimo na eisa-ggoume mia kainoÔrgia (bohjhtik ) metablht . Gia pardeigma, sth lÔsh thc omogenoÔcdiaforik c exÐswshc y′ = f( yx) jèsame y

x = z, gia th lÔsh thc exÐswshc tou Bernoulli:y′ + g(x)y + h(x)yρ = 0 jèsame y = z1/(1−ρ), k.lp. 'Allote pli, ìpwc sthn perÐptwsh thcexÐswshc (27) thc Enìthtac §0.2.1, antÐ twn metablht¸n x kai y qrhsimopoi same dÔo nèecmetablhtèc z kai ω antÐstoiqa, ìpou jewr same ω = ω(z).

Opwsd pote den uprqoun kajorismènec suntagèc gia to pìte eÐnai skìpimo anagkaÐona qrhsimopoioÔme antÐ twn Kartesian¸n suntetagmènwn (x, y) kpoio llo zeÔgoc sunte-tagmènwn (u,w), ìpwc p.q. polikèc suntetagmènec. EÐnai jèma krÐshc kai exoikeÐwshc me thsugkekrimènh D.E. pou epiqeiroÔme na lÔsoume kai me th morf twn lÔsewn pou endeqomènwcanamènoume.

Gia pardeigma, h diaforik exÐswsh qwrizìmenwn metablht¸n

y′ =x

y(82)

èqei lÔseic thn oikogèneia twn omìkentrwn kÔklwn x2 + y2 = c (me c > 0). Oi lÔseic autècse polikèc suntetagmènec (r, θ) ja perigrfontan apì thn apl sqèsh r = c pou prokÔpteiapì thn exÐswsh

dr

dθ= 0. (83)

Prgmati, an sth D.E. (82) antikatast soume tic Kartesianèc suntetagmènec (x, y) me ticpolikèc (r, θ), qrhsimopoi¸ntac tic sqèseic metasqhmatismoÔ

x = r cos θ, y = r sin θ (84)

kaidx = cos θdr − r sin θdθ, dy = sin θdr + r cos θdθ,

katal goume sthn exÐswsh (83).Sth sunèqeia parajètoume dÔo pio sÔnjeta paradeÐgmata.Pardeigma : Na lujeÐ h D.E.

(xdx+ ydy)− xy(xdy − ydx) = 0. (85)

H emfnish twn sumplegmtwn xdx+ ydy kai xdy − ydx mac odhgeÐ sth skèyh na qrhsimo-poi soume polikèc suntetagmènec. 'Eqoume tic sqèseic metasqhmatismoÔ (84) all, epÐshc,apì th x2 + y2 = r2 paÐrnoume thn

x dx+ y dy = r dr,

kai apì thn tan θ = y/x paÐrnoume thn

x dy − y dx = r2 dθ.

0.6. ALLAGH METABLHTWN 31

'Etsi h (85) grfetairdr − r4 sin θ cos θdθ = 0, r 6= 0. (86)

H (86) eÐnai qwrizìmenwn metablht¸n kai h olokl rws thc mac dÐnei

1

2r2= −sin2 θ

2+ c

, se Kartesianèc suntetagmènec,

c x2 + (c− 1)y2 = 1. (87)

H (87) paristnei genik mia monoparametrik oikogèneia kwnik¸n tom¸n. Pio sugkekrimèna,gia c > 1 h (87) orÐzei oikogèneia elleÐyewn kai gia 0 < c < 1 mia oikogèneia uperbol¸n.Eidik gia c = 1, brÐskontai apì th (87) wc lÔseic oi eujeÐec x = ±1, en¸ gia c < 0 denorÐzetai kami pragmatik kampÔlh. J

Pardeigma : Na lujeÐ h D.E.

(1 + xy)x

yy′ = 1, y 6= 0. (88)

Epeid sth (88) emfanÐzontai ta sumplègmata xy kai xy , jètoume14

x y = u,x

y= w. (89)

Prèpei t¸ra na ekfrsoume thn y′ wc sunrthsh twn nèwn metablht¸n u,w kai thc dudw .

ParagwgÐzontac tic (89) wc proc x, paÐrnoume

y + x y′ =du

dw

dw

dxkai 1 = y′w + y

dw

dx. (90)

ApaleÐfontac to dwdx apì tic (90), brÐskoume

y′ =

dudw− uw

u+ w dudw

. (91)

SÔmfwna me tic (89) kai (91), h D.E. (88) grfetai

w(1 + u)(du

dw− u

w) = u+ w

du

dw⇒ du

u+ 2=dw

w. (92)

Apì th (92) paÐrnoume

(2 + u) = cw x y + 2 =c x

y, c 6= 0.

H sqèsh aut apoteleÐ th genik lÔsh thc (88). J

14Αυτό φυσικά δεν σημαίνει ότι η μέθοδος οπωσδήποτε θα λειτουργήσει. Για καμία Δ.Ε. δεν θα

μπορούσαμε να βεβαιώσουμε εκ των προτέρων κάτι τέτοιο. Ούτε σημαίνει ότι οι μετασχηματισμοί (89)

είναι το μόνο που μπορούμε να κάνουμε. Για παράδειγμα, θα μπορούσε κανείς να θεωρήσει τη (88)

ως γραμμική Δ.Ε. με άγνωστη τη συνάρτηση x = x(y).

32

0.7 Diaforikèc exis¸seic anwtèrou neu

bajmoÔ

SuneqÐzoume kai se aut thn enìthta na meletoÔme D.E. pr¸thc txhc, dhlad exi-s¸seic thc morf c

F (x, y, y′) = 0 (93)

pou ìmwc den mporoÔn na grafoÔn se lumènh morf 15. Tètoiou eÐdouc exis¸seic eÐnai anwtèrou bajmoÔ den èqoun bajmì.

0.7.1 Diaforikèc anwtèrou bajmoÔ

Mia D.E. (93) opoÐa eÐnai algebrik c morf c k bajmoÔ (k 6= 1) wc proc thn y′,mporeÐ na grafeÐ wc ginìmeno paragìntwn sth morf 16

(y′ − f1(x, y)) (y′ − f2(x, y)) . . . (y′ − fk(x, y)) = 0.

Gia aplìthta, ac upojèsoume ìti h (93) eÐnai deutèrou bajmoÔ kai mporeÐ na grafeÐsth morf

(y′ − f1(x, y))(y′ − f2(x, y)) = 0, (94)

y′ = f1(x, y) (a)y′ = f2(x, y). (b)

(95)

Oi parapnw exis¸seic eÐnai pr¸thc txhc kai pr¸tou bajmoÔ. Genik prìkeitai giamorfèc aploÔsterec apì th (93). 'Estw ìti

Φ1(x, y, c) = 0 (a)Φ2(x, y, c) = 0, (b)

(96)

eÐnai oi genikèc lÔseic twn (95a) kai (95b) antÐstoiqa. 'Omwc, tìso h monoparametrik oikogèneia kampul¸n (96a) ìso kai h oikogèneia (96b) apoteloÔn lÔseic thc (94) thc arqik c diaforik c (93). Oi dÔo parapnw oikogèneiec mporoÔn na grafoÔn semÐa parstash wc

Φ1(x, y, c) Φ2(x, y, c) = 0. (97)

H (97) apoteleÐ th genik lÔsh thc (93). Sth morf aut emfanÐzetai bèbaia mÐa mìnostajer diìti h D.E. (93) eÐnai pr¸thc txhc.

Pardeigma : H D.E

y′2 − x(x+ y)y′ + x3y = 0 (98)

15Θυμίζουμε ότι μια Δ.Ε. 1ης τάξης είναι λυμένη όταν, σε κάποιον τόπο T του επιπέδου Οxy,

μπορεί να γραφεί ως y′ = f(x, y), με την f(x, y) μονότιμη συνάρτηση.16Χωρίς να αποκλείεται κάποιες από τις fi(x, y), i = 1 . . . k να μην είναι πραγματικές.

0.7. DIAFORIKES EXISWSEIS ANWTEROU H ANEU BAJMOU 33

eÐnai pr¸thc txhc all deutèrou bajmoÔ. Grfetai wc

(y′ − x2)(y′ − xy) = 0

kai epomènwc h lÔsh thc angetai sth lÔsh twn D.E.

y′ = x2

y′ = x y.

LÔseic twn dÔo aut¸n D.E. eÐnai, antÐstoiqa, oi

y = (x3 + c)/3 (a)y = cex

2/2. (b)

H genik lÔsh mporeÐ loipìn na grafeÐ sth morf

(3y − x3 − c)(y − c ex2/2) = 0.

Apì kje shmeÐo (x0, y0) tou epipèdou pernoÔn dÔo lÔseic thc (98) pou dÐnontai apìtic sqèseic (0.7.1a) kai (0.7.1(b gia kpoia tim tou c (Sq ma 12a). Sto shmeÐo tom ctouc (x0, y0), oi lÔseic autèc èqoun genik diaforetik efaptomènh, dhlad tèmnontaiegkrsia. ExaÐresh apoteloÔn ta shmeÐa sta opoÐa sumbaÐnei na eÐnai x2

0 = x0y0

(dhlad o xonac twn y kai h eujeÐa y = x). Apì aut pernoÔn pli dÔo lÔseicall me koin efaptomènh. Aut h parat rhsh èqei to ex c nìhma: Ac proume giapardeigma tic lÔseic pou pernoÔn apì to shmeÐo A(1, 1) thc eujeÐac y = x (Sq ma12b). Autèc eÐnai oi

y =1

3(x3 + 2) (I) kai y = e(x2−1)/2 (II),

ìpou −∞ < x <∞.Akrib¸c epeid autèc oi lÔseic èqoun sto shmeÐo A koin efaptomènh, mporoÔme

na poÔme ìti mÐa lÔsh thc arqik c D.E. (98), pou pernei apì to shmeÐo (1, 1), eÐnaikai h

y =

13(x3 + 2) −∞ < x ≤ 1

e(x2−1)/2 1 ≤ x <∞,

dhlad h kampÔlh pou dhmiourgeÐtai apì thn ènwsh tou tm matoc thc kampÔlhc (I)arister apì to A me to tm ma thc kampÔlhc (II) dexi apì to A. Aut h ènwshtmhmtwn twn kampul¸n den mporeÐ na gÐnei sto shmeÐo B, ìpou epÐshc tèmnontai oikampÔlec (I) kai (II). Kai toÔto diìti h sunrthsh pou ja proèkupte den ja tanparagwgÐsimh se autì to shmeÐo. (MporeÐte na aitiolog sete se ti ja enoqloÔseautì?)J

34

Sq ma 12: a) Oi monoparametrikèc oikogèneiec kampul¸n-lÔsewn (i) kai (ii) thc dia-forik c exÐswshc (98), pou dÐnontai apì tic (0.7.1a) kai (0.7.1b). b) Oi merikèc lÔseic(I) kai (II) thc diaforik c exÐswshc (98). Oi dÔo kampÔlec sunant¸ntai efaptomeniksto shmeÐo A kai tèmnontai egkrsia sto B.

ShmeÐwsh : Ap ìsa anafèrame wc ed¸, prokÔptei ìti mia D.E. ìpwc h (93),efìson eÐnai kpoiou bajmoÔ wc proc y′, th metaqeirizìmaste san na eÐnai sun jhcalgebrik exÐswsh wc proc y′ kai brÐskoume ìlec tic pragmatikèc rÐzec thc. 'Etsip.q., efìson endiaferìmaste gia pragmatikèc lÔseic, h D.E.

(y′2 − xy′ + x2)(y′ − xy) = 0,

sth morf

y′3 − x(1 + y)y′2 + x2(y + 1)y′ − x3y = 0,

eÐnai isodÔnamh me th D.E.

y′ − xy = 0.

Autì isqÔei epeid h

y′2 − x y′ + x2 = 0

den èqei pragmatikèc lÔseic wc proc y′.

0.7.2 Eidikèc morfèc diaforik¸n exis¸sewn neu baj-

moÔ

Met apì autèc tic genikèc parathr seic, erqìmaste na exetsoume dÔo eidikìterec morfèct c D.E. (93). Se ìsa akoloujoÔn, den jewroÔme aparaÐthto to pr¸to mèloc thc (93) naeÐnai èkfrash poluwnumik (kpoiou bajmoÔ) wc proc y′.


(a) LeÐpei apì thn exÐswsh h exarthmènh metablht y

'Eqoume dhlad f(x, y′) = 0. (99)

EÐnai fusikì na epiqeir soume amèswc na lÔsoume wc proc y′, dhlad

y′ = φ(x),

opìte ftnoume en gènei se mia apl , mesa oloklhr¸simh D.E.An h epÐlush wc proc y′ den eÐnai eÔkolh dunat , epiqeiroÔme na lÔsoume th (99) wc

proc x, brÐskontac ètsi thn exÐswsh

x = g(y′). (100)

Jètoume y′ = p kai h (100) grfetai

x = g(p). (101)

EpÐshc, èqoume

y =

∫dy

dxdx =

∫pdg(p)

dpdp = φ(p) + c. (102)

Oi sunart seic (101) kai (102) dÐnoun th genik lÔsh thc (99) se parametrik morf

x = g(p) kai y = φ(p) + c.

Parmetroc eÐnai h p kai aujaÐreth stajer h c.Pardeigma : Ac proume gia pardeigma thn exÐswsh

y′ + sin y′ = x (103)

ki ac parathr soume ed¸ ìti h exÐswsh aut eÐnai pr¸thc txewc all ìqi kpoiou bajmoÔ.Jètoume y′ = p, opìte

x = p+ sin p (104)

kai

y =

∫dy

dx

dx

dpdp =

∫p(1 + cos p)dp =

p2

2+ p sin p+ cos p+ c. (105)

Oi exis¸seic (104) kai (105) dÐnoun se parametrik morf th lÔsh thc (103). J

(b) LeÐpei apì thn exÐswsh h anexrthth metablht x

'Eqoume dhlad f(y, y′) = 0. (106)

An h (106) mporeÐ na lujeÐ wc proc y′, an dhlad y′ = φ(y), èqoume mia D.E. me qwrizìmenecmetablhtèc. Diaforetik lÔnoume th (106) wc proc y, dhlad

y = g(y′).

36

Jètoume y′ = p, opìtey = g(p) (107)

kai

x =

∫dx

dydy =

∫1

p

d g(p)

dpdp+ c. (108)

Oi sunart seic (107) kai (108) dÐnoun se parametrik morf th genik lÔsh thc D.E. (106).Pardeigma :

y − 2y′ =√

1 + y′2 (109)

Jètoume y′ = p, opìte

y = 2p+√

1 + p2 (110)

kaix =

∫dx =

∫ dyp =

∫1p(2 + p√

1+p2)dp = . . .

= ln [p2(p+√

1 + p2)] + c.(111)

Oi kampÔlec pou dÐnontai apì tic parametrikèc exis¸seic (110) kai (111), me parmetro thnp, apoteloÔn th genik lÔsh thc D.E. (109). J

(g) H diaforik exÐswsh tou Clairaut

Prìkeitai gia th D.E.y = xy′ + f(y′). (112)

Upojètoume ìti h sunrthsh f(y′) eÐnai suneq c kai paragwgÐsimh wc proc y′. H lÔshepitugqnetai me th qr sh tou sumbolismoÔ y′ = p, opìte h (112) grfetai sth morf

y = xp+ f(p). (113)

Ac paragwgÐsoume wc proc x ta dÔo mèlh thc (113). PaÐrnoume

p = p+ xdp

dx+df(p)

dp

dp

dx⇒ dp

dx(x+

df(p)

dp) = 0.

Dhlad , eÐtedp

dx= 0 (114)

eÐte

x = −df(p)

dp= φ(p). (115)

Apì th (114) prokÔptei p = dydx = c, ìpou c aujaÐreth stajer. 'Ara, lìgw thc (113),

eÐnai17

y = c x+ f(c). (116)

17Μια οποιαδήποτε σταθερά στη θέση της f(c) θα ήταν αρκετή για να ικανοποιήσει την dy

dx = cαφού η ολοκλήρωσή της μας δίνει y = cx + c′, όπου c′ μια νέα αυθαίρετη σταθερά. Πρέπει όμως ησταθερά αυτή να επιλεγεί έτσι ώστε να ικανοποιείται η αρχική εξίσωση (112) και αυτό συμβαίνει μόνο

αν c′ = f(c).


H (116) eÐnai genik lÔsh thc exÐswshc (113) afoÔ parist mia monoparametrik oikogèneiakampul¸n kai mlista eujei¸n.

38

An t¸ra antikatast soume sth (113) thn x = φ(p), ìpwc dÐnetai apì th sqèsh (115),paÐrnoume

y = xφ(p) + f(p) (117)

Oi exis¸seic (115) kai (117) dÐnoun se parametrik morf mia lÔsh thc exÐswshc Clairaut,h opoÐa den sumperilambnetai sth genik lÔsh. Prìkeitai gia mia idizousa lÔsh pou eÐnaih peribllousa18 thc oikogèneiac eujei¸n (116).

Pardeigma : DÐnetai h exÐswsh Clairaut

y = xy′ + 2y′2. (118)

Jètoume y′ = p kai paragwgÐzoume th (118) wc proc x. 'Etsi paÐrnoume

y = x p+ 2p2 ⇒ y′(= p) = p+ x p′ + 4p p′ ⇒ p′(x+ 4p) = 0,

dhlad dp

dx= 0 (119)

x+ 4p = 0. (120)

Apì th (119) paÐrnoume p = c

y′ = c ⇒ y = c x+ c′. (121)

An antikatast soume th (121) sth (118), tìte prokÔptei ìti

c x+ c′ = x c+ 2c2 ⇒ c′ = 2c2

kai h genik lÔsh thc (118) ja eÐnai h oikogèneia eujei¸n

y = c x+ 2c2. (122)

EpÐshc, apì th (120) prokÔptei me olokl rwsh

p = y′ = −x4⇒ y = −x

2

8+ c′′, (123)

ìpou c′′ h stajer thc olokl rwshc. 'Omwc, h (123) ikanopoieÐ th (118) mìno an c′′ = 0.'Ara, lÔsh (idizousa) thc (118) eÐnai kai h

y = −x2

8. J

18Περιβάλλουσα μιας οικογένειας καμπυλών f(x, y; c) = 0 είναι μια καμπύλη που σε κάθε σημείο

της εφάπτεται και σε ένα μέλος της οικογένειας. Αποδεικνύεται ότι αν υπάρχει περιβάλλουσα, αυτή

δίνεται από την απαλοιφή της σταθεράς c μεταξύ των εξισώσεων f(x, y; c) = 0 και ∂f(x, y; c)/∂c = 0.


Sq ma 13: LÔseic y = cx + c2 thc exÐswshc Clairaut (118). H idizousa lÔshy = −x2/8 eÐnai h kampÔlh me thn èntonh gramm .

(d) H diaforik exÐswsh tou Lagrange

EÐnai D.E. thc morf cy = xφ(y′) + f(y′), (124)

ìpou oi sunart seic φ(y′) kai f(y′) eÐnai suneqeÐc kai paragwgÐsimec wc proc y′. Gia φ(y′) =y′ h (124) eÐnai tÔpou Clairaut.

Gia th lÔsh t c (124) jètoume y′ = p kai grfoume th sqèsh sth morf

y = xφ(p) + f(p). (125)

ParagwgÐzontac th (125) wc proc x, èqoume

p = φ(p) + xdφ(p)

dp

dp

dx+df(p)

dp

dp

dx. (126)

Pollaplasizontac ta dÔo mèlh thc (126) epÐ dxdp , paÐrnoume th D.E.

(p− φ(p))dx

dp− xdφ(p)

dp− df(p)

dp= 0. (127)

H (127) eÐnai mia grammik D.E. gia thn x = x(p) kai h lÔsh thc, bsei tou tÔpou (39), jaeÐnai thc morf c

x = x(p, c). (128)

40

Oi exis¸seic (125) kai (128) dÐnoun th genik lÔsh thc (124) se parametrik morf . AnmporoÔme na apaleÐyoume to p apì tic parametrikèc exis¸seic thc lÔshc, paÐrnoume th lÔshsth morf

Φ(x, y, c) = 0 y = y(x, c).

An uprqoun idizousec lÔseic thc (124), mporoÔme na tic broÔme apì th (126) sÔmfwna meìsa eÐpame sthn prohgoÔmenh enìthta.

Pardeigma : H exÐswshy = x y′2 + y′2 (129)

eÐnai tÔpou Lagrange19. Jètoume y′ = p kai th grfoume wc

y = xp2 + p2. (130)

ParagwgÐzontac wc proc x thn teleutaÐa aut sqèsh kai pollaplasizontac me dx/dp,katal goume sth grammik D.E. gia th x = x(p)

dx

dp+

2

p− 1x+

2

p− 1= 0, (p 6= 0, p 6= 1), (131)

me genik lÔsh th

x =c

(p− 1)2− 1. (132)

ApaleÐfontac to p metaxÔ twn (130) kai (132), brÐskoume

y = (√c+

√(x+ 1))2, (x ≥ −1, c ≥ 0). (133)

EÐnai aparaÐthth kpoia xeqwrist melèth twn peript¸sewn p = 0 kai p = 1, dhlad twntim¸n pou exairoÔntai sth (131). H p = dy

dx = 0 dÐnei y = c′ = 0 (idizousa lÔsh), en¸ h

p = dydx = 1 dÐnei y = x+ c′ = x+ 1, pou prokÔptei apì th genik lÔsh (133) gia c = 0. J

19Η (129) μπορεί να λυθεί και ως Δ.Ε. με χωριζόμενες μεταβλητές.

0.8. ASKHSEIS 41

0.8 Ask seic

1. Na brejeÐ h genik lÔsh twn D.E. qwrizìmenwn metablht¸n:

a) y′ = x2/y, y 6= 0

[ y2

2− x3

3= c y = ±

√23x3 + c ]

b) y′ = − x y2

1− x, x 6= 1, y 6= 0.

[ y = −(ln |1− x|+ x− c)−1 ]

g) y′ = x(y2 + 1).

[ y = tan(x2

2+ c)

]

d) y′ = y2 + 1x2 + 1

[ arctan(x) = arctan(y) + c y = x−c1+c x

]

2. Na brejeÐ h genik lÔsh twn D.E. :

a) y′ = −(x− y)2

[ y = x− tan(x+ c) ]

b) y′ = ex+y

[ y = − ln |c− ex| ]

3. Na brejeÐ h merik lÔsh gia ta paraktw probl mata arqik¸n tim¸n:

a) y′ + y = 0, y(1) = 1

[ y = e1−x ]

b) y′ = x√

1− y, y(√

2) = 34

[ y = x2

2− x4

16me −2 ≤ x ≤ 2 ]

42

4. Na brejeÐ h genik lÔsh twn omogen¸n D.E. :

a) y′ = yx −

√1 +

(yx

)2

, x > 0

[ y = c2−x2

2c]

b) y′ = 1x

(y +√xy), x > 0, y > 0

[ y = x4

ln2(c x), c > 0 ]

g) y′ = x3 + y3

x y2

[ y = x 3√

3 ln(c x), c x > 0 ]

5. Na brejeÐ h merik lÔsh thc D.E. y′ =x2 + y2

x y pou tèmnei kjeta thn eujeÐa

y = −12x+ 1.

[ y2 = 2x2(ln∣∣3x

2

∣∣+ 12

)]

6. Na brejoÔn oi orjog¸niec troqièc thc monoparametrik c oikogèneiac twn kÔ-klwn x2 + y2 = a x, a ∈ R.

[ oi kÔkloi x2 + y2 = b y, b ∈ R ]

7. Na brejeÐ h genik lÔsh thc y′ = −2x+ 3y + 13x+ 4y + 1 .

[ x2 + 2y2 + 3xy + x+ y = c ]

8. Na brejeÐ h genik lÔsh twn grammik¸n D.E. :

a) xy′ − y − x2 = 0, x 6= 0

[ y = c x+ x2 ]

b) y′ + 1− 2xx2 y = 1, x 6= 0

[ y = x2(1 + c e1/x) ]

0.8. ASKHSEIS 43

9. Na brejeÐ h merik lÔsh thc grammik c D.E. xy′ − 2y + x = 0, x 6= 0, pouefptetai sthn eujeÐa x+ y − 1 = 0. Se poio shmeÐo gÐnetai h epaf ?

[ y = −x2 + x, shmeÐo epaf c to (1,0) ]

10. BreÐte th genik lÔsh twn paraktw exis¸sewn Bernoulli:

a) y′ + xy + xy2 = 0

[ y =(c ex

2/2 − 1)−1

]

b) y′ − y = exy

[ y = ±√c e2x − 2ex ]

11. BreÐte th genik lÔsh twn paraktw exis¸sewn Riccati :

a) y′ = x2 − 2 + 2xy + y2, me merik lÔsh ψ(x) = 1− x,

[ y = 1− x− 2ce2x12

+ce2x]

b) y′ + y2 = 1x4 , x 6= 0, me merik lÔsh thc morf c ψ(x) = α

x+ β

x2

[ ψ(x) = 1x

+ −1x2 , y = 1

x+ 1

x22ce−2/x+12ce−2/x−1

]

12. DeÐxte ìti oi paraktw D.E. eÐnai pl reic kai breÐte th genik lÔsh touc:

a) 2xy dx+ (x2 + y2)dy = 0

[ 3x2y + y3 = c ]

b) xy + 1y dy +

2y − xy2 dy = 0, y 6= 0

[ x2

2+ 2 log |y|+ x

y= c ]

g) (y sinx− x2)dx− cosxdy = 0, x 6= kπ ± π2

[ y = c−x3/3cosx

]

44

13. BreÐte ènan pollaplasiastikì pargonta gia tic paraktw D.E. pou na tic kneipl reic kai sth sunèqeia breÐte th genik lÔsh touc:

a) xydx+ (1 + x2)dy = 0

[ R(y) = y, y2(1 + x2) = c ]

b) (x2 + y2 + x) dx+ xy dy = 0

[ R(x) = x, x4 + 2x2y2 + 4x3/3 = c ]

g) (x4y2 − y)dx+ (x2y4 − x)dy = 0

[ R(x, y) = (xy)−2, x4y + xy4 − cxy + 3 = 0 ]

14. LÔste ta paraktw probl mata arqik¸n tim¸n:

a) (1 + ex)yy′ = ex, y(0) = 1

[ y2 = 1 + ln(

1+ex

2

)2]

b) (x+ y)2y′ = 1, y(0) = 0

[ x = tan(y)− y ]

g) y′ = (x− y)/(x+ y), y(1) = 0

[ y = −x+√−1 + 2x2 ]

d) y′ − y cosx = sin 2x, y(0) = 12

[ y = 52esinx − 2(1 + sin x) ]

e) xy′ + y = y2 lnx, y(1) = 1

[ y = (1 + lnx)−1 ]

st) 2xy ln y dx+ (x2 + y2√y2 + 1) dy = 0, y(0) =

√3

[ 3x2 ln y + (y2 + 1)3/2 = 8 ]

z) y′ = (x+ y2)/(2xy), y(1) = 1

0.8. ASKHSEIS 45

[ y =√x(lnx+ 1), x > 1

e]

15. Me katllhlo metasqhmatismì, metatrèyte th D.E. xy′+y+x(1−x2y2)1/2 = 0se qwrizìmenwn metablht¸n kai breÐte th genik lÔsh thc (parathr ste ìti hparstash xy′ + y eÐnai h pargwgoc tou ginomènou xy).

[ y = sin(c− 12x2)/x ]

16. BreÐte th genik lÔsh thc D.E. y′+ sin y+x cos y+x = 0 kaj¸c kai th merik lÔsh pou pernei apì to shmeÐo (0, π

2) (bo jeia: gryte tic trigwnometrikèc

sunart seic me ìrisma y2kai qrhsimopoi sete ton metasqhmatismì z = tan

(y2

).

[ tan y2

= 1− x+ ce−x, yµ = 2 arctan(1− x) ]

17. DÐnetai h D.E. (x3 − y)dx+ (x2y + x)dy = 0. AnaptÔxte th D.E. se morf meìrouc xdx + ydy kai xdy − ydx kai, qrhsimopoi¸ntac polikèc suntetagmènec,breÐte th genik lÔsh.

[ x2 + y2 = −2 yx

+ c ]

18. BreÐte tic lÔseic twn paraktw problhmtwn arqik¸n tim¸n:

a) y′ = |x− y|, y(0) = 1

[ y = x+ 1 ]

b) y′ = max(x, y), y(0) = 0

[ y = x2

2, 0 ≤ x ≤ 2 ]

19. Na lujoÔn oi exis¸seic tou Clairaut

a) y = x y′ + αy′2

[ y = c x+ αc2, kai y3 = 27α

4x2 ]

b) y = x y′ + α√

1 + y′2

[ y = c x+ α√

1 + c2, kai y2 + x2 = α2 ]

46

20. Na brejeÐ h lÔsh x = x(p), y = y(p) twn paraktw problhmtwn arqik¸ntim¸n:

a) y′ + ln |y′| = y, y(1) = 1

[ x = ln |p| − 1p

+ 2, y = p+ ln |p| ]

b) y′ e−y′ − x = 0, y(0) = 0

[ x = p e−p, y = e−p(p2 + p+ 1) + c, c = −1, 0 ]

21. O rujmìc me ton opoÐo diasp¸ntai oi pur nec miac radienergoÔc ousÐac eÐnaianlogoc (me suntelest analogÐac Ðso me k > 0)20 tou arijmoÔ twn pur nwnpou den èqoun akìmh diaspasteÐ. An se èna deÐgma èqoun diaspasteÐ oi misoÐpur nec se qrìno t = 15 ai¸nec (qrìnoc hmÐseiac zw c), ti posostì tou ulikoÔja èqei apomeÐnei se 45 ai¸nec?

[ 18 to 12.5% (N = N0e

−kt) ]

22. 'Ena sfairikì kommti nafjalÐnhc exaqn¸netai me taqÔthta (metroÔmenh segr/sec) anlogh thc epifneiac tou sfairikoÔ kommatioÔ se kje qronik stig-m . An se qrìno t = T0 exaqn¸netai h mis mza, se pìso qrìno ja exaqnwjeÐìlh h nafjalÐnh?

[ T = 2T0

2−22/3 ]

23. H taqÔthta me thn opoÐa yÔqetai jermaÐnetai èna s¸ma eÐnai anlogh thc dia-forc jermokrasÐac tou s¸matoc kai tou peribllontoc (Nìmoc jèrmanshc -yÔxhc tou NeÔtwna). a) An θ0 eÐnai h arqik jermokrasÐa kai θπ h jermokrasÐatou peribllontoc, gryte th D.E. pou perigrfei th metabol thc jermokra-sÐac θ kai breÐte th lÔsh thc. b) 'Ena s¸ma jermokrasÐac 80 topojeteÐtaise peribllon meglhc jermoqwrhtikìthtac pou diathreÐ stajer jermokrasÐa50. Sta pr¸ta 5min h jermokrasÐa tou s¸matoc èpese stouc 70. Pìsh jagÐnei h jermokrasÐa tou sto tèloc tou epìmenou pentleptou?

[ a) θ = −k(θ − θπ), θ = θπ + (θπ − θπ)e−kt, b)=63.3 ]

20Στη συνέχεια το σύμβολο k θα χρησιμοποιείται πάντα ως η απόλυτη τιμή ενός συντελεστή ανα-

λογίας

0.8. ASKHSEIS 47

24. Mia dexamen perièqei arqik 50 m3 kajarì nerì. Se qrìno t = 0 arqÐzei naqÔnetai sth dexamen alatìnero, periektikìthtac se alti 2 Kgr/m3, me rujmì3 m3/min. Sugqrìnwc, apì swl na sth bsh thc dexamen c, feÔgei alatìnerome ton Ðdio rujmì (3 m3/min) ètsi ¸ste o ìgkoc tou alatìnerou sth dexamen na eÐnai stajerìc (50 m3). To mÐgma anakateÔetai suneq¸c sth dexamen ètsi¸ste na diathreÐtai omogenèc. Na brejeÐ h posìthta x tou alatioÔ pou uprqeisth dexamen wc sunrthsh tou qrìnou t.

[ x = 100(

1− e− 350t)

]

25. 'Ena ulikì shmeÐo me mza m ektoxeÔetai apì to èdafoc katakìrufa proc tapnw me arqik taqÔthta ~υ0. An sth diadrom dèqetai antÐstash anlogh thc

taqÔthtac, ~R = −k~υ, ìpou k > 0 o suntelest c analogÐac, breÐte to qrìno pouja qreiasteÐ to shmeÐo gia na ftsei sto mègisto Ôyoc. Ti dÐnei to apotèlesmaautì gia k → 0?

[ t = mk

ln(

1 + kmgυ0

), limk→0 t = υ0

g]

26. Mia stagìna broq c, me arqik mza m0 pèftei mèsa se èna sÔnnefo kai h mzathc auxnei me rujmì anlogo thc mzac thc se kje stigm . Agno¸ntac thnantÐstash tou aèra kai lambnontac upìyh mìno to omogenèc pedÐo barÔthtac,breÐte thn taqÔthta me thn opoÐa pèftei h stagìna.

[ υ = gk(1− e−kt) ]

27. ZhteÐtai mÐa kampÔlh y = y(x), x > 0, pou na pernei apì to shmeÐo (1,0) kaina èqei thn idiìthta : Se kje shmeÐo thc (x, y), h apìstash |~r| =

√x2 + y2 na

isoÔtai me to kommti tou xona Oy pou kìbei h efaptomènh thc kampÔlhc stoshmeÐo (x, y).

[ y = 1−x2

2, lÔsh thc y′ = y

x−√

1 +(yx

)2]

28. ZhteÐtai fjÐnousa kampÔlh y = y(x), x > 0, pou na pernei apì to shmeÐo (0,1)kai na èqei thn idiìthta : H efaptomènh thc se tuqìn shmeÐo thc A(x, y) tèmneiton xona Ox sto shmeÐo B ètsi ¸ste to trÐgwno ABΓ, me Γ to shmeÐo (x, 0),na èqei stajerì embadìn Ðso me 1.

[ y = 1x/2+1

]

48

29. Mèsa se èna qhmikì diluma, mia ousÐa A antidr me thn ousÐa B kai pargetaih ousÐa G sÔmfwna me th qhmik antÐdrash A+2B→ Γ. An x eÐnai h posìthta(se moles) thc ousÐac G th qronik stigm t tìte h taqÔthta metabol c thckat thn antÐdrash eÐnai anlogh twn posot twn twn ousi¸n A kai B poubrÐskontai ekeÐnh th stigm sto diluma. a) An arqik (t = 0) uprqoun stodiluma XA kai XB moles twn ousi¸n A kai B, antÐstoiqa, poia diaforik exÐswsh perigrfei thn posìthta x = x(t) thc ousÐac G? b) An XA = 1 kaiXB = 1 kai met apì qrìno t = 1 min èqoun paraqjeÐ 1/4 moles thc ousÐacG, se pìso qrìno ja paraqjoÔn 6/15 moles? g) se pìso qrìno h antÐdrash jaoloklhrwjeÐ?

[ a)dxdt

= k(XA − x)(XB − 2x), b) t = ln 32, g) ∞ ]

30. Se mia pìlh plhjusmoÔ P epikrateÐ mia epidhmÐa. O rujmìc aÔxhshc twn ka-toÐkwn pou prosbllontai eÐnai anlogoc tìso tou arijmoÔ twn katoÐkwn pouèqoun prosblhjeÐ ìso kai tou arijmoÔ twn katoÐkwn pou den èqoun prosblh-jeÐ. Kpoia qronik stigm (t = 0) katagrfetai ìti èqei prosblhjeÐ to 1/9tou plhjusmoÔ kai met apì 3 m nec èqei prosblhjeÐ o misìc plhjusmìc. NabrejeÐ kai na sqediasteÐ pwc exelÐssetai sto qrìno o arijmìc x twn atìmwnpou èqoun prosblhjeÐ.

[ x(t) = 2t

8+2tP ]

31. 'Ena kulindrikì barèli me diatom S kai Ôyoc h èqei sto pujmèna tou mia mikr kuklik op me diatom s. Sthn arq to barèli eÐnai gemto me nerì. Se pìsoqrìno ja adeisei apì th stigm pou arqÐzei h ekro ? (h taqÔthta me thn opoÐafeÔgei to nerì apì thn op eÐnai υ =

√2gz, ìpou z to Ôyoc thc stjmhc tou

neroÔ apì to pujmèna)

[ tα = Ss

√2hg

]

32. H arqaÐa kleyÔdra tan èna doqeÐo (se sq ma epifneiac ek peristrof c perÐton katakìrufo xona Oz) me mia mikr trÔpa sto ktw mèroc O ap' ìpou ètreqeto nerì. Poia ja èprepe na eÐnai h exÐswsh thc epifneiac aut c an sto ìrganoautì h stjmh tou neroÔ katèbaine omal?

[ z = k(x2 + y2)2, k =staj. ]

33. Sto pr¸to tetarthmìrio tou epipèdou (x > 0, y > 0) zhtoÔme mia kampÔlhy = y(x) me thn ex c idiìthta : to kommti T1T2 thc efaptomènhc thc kampÔlhcsto tuqìn shmeÐo thc (x, y) metaxÔ twn jetik¸n hmiaxìnwn Ox kai Oy èqeistajerì m koc Ðso me l. ApodeÐxte ìti odhgoÔmaste se mia exÐswsh Clairautkai breÐte a) th genik lÔsh kai b) thn idizousa lÔsh.

0.8. ASKHSEIS 49

[ a) y = c x− l c√1+c2

, c =staj. b) x2/3 + y2/3 = l2/3 ]

prìqeirh èkdosh tou 2ou kef apì to biblÐo g. bougiatz c, g...

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