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BERNULLI DIFFERENSIAL TENGLAMASI.
Bernulli differensial tenglamasi deb,
π¦β² + π(π₯) β π¦ = π(π₯) β π¦π
koΚ»rinishdagi differensial tenglamaga aytiladi.
KoΚ»rinib turibtiki Bernulli differensial tenglamasi tuzilishi boΚ»yicha chiziqli bir jinsli
boΚ»lmagan birinchi tartibli differensial tenglamani eslatayapti. Differensial tenglama
Bernulli differensial tenglamasi ekanligini aniqlash uchun oΚ»ng tomonda y ning a-
darajasi qatnashganligidir.
π = 0 boΚ»lganda π¦β² + π(π₯) β π¦ = π(π₯)
π = 1 boΚ»lganda π¦β² + (π(π₯) β π(π₯)) β π¦ = 0
koΚ»rinishdagi differensial tenglamalarga keladi, ularni qanday qilib yechishni esa
koΚ»rib chiqdik.
y ning darajasidagi a βmusbat ham (a>0), manfiy ham (a<0), kasr son ham
(π =
1
2 βΉ π¦
1
2 = βπ¦ ) boΚ»lishi mumkin.
Bernulli tenglamasi turli xil koΚ»rinishlarda berilishi mumkin:
π(π₯) β π¦β² + π(π₯) β π¦ = π(π₯) β π¦π
π(π₯) β π¦β² + π¦ = π(π₯) β π¦π
π¦β² + π¦ = π(π₯) β π¦π
π¦β² + π(π₯) β π¦ = π¦π
Muhimi y ning birdan farqli darajasi qatnashsa boΚ»lgani. a>0 boΚ»lganda y=0 yechim
Bernulli tenglamasining xususiy yechimi boΚ»ladi.
Shunday qilib Bernulli tenglamasini yechish algoritmi quyidagicha:
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1. OΚ»ng tomondagi π¦π dan qutulish lozim. Buning uchun tenglamani ikkala
tomonini π¦π ga boΚ»lamiz.
π¦β²
π¦π + π(π₯) β π¦1βπ = π(π₯)
2. π¦1βπ dan qutulish lozim, buning uchun π¦1βπ = π§ deb belgilash kiritamiz.
3. π§β² = (1 β π) π¦β²
π¦π βΉ π¦β² =
π¦π
1βπ π§β² βΉ
1
1βπ π§β² + π(π₯) β π§ = π(π₯)
koΚ»rinishdagi chiziqli bir jinsli boΚ»lmagan 1-tartibli differensial tenglamaga
kelamiz. Uni yechish algoritmini esa bilamiz.
Misol 3. β1 β π₯2 β π¦β² + π¦ = ππππ πππ₯ β π¦2, π¦(0) = β1
1. OΚ»ng tomonda y dan qutulish kerak.
2. Konturga olingan qoΚ»shiluvchida y dan qutulish kerak, buning uchun
1
π¦ = π§
almashtirish bajaramiz.
3. π§β² = β
π¦β²
π¦2 βΉ π¦β² = βπ¦2 β π§β² βΉ ββ1 β π₯2 β π§β² + π§ = ππππ πππ₯ βΉ
π§β² β
1
β1 β π₯2 π§ = β ππππ πππ₯
β1 β π₯2
Natijada Bernulli differensial tenglamasidan chiziqli bir jinsli boΚ»lmagan birinchi
tartibli differensial tenglamaga kelamiz. Bunday tenglamalarni yechish usullarini
esa bilamiz.
RIKKATI TENGLAMALARI.
Rikkati tenglamasi birinchi tartibli chiziqli boΚ»lmagan differensial tenglamalarning
eng qiziqarlilaridan hisoblanadi. Umumiy holda Rikkati tenglamasi quyidagi
koΚ»rinishda yoziladi:
π¦β² = π(π₯)π¦ + π(π₯)π¦2 + π(π₯)
Bunda π(π₯), π(π₯), π(π₯)-oΚ»zgaruvchi x ga bogΚ»liq boΚ»lgan uzluksiz funksiyalar.
Rikkati tenglamasi matematikaning turli sohalari (masalan algebraik geometriya va
conform akslantirishlar nazariyasida) va fizikada keng qoΚ»llaniladi. Amaliy
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matematika masalalarida ham koΚ»p hollarda uchrab turadi. Rikkati tenglamasining
yechimi quyidagi teoremaga asoslangan:
Teorema. Agar Rikkati tenglamasining xususiy yechimi π¦1 aniq boΚ»lsa, u holda
uning yechimi quyidagicha formula bilan aniqlanadi:
π¦ = π¦1 + π’
Haqiqatdan ham ushbu yechimni Rikkati tenglamasiga qoΚ»yilsa, quyidagiga ega
boΚ»lamiz:
(π¦1 + π’)β² = π(π₯)(π¦1 + π’) + π(π₯)(π¦1 + π’)2 + π(π₯)
π¦1
β² + π’β² = π(π₯)π¦1 + π(π₯)π’ + π(π₯)π¦1
2 + 2π(π₯)π¦1π’ + π(π₯)π’2 + π(π₯)
Chap va oΚ»ng tomondagi tagi chizilgan ifodalarni qisqartirish mumkin, chunki π¦1
tenglamani qanoatlantiruvchi xususiy yechim. Natijada u(x) funksiya uchun
differensial tenglamaga ega boΚ»lamiz:
π’β² = π(π₯)π’2 + [2π(π₯)π¦1 + π(π₯)]π’
Ushbu tenglama esa Bernulli tenglamasi hisoblanadi. π§ =
1
π’ oΚ»zgaruvchi
almashtirish Bernulli tenglamasini integrallash mumkin boΚ»lgan chiziqli differensial
tenglamaga aylantiradi.
π(π₯), π(π₯), π(π₯)-larning koΚ»rinishiga qarab Rikkati tenglamasi turli xil
koΚ»rinishlarga ega boΚ»lib, ularning koΚ»pchiligi integrallanuvchi yechimga ega. Lekin
afsuski umumiy holda xususiy yechimni topish uchun qatΚΌiy algoritm yoΚ»q.
Rikkarti tenglamasining xususiy hollarini koΚ»rib chiqamiz:
Xususiy hol β1. π, π, π β koeffitsiyentlarning barchasi konstanta boΚ»lgan hol:
Agar Rikkati tenglamasidagi koeffitsiyentlar oΚ»zgarmas boΚ»lsa, u holda uni
oΚ»zgaruvchilari ajraladigan tenglamalarga olib kelish mumkin
π¦β² = ππ¦ + ππ¦2 + π; βΉ ππ¦
ππ₯ = ππ¦ + ππ¦2 + π; βΉ β«
ππ¦
ππ¦ + ππ¦2 + π = β« ππ₯
Ushbu integral esa π, π, π β larning ixtiyoriy qiymatlarida oson hisoblanadi.
Xususiy hol β2. π(π₯) = 0, π(π₯) = π, π(π₯) = π β π₯π; βΉ π¦β² = π β π¦2 + π β π₯π
Ilmiybaza.uz
Rikkati tenglamasining ushbu xusuaiy holi ajoyib yechimga ega. Agar n=0 boΚ»lsa,
biz yana 1-holga kelamiz. Agar n=-2 boΚ»lsa, Rikkati tenglamasi π¦ =
1
π§ almashtirish
orqali bir jinsli tengalamaga aylanadi.
Ushbu differensial tenglamani π =
4π
1β2π ; π = Β±1; Β±2; Β±3; β¦ holler uchun ham
yechish mumkin boΚ»ladi, bunda umumiy yechimlar silindrik funksiyalar orqali
ifodalanadi.
n ning boshqa qiymatlarida Rikkati tenglamasini elementar funksiyalarning
integrallari orqali ifodalash mumkin. Ushbu fakt fransuz matematigi J.Liuvill
tomonidan aniqlangan.
Rikkati
tenglamasining
boshqa
koΚ»pgina
xususiy
hollarini
http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc1.htm saytda koΚ»rish mumkin.
TOΚ»LA DIFFERENSIALLI TENGLAMA.
Agar π(π₯, π¦)ππ₯ + π(π₯, π¦)ππ¦ = 0 differensial tenglamada βπ§ = π(π₯, π¦) funksiya
topilsaki,
ππ(π₯,π¦)
ππ₯
= π(π₯, π¦),
ππ(π₯,π¦)
ππ¦
= π(π₯, π¦) (*)
boΚ»lsa,
π(π₯, π¦)ππ₯ + π(π₯, π¦)ππ¦ = ππ(π₯, π¦)
ππ₯
ππ₯ + ππ(π₯, π¦)
ππ¦
ππ¦ = ππ(π₯, π¦) = 0 βΉ
boΚ»lib, umumiy yechim
π(π₯, π¦) = πΆ
koΚ»rinishda boΚ»ladi.
Agar differensial tenglama uchun
ππ(π₯, π¦)
ππ¦
= ππ(π₯, π¦)
ππ₯
shart bajarilsa, u holda differensial tenglama toΚ»la differensialga keladi va π(π₯, π¦)
funksiya quyidagicha koΚ»rinishda qidiriladi:
1-usul:
ππ(π₯,π¦)
ππ₯
= π(π₯, π¦) shartdan y ni oΚ»zgarmas deb olib, x boΚ»yicha integral
olamiz, constantani y ga bogΚ»liq funksiya qilib olib
Ilmiybaza.uz
π(π₯, π¦) = β« π(π₯, π¦)ππ₯ + π½(π¦)
ikkinchi shartni
ππ(π₯,π¦)
ππ¦
= π(π₯, π¦) ham bajarilishini talab qilib, π½(π¦) ni ham
koΚ»rinishini aniqlaymiz.
2-usul:
ππ(π₯,π¦)
ππ¦
= π(π₯, π¦) shartdan x ni oΚ»zgarmas deb olib, y boΚ»yicha integral
olamiz, constantani x ga bogΚ»liq funksiya qilib olib
π(π₯, π¦) = β« π(π₯, π¦)ππ¦ + Ξ³(π₯)
birinchi shartni
ππ(π₯,π¦)
ππ₯
= π(π₯, π¦) ham bajarilishini talab qilib, Ξ³(π₯) ni ham
koΚ»rinishini aniqlaymiz.
Misol. (2π₯ + 3π₯2π¦)ππ₯ + (π₯3 β 3π¦2)ππ¦ = 0
Differensial tenglama toΚ»la differensialga keltirilish shartini tekshiramiz
ππ(π₯, π¦)
ππ¦
= π(2π₯ + 3π₯2π¦)
ππ¦
= 3π₯2
ππ(π₯, π¦)
ππ₯
= π(π₯3 β 3π¦2)
ππ₯
= 3π₯2
ππ(π₯,π¦)
ππ¦
=
ππ(π₯,π¦)
ππ₯
shart bajarildi, u holda
1-usul
π(π₯, π¦) = β« π(π₯, π¦)ππ₯ + π½(π¦) = β«(2π₯ + 3π₯2π¦)ππ₯ + π½(π¦)=π₯2 + π₯3π¦ + π½(π¦)
π½(π¦) ni topish uchun ikkinchi shartni bajarilishini talab qilamiz:
ππ(π₯,π¦)
ππ¦
= π(π₯, π¦) β
ππ(π₯,π¦)
ππ¦
=
π(π₯2+π₯3π¦+π½(π¦))
ππ¦
= π₯3 + π½β²(π¦) = π₯3 β 3π¦2 β
π½β²(π¦) = β3π¦2 β π½(π¦) = βπ¦3 + ππππ π‘ β π(π, π) = ππ + πππ β ππ = πͺ
2-usul
π(π₯, π¦) = β« π(π₯, π¦)ππ¦ + Ξ³(π₯) = β«(π₯3 β 3π¦2)ππ¦ + Ξ³(π₯) = π₯3π¦ β π¦3 + Ξ³(π₯)
Ξ³(π₯) ni topish uchun birinchi shartni bajarilishini talab qilamiz:
ππ(π₯, π¦)
ππ₯
= π(π₯, π¦) β ππ(π₯, π¦)
ππ₯
= π(π₯3π¦ β π¦3 + Ξ³(π₯))
ππ₯
= 3π₯2π¦ + Ξ³β²(π₯)
= 2π₯ + 3π₯2π¦ β
Ilmiybaza.uz
Ξ³β²(π₯) = 2π₯ β Ξ³(π₯) = π₯2 + ππππ π‘
π(π, π) = πππ β ππ + ππ = ππ + πππ β ππ = πͺ
Eslatma: Agar π(π₯, π¦)ππ₯ + π(π₯, π¦)ππ¦ = 0 ni m(x,y)β’ 0 funksiyaga koΚ»paytirish
natijasida toΚ»la differensialga aylansa, m(x,y) ga integrallovchi koΚ»paytuvchi
deyiladi. Agar M(x,y) va N(x,y) funksiyalar uzluksiz xususiy hosilalarga ega va bir
vaqtni oΚ»zida nolga aylanmasa, u holda integrallovchi koΚ»paytuvchi mavjud. Lekin
uni qidirishning umumiy usuli mavjud emas.
INTEGRALLOVCHI KOΚ»PAYTUVCHI VA UNI TANLASH USULLARI.
Aytaylik quyidagicha differensial tenglama
π(π₯, π¦)ππ₯ + π(π₯, π¦)ππ¦ = 0 (*)
berilgan boΚ»lib, P(x,y) va Q(x,y) lar ikkita oΚ»zgaruvchi x va y larning funksiyasi
boΚ»lib, biror bir D sohada uzluksiz boΚ»lsin. Agar
ππ
ππ₯ β ππ
ππ¦
boΚ»lsa, u holda tenglama toΚ»la differensialli tenglama boΚ»lmaydi. Biroq
integrallovchi koΚ»paytuvchini tanlashga urinib koΚ»rishimiz mumkin. Agar (*)
tenglamani m(x,y)β’ 0 funksiyaga koΚ»paytirish natijasida toΚ»la differensialga
aylansa, m(x,y) ga integrallovchi koΚ»paytuvchi deyiladi. U holda quyidagicha
tenglik oΚ»rinli boΚ»ladi:
π(π(π₯, π¦)π(π₯, π¦))
ππ₯
= π(π(π₯, π¦)π(π₯, π¦))
ππ¦
Ushbu shartni quyidagicha koΚ»rinishda yozish mumkin:
π β ππ
ππ₯ + π β ππ
ππ₯ = π β ππ
ππ¦ + π β ππ
ππ¦ β
π β ππ
ππ₯ β π β ππ
ππ¦ = π β (ππ
ππ¦ β ππ
ππ₯) β
Oxirgi ifoda birinchi tartibli xususiy hosilali tenglama boΚ»lib, integrallovchi
koΚ»paytuvchi m(x,y) ni aninqlaydi. Integrallovchi koΚ»paytuvchini topishning
umumiy usuli mavjud emas, lekin ayrim xususiy hollarda olingan xususiy hosilali
tenglamani yechib, natijada integrallovchi koΚ»paytuvchini aniqlash mumkin.
1. Integrallovchi koΚ»paytuvchi x oΚ»zgaruvchiga bogΚ»liq boΚ»lsa: m=m(x)
Ilmiybaza.uz
Bunday holda
ππ
ππ¦ = 0 boΚ»lib, shuning uchun ham m(x,y) uchun tenglamani
quyidagicha yozish mumkin:
π β ππ
ππ₯ = π β (ππ
ππ₯ β ππ
ππ¦) β 1
π β ππ
ππ₯ = 1
π β (ππ
ππ¦ β ππ
ππ₯) β
Ushbu tenglamaning oΚ»ng tomoni faqat x ning funksiyasi boΚ»lsa, u holda m(x)
funksiyani oxirgi tenglamani integrallash orqali topish mumkin.
2. Integrallovchi koΚ»paytuvchi y oΚ»zgaruvchiga bogΚ»liq boΚ»lsa: m=m(y)
Oldingi holatdagidek, bu holda
ππ
ππ₯ = 0 boΚ»lib, shuning uchun ham m(x,y) uchun
tenglamani quyidagicha yozish mumkin:
1
π β ππ
ππ¦ = β 1
π β (ππ
ππ¦ β ππ
ππ₯)
Ushbu tenglamaning oΚ»ng tomoni faqat y ning funksiyasi boΚ»lsa, u holda m(y)
funksiyani oxirgi tenglamani integrallash orqali topish mumkin.
3. Integrallovchi koΚ»paytuvchi x va y oΚ»zgaruvchilarning aniq bir
kombinatsiyalariga bogΚ»liq bolsa: m=m(z(x,y))
Yangi z(x,y) funksiya masalan quyidagicha koΚ»rinishlarda boΚ»lishi mumkin:
π§ = π₯
π¦ ; π§ = π₯ β π¦; π§ = π₯ + π¦; π§ = π₯2 + π¦2; β¦
Bunda muhimi shundaki, integrallovchi koΚ»paytuvchi m(x,y) bitta z oΚ»zgaruvchining
funksiyasi sifatida keladi:
m(x,y)=m(z)
va quyidagicha differensial tenglamadan topiladi:
1
π β ππ
ππ§ =
ππ
ππ¦ β ππ
ππ₯
π ππ§
ππ₯ β π ππ§
ππ¦
OΚ»ng tomon faqat z ga bogΚ»liq va maxraj nolga teng emas deb faraz qilinadi.
Misol. (1 + π¦2)ππ₯ + π₯π¦ππ¦ = 0 differensial tenglama yechhilsin.
Boshlanishida tenglama toΚ»la differensialli tenglama ekanligini tekshiramiz:
ππ
ππ₯ = π
ππ₯ (π₯π¦) = π¦; ππ
ππ¦ = π
ππ¦ (1 + π¦2) = 2π¦
Ilmiybaza.uz
KoΚ»rinib turibtiki xususiy hosilalar bir-biriga teng emas, demak tenglama toΚ»la
differensiallanuvchi tenglamaga kelmaydi. ToΚ»la differensiallanuvchi koΚ»rinishga
olib kelish uchun integrallovchi koΚ»paytuvchini tanlashga harakat qilib koΚ»ramiz:
Quyidagicha funksiyani hisoblaymiz:
ππ
ππ¦ β ππ
ππ₯ = 2π¦ β π¦ = π¦
KoΚ»rinib turibtiki:
1
π β (
ππ
ππ¦ β
ππ
ππ₯) =
1
π₯π¦ β π¦ =
1
π₯
Ifoda faqat x oΚ»zgaruvchiga bogΚ»liq, demak integrallovchi koΚ»paytuvchi ham faqat
x ga bogΚ»liq boΚ»ladi: m=m(x), uni esa quyidagi tenglamadan topamiz:
1
π β ππ
ππ₯ = 1
π β (ππ
ππ¦ β ππ
ππ₯) = 1
π₯ βΉ ππ
π = ππ₯
π₯ βΉ ππ|π| = ππ|π₯| β π = Β±π₯
m=x ni tanlaymiz va berilgan differensial tenglamani x ga koΚ»paytiramiz. Natijada
toΚ»la differensialli tenglamaga kelamiz:
(π₯ + π₯π¦2)ππ₯ + π₯2π¦ππ¦ = 0
Endi toΚ»la differensiallilik sharti bajariladi:
ππ
ππ₯ = π
ππ₯ (π₯2π¦) = 2π₯π¦; ππ
ππ¦ = π
ππ¦ (π₯ + π₯π¦2) = 2π₯π¦
u(x,y) funksiyani tenglamalar sistemasidan aniqlash mumkin:
{
ππ’
ππ₯ = π₯ + π₯π¦2
ππ’
ππ¦ = π₯2π¦
Birinchi tenglamadan
π’(π₯, π¦) = β«(π₯ + π₯π¦2) ππ₯ = π₯2
2 + π₯2π¦2
2
+ π(π¦)
Ushbu ifodani ikkinchi tenglamaga qoΚ»yib π(π¦) ni aniqlaymiz:
ππ’
ππ¦ = π
ππ¦ [π₯2
2 + π₯2π¦2
2
+ π(π¦)] = π₯2π¦; βΉ π₯2π¦ + πβ²(π¦) = π₯2π¦ βΉ
πβ²(π¦) = 0 βΉ π(π¦) = πΆ, bunda C β ixtiyoriy konstanta
Shunday qilib, differensial tenglamaning umumiy yechimi
π₯2
2 + π₯2π¦2
2
+ πΆ = 0
Ilmiybaza.uz
EGRI CHIZIQLAR OILASI. BIRINCHI TARTIBLI DIFFERENSIAL
TENGLAMALARNING MAXSUS YECHIMLARI.
Egri chiziq tenglamasi umumiy holda f(x,y)=0 tenglama bilan aniqlanadi. Ayrim
hollarda egri chiziqlarga bitta yoki bir nechta parametrlar qoΚ»shishadi. Parametrlarga
turli xil qiymatlar berish mumkin. Egri chiziq ifodasida bitta parameter kirgan holni
koΚ»rib chiqamiz:
π(π₯, π¦, π) = 0 (1)
Parametr ixtiyoriy π0 qiymat qabul qiladi va qandaydir π(π₯, π¦, π0) = 0 egri chiziqni
aniqlaydi. C parameter oΚ»zgarganda egri chiziq shaklini va XOY tekislikdagi
joylashuvini oΚ»zgartiradi.
TaΚΌrif 1. π(π₯, π¦, π) = 0 tenglama bilan aniqlanadigan barcha egri chiziqlar
majmuasiga bir parametrli egri chiziqlar oilasi deyiladi, tenglamaning oΚ»ziga esa
egri chiziqlar oilasining tenglamasi deyiladi.
TaΚΌrif 2. Har bir nuqtasida egri chiziqlar oilasining bitta egri chizigΚ»i bilan
urinadigan egri chiziqqa berilgan egri chiziqlar oilasining gardishi deyiladi.
Gardishning parametrik tenglamasi
{ π(π₯, π¦, π) = 0
ππ
β²(π₯, π¦, π) = 0
tenglamalar
sistemasi
orqali
aniqlanadi.
Tenglamalar
sistemasidan
c
ni
yoΚ»qotsak, u holda gardishning aniq yoki noaniq
koΚ»rinishini topib olamiz.
Ushbu sistema gardish mavjudligining zaruriy
sharti hisoblanadi. Ushbu sistemaning yechimi
gardishdan tashqari, gardishga tegishli boΚ»lmagan egri chiziqlar oilasining maxsus
nuqtalarini oΚ»z ichiga olishi mumkin. Sistemaning barcha yechimlar toΚ»plami
diskriminant egri chiziq deyiladi. Shunday qilib gardish β diskriminant egri
chiziqning bir qismini tashkil qiladi. Gardishni bir qiymatli topish uchun uning
mavjudligining yetarli shartidan foydalaniladi. Ushbu shart quyidagicha shartni
bajarilishi talab qiladi: