to’la differensial tenglama. hosilaga nisbatan yechilmagan birinchi tartibli tenglamalar

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1662926777.doc ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ u x y m x y dx y x x ( , ) ( , ) ( ) = + ò j 0 ) , ( ) ( 0 x n y m y x n y dx y m dy du x x ¶ ¶ = ¶ ¶ = ¢ + ¶ ¶ = ò j ). , ( ) ( ) , ( , ) , ( ) ( 0 0 y x n х y x n yani у х n y dx x n x x э x x = + = + ¶ ¶ ò j j n x y n x y y n x y ( , ) ( , ) ' ( ) ( , ). - + = 0 j ò + = …

to’la differensial tenglama deyiladi, bunda m/ y, n/ x - uzluksiz funksiyalar. (3.1) tenglamani integrallashga o’tamiz. (3.1) tenglamaning chap tomoni biror u(x,y) funksiyaning to’la differensiali bo’lsin deb faraz qilamiz, ya’ni m(x,y)dx+n(x,y)dy=du(x,y), du/dx =( u/ x)dx+( u/ y)dy u holda m= u/ x, n= u/ y (3.3) u/ x=m munosabatdan ni topamiz. bu tenglikni har ikki tomonini u bo’yicha differensiallab natijani n (x,y) ga tenglaymiz: bo’lgani uchun yoki demak shunday qilib ko’rinishda bo’ladi. du=0 bo’lganda , u(x,y)=c. demak, umumiy integral (3.4) integrallovchi ko’paytuvchi (3.1) tenglamada (3.2) munosabat bajarilmasin. ba’zan shunday funksiyani tanlab olish mumkinki, (3.1) tenglamani shu funksiyaga ko’paytirganda tenglamaning chap tomoni biror funksiyaning to’la differensialini ifodalaydi. bunday tanlangan ((x,y) funksiyaga (3.1) tenglamaning integrallovchi ko’paytuvchisi deyiladi. ((x,y) ni topish usuli: (3.1) ni ((x,y) ga ko’paytiramiz (mdx+(ndy=0 keyingi tenglama to’la differensialli tenglama bo’lishi uchun (3.2) munosabat bajarilishi zarur va etarli: oxirgi tenglamaning har ikki qismini ( ga bo’lib (3.5) munosabatni hosil qilamiz. …

(i=1,2...) bu tenglamalarni integrallab, (3.6) tenglama yechimlarini hosil qilamiz. lekin (3.6) tenglamani har doim ga nisbatan oson yechilmaydi va ga nisbatan tenglamalar sodda integrallanmasligi mumkin. shuning uchun (3.6) tenglamani boshqa usullarda integrallash qulay bo’ladi. quyidagi hollarni qaraymiz. 1. f( )=0, bunda hech bo’lmaganda tenglamaning bitta =ki yechimi mavjud bo’lsin. tenglama x va y o’zgaruvchilarga bog’liq bo’lmaganligi sababli, ki=const. y’=ki ni integrallab y=kix+c yoki ki=(y-c)/x. ki berilgan tenglama yechimi ekanligidan f((y-c)/x)=0 qaralayotgan tenglama yechimi bo’ladi. misol ( )7 - ( )5+ +3=0 tenglama integrali ((y-c)/x)7-((y-c)/x)5+(y-c)/x+3=0 2. (3.6) tenglama quyidagi ko’rinishda bo’lsin. f(x, )=0 (3.7) agar tenglamani y’ ga nisbatan yechish qiyin bo’lsa, u holda t parametr kiritish bilan (3.7) ikkita tenglamaga keltiriladi: x=((t) va =((t) dy= dx ekanligidan, dy =((t) (’(t)dt, bundan demak, (3.7) tenglama yechimlari parametrik holda quyidagi ko’rinishda bo’ladi x=((t) agar (3.7) x ga nisbatan yechilsa, ya’ni x=(( ), u holda deyarli har doim =t deb parametr kiritish qulay. …

i x ga nisbatan differensiallab (3.10) hosil bo’lgan tenglama x(() va dx/d( ga nisbatan chiziqli tenglamadir. uni yechib f(x, ,c)=0 ni hosil qilamiz. demak, lagranj tenglamasini yechimi parametrik ko’rinishda bo’ladi. (3.10) tenglamani hosil qilishda deb qaralgan edi. demak, bunda =const yechimlar, agar ular mavjud bo’lsa, yo’qotilgan edi. =const bo’lsa, u holda (3.101) tenglama faqat , bo’lganda bajariladi. demak, agar tenglama haqiqiy r=ri ildizlarga ega bo’lsa, yuqoridagi yechimlarga yana y=x ((()+(((), (=(i yechimlarni ham qo’shish kerak bo’ladi. klero tenglamasi (-((()(0 bo’lsin. d(/dx ga bo’lishdan (=с, с=const yechimlar yo’qotilgan bo’ladi. bu holda (( )= bo’lib, (3.9) tenglama y=x +(( ) (3.11) ko’rinishiga keladi va bu tenglama- klero tenglamasi deyiladi. bu teglamani yechish uchun =( deb belgilash kiritamiz. natijada y=x(+((() ni hosil qilamiz. bu tenglamani x bo’yicha differensiallab (=(+xd(/dx+(’(()d(/dx yoki tenglamani hosil qilamiz. bundan d(/dx=0, demak (=c yoki x+(’(()=0. (=с da yechimdan y=cx+((c) ikkinchi holda esa yechim ko’rinishda bo’ladi. a d a b …

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1662926777.doc ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ u x y m x y dx y x x ( , ) ( , ) ( ) = + ò j 0 ) , ( ) ( 0 x n y m y x n y dx y m dy du x x ¶ ¶ = ¶ ¶ = ¢ + ¶ ¶ = ò j ). , ( ) ( ) , ( , ) , ( ) ( 0 0 y x n х y x n yani у х n y dx x n x x э x x = + = + ¶ ¶ ò j j …

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