СИСТЕМА РЕКУРРЕНТНО-ДИФФЕРЕНЦИАЛЬНЫХ УРАВНЕНИЙ ДЛЯ ПРОИЗВОДНЫХ РЕШЕНИЯ УРАВНЕНИЯ РИКАТТИ

Аннотация

This is 𝑦''+𝑥𝑦 = 0 equivalent to the Liouville 𝑧'(𝑥) = 𝑥 +𝑧2(𝑥) equation. The system of recurrent-differential equations for the derivatives of the solution of the Riccati equation is studied. Key words: recurrent-differential equation, polynomial triangle, infinite system. Annotatsiya: Ushbu 𝑦''+𝑥𝑦 = 0 Liuvill tenglamasiga 𝑧'(𝑥) = 𝑥+𝑧2(𝑥) tengkuchli. Rikkati tenglamasi yechimining hosilalari uchun rekurrent-differensial tenglamalar sistemasi o‘rganiladi. Kalit so’zlar: rekurrent-differensial tenglama, polinomial uchburchak, cheksiz sistema. Аннотация: Это эквивалентно 𝑦''+𝑥𝑦 = 0 уравнению Лиувилля. Исследована 𝑧'(𝑥) = 𝑥+𝑧2(𝑥) система рекуррентно-дифференциальных уравнений для производных решения уравнения Риккати. Ключевые слова: рекуррентно-дифференциальное уравнение, полиномиальный треугольник, бесконечная система. Language: Uzbek 77 Universaljurnal.uz Universal International Scientific Journal 2024, 1(6) Citation: Kurbanov , A. (2024). SYSTEM OF RECURRENT-DIFFERENTIAL EQUATIONS FOR THE DERIVATIVES OF THE SOLUTION OF THE RICATTI EQUATION. Universal International Scientific Journal, 1(6), 77–80. Retrieved from https://universaljurnal.uz/index.php/jurnal/article/view/1113 Doi: https://doi.org/10.5281/zenodo.13929631 Parametrga bog‘liq chiziqli tenglamalarni kichik parametr usuli bilan 𝑦' (𝑥)=𝑎 (𝑥)𝑦 (𝑥)+𝑏 (𝑥)𝑦 (𝑥)+𝑐 (𝑥),𝑛= 𝑛 𝑛 𝑛 𝑛 𝑛−1 𝑛 0,1,2,... (1) (2) munosabatdan bevosita quyidagi ko’rinishidagi rekurrent-differensial xossalar kelib chiqadi: 1) Agar 𝑘− juft tenglamalar cheksiz sistemasiga keltirish 𝑘 bo‘lsa 𝑃 (𝑥)bu ( − 1) -darajali, agar toq 𝑛,𝑘 2 mumkinligi yaxshi ma’lum, bunda 𝑦 (𝑥) 0 𝑘 bo’lsa ([ ] + 1) -darajali ko‘phad; 2) 𝑑 𝑘 = berilgan deb qaraladi. Bu sistema aslida sof 2 𝑛 rekurrent tabiatli bo‘lib, no’malumlar 𝑑𝑒𝑔𝑃 𝑘(𝑥) deyilsa, 𝑑𝑘+2 = 𝑑𝑘 +1, 𝑛 = 𝑛 𝑛+1 𝑛 2,3,...,𝑛−1;𝑘 = 1,2,...,𝑛−1. birin-ketin integrallash yo‘li hisoblanadi. Amaliyotda (1) sistemadan farqli Endi 𝑦''+𝑥𝑦 = 0 tenglamani qaraymiz. Uning yechimi elementar funksiyalar 𝑃 (𝑥) = 𝑎 (𝑥)𝑃 (𝑥)+𝑏 (𝑥)(𝑃 (𝑥))′ 𝑛+1,𝑘 𝑛,𝑘 𝑛,𝑘 𝑛,𝑘 𝑛,𝑘−1 +𝑐 (𝑥)𝑃 (𝑥)(2) orqali, ular ustida algebraik amallar va 𝑛,𝑘 𝑛,𝑘−2 integrallash amali bilan ifodalanmaydi ko‘rinishdagi tenglamalar sistemasiga (J.Liuvil teoremasi) [1]. U keltiriladigan masalalar uchraydi. Xususan, 𝑎 (𝑥)= 𝑛−𝑘,𝑏 (𝑥) = 1,𝑐 (𝑥) = (𝑛−𝑘+ 𝑧'(𝑥)= 𝑥+𝑧2(𝑥) 𝑛,𝑘 𝑛,𝑘 𝑛,𝑘 2)𝑥 bo’lgan hol Liuvill tenglamasi uchun (3) kombinatorik masalaga tengkuchli. Rikkati tenglamasiga teng kuchli. Boshlang‘ich shart 𝑃 = 𝑛! bo’lgan holni 𝑛,0 Teorema. (3) tenglama uchun 𝑧(0)= qaraymiz (bu yerda 𝑘 < 𝑛). Shunda 𝑃 (𝑥) 𝑛,𝑘 𝑧 Koshi masalasining yechimi 𝑧(𝑥) 0 ko‘phadlardan iborat bo‘lib, u binomial 1 1 𝑧(𝑛)(𝑥)= 𝑛!𝑧𝑛+1+ (𝑛+1)!𝑥𝑧𝑛−1+ (𝑛 koeffitsientlar kabi uchburchak hosil 3 12 +1)!𝑧𝑛−2+ qiladi. 78 Universaljurnal.uz Universal International Scientific Journal 2024, 1(6) 𝑛 2)𝑃 + (𝑃 )′, va 𝑘 = 3,...,𝑛, 𝑃 = +∑[(𝑛−𝑘+1)𝑥𝑃 +(𝑃 )′+(𝑛−1 𝑛,2 𝑛,1 𝑛+1,𝑘 𝑛−1,𝑘−2 𝑛−1,𝑘−1 𝑘=3 (𝑛−𝑘+2)𝑥𝑃 +(𝑃 )′+(𝑛−𝑘)𝑃 . 𝑛,𝑘−2 𝑛,𝑘−1 𝑛,𝑘 −𝑘)𝑃 ]𝑧𝑛−𝑘 𝑛−1,𝑘 xossaga ega. Xususan, 𝑃 = 𝑥𝑃 +(𝑃 )′. 𝑛+1,𝑛+1 𝑛,𝑛−1 𝑛,𝑛 Isbot. (3) tenglamadan 𝑧″ = 2𝑧3 + Boshlang‘ich shart 𝑃 =𝑛! bo‘lgan 𝑛,0 2𝑥𝑧 + 1,𝑧‴ = 6𝑧4 + 8𝑞𝑧2 + 2𝑧 + 2𝑥2 holdan keyingi 𝑃 ko‘phadni topamiz. 𝑛,1 bo‘lishi ravshan[2]. Yuqorida keltirilgan So‘ng 𝑃 = 𝑛!𝑥+(𝑛−2)𝑃 𝑛,1 𝑛−1,1 𝑧(𝑛) = 𝑛!𝑧𝑛+1+𝑃 𝑛,1 𝑧𝑛−1+𝑃 𝑛,2 𝑧𝑛−2+ (6) 𝑃 𝑧𝑛−3+...+𝑃 𝑧2+𝑃 𝑧1+𝑃 𝑛,3 𝑛,𝑛−2 𝑛,𝑛−1 𝑛,𝑛 tenglamaning bir jinsli qismini quyidagicha (4) yechamiz: 𝑃 = (𝑛−2)𝑃 tenglamada 𝑛,1 𝑛−1,1 o‘rinli bo‘lsin deb faraz qilaylik. U holda 𝑃 = 𝑦 kabi belgilash kiritsak,𝑦 = (𝑛− 𝑛,1 𝑛 𝑛 𝑧(𝑛+1) = (𝑛+1)!𝑧𝑛+2+[(𝑛−1)𝑃 +(𝑛 𝑛,1 2)𝑦 ga ega bo’lamiz. Uning yechimi𝑦 = +1)!𝑥]𝑧𝑛+[(𝑛−2)𝑃 𝑛−1 𝑛 𝑛,2 +(𝑃 )']𝑧𝑛−1+ (𝑛−2)!𝐶ko‘rinishda bo‘ladi. O‘zgarmasni 𝑛,2 +[(𝑛−1)𝑥𝑃 +(𝑛−3)𝑃 +(𝑃 )']𝑧𝑛−2+[(𝑛 variatsiyalash yo‘li orqali 𝑛,1 𝑛,3 𝑛,2 −2)𝑥𝑃 𝑛,2 +(𝑛−4)𝑃 𝑛,4 𝑦 𝑛 = (𝑛−2)!𝐶 𝑛 +(𝑃 )']𝑧𝑛−3+(5) (7) 𝑛,3 +[(𝑛−3)𝑥𝑃 +(𝑛−5)𝑃 +(𝑃 )']𝑧𝑛−4+[(𝑛 𝑛,3 𝑛,5 𝑛,4 ifodani (6) tenglamaga qo‘yamiz va 𝐶 ga 𝑛 −4)𝑥𝑃 +(𝑛−6)𝑃 𝑛,4 𝑛,6 nisbatan 𝐶 = 𝐶 + (𝑛 − 1)𝑛𝑥 𝑛 𝑛−1 +(𝑃 )']𝑧𝑛−5+ 𝑛,5 rekurrent formulaga ega bo‘lamiz. Keyingi +[(𝑛−𝑘+2)𝑥𝑃 +(𝑛−𝑘)𝑃 𝑛,𝑘−2 𝑛,𝑘 +(𝑃 )']𝑧𝑛−𝑘+1+[(𝑛−𝑘 qadamda 𝑛 ga 1,2,...,𝑛 qiymatlarni berib 𝑛,𝑘−1 +1)𝑥𝑃 +(𝑛−𝑘−1)𝑃 + tenglamalar sistemasini hosil qilamiz va 𝑛,𝑘−1 𝑛,𝑘+1 +(𝑃 )']𝑧𝑛−𝑘+...+[2𝑥𝑃 +(𝑃 )']𝑧 𝑛,𝑘 𝑛,𝑛−2 𝑛,𝑛−1 uni yechamiz:𝐶 = (1⋅2+2⋅3+...+(𝑛−1)⋅ 𝑛 +𝑥𝑃 +(𝑃 )'. 𝑛,𝑛−1 𝑛,𝑛 𝑛)𝑥 = 1 𝑛(𝑛−1)(𝑛+1)𝑥. 𝐶 uchun topilgan 𝑛 Bu ifodani (4) formulada 𝑛 o‘rniga 3 ifodani (7) ga qo‘yamiz va quyidagiga ega 𝑛 + 1 qo‘yilgani bilan taqqoslab, quyidagi bo‘lamiz : tengliklarga ega bo‘lamiz: 𝑃 = (𝑛+1)!, 𝑛+1,0 1 𝑃 = 𝑦 = (𝑛−2)!𝐶 = (𝑛−2)!𝑛(𝑛−1)(𝑛+ 𝑛,1 𝑛 𝑛 P =(n+1)!x+(n−1)P ,𝑃 = (𝑛 − 3 n+1,1 n,1 𝑛+1,2 1)𝑥 = 1 (𝑛+1)!𝑥. 3 Yuqoridagi qadamlarni davom ettirib, 79 Universaljurnal.uz Universal International Scientific Journal 2024, 1(6) 1 1 P n,2 = 12 (n+1)!,𝑃 𝑛,3 = 9 1 0 (𝑛+1)!(5𝑛−8)𝑥2, + 12 (𝑛+1)!𝑧𝑛−2 𝑛 1 +∑[(𝑛−𝑘+1)𝑥𝑃 P = (n+1)!(5n−12)x,𝑃 = 1 (𝑛+ 𝑛−1,𝑘−2 n,4 180 𝑛,5 5670 𝑘=3 +(𝑃 )′+(𝑛−1 𝑛−1,𝑘−1 1)!(35𝑛2−203𝑛+264)𝑥3+ 1 (𝑛+ 10080 −𝑘)𝑃 ]𝑧𝑛−𝑘. 𝑛−1,𝑘 1)!(35𝑛−106),... Xulosa. larni hosil qilamiz. Ushbu maqolada Rikkati tenglamasi Ushbu