2-mustaqil ishi Differentsial tenglamalarni qatorlar yordamida yechish, besselp tenglamasi

u=a₀ +a₁x+a₂x²+...+a_nxⁿ+... (7.2.7)

Bundan ikki marta hosila olsak

y^’=a₁ +2a₂x+3a₃x²+4a₄x³+...+na_nx^n-1...

u^’’=2a₂ +6a₃x+12a₄x²+...+ n(n-1) a_n x^n-2...

Boshlang’ich shartlarni hisobga olgan holda a₀=1; a₁=0 ekanligini aniqlaymiz. a₀ va a₁ ni (7.2.7) ga qo’ysak

u=1+a₂x²+a₃x³+a₄x⁴...+a_nxⁿ

Bu qatorni qolgan koeffitsiyentlarini topish uchun berilgan tenglamadan y^’’-x²y =0 foydalanamiz:

2a₂ +6a₃x+12a₄x²+20a₅x³+30a₆x⁴+...+ n(n-1) a_n x^n-2–

–x²(1+a₂x²+a₃x³+a₄x⁴...+ a_nxⁿ+...)=0.

Bu tenglikni “x” ning darajalari bo’yicha guruhlarga ajratamiz

2a₂+6a₃x+(12a₄–1)x²+20a₅x³+(30a₆ –a₂)x⁴+(42a₇–a₃)x⁵+

+(56a₈–a₄)x⁵...=0.

Biz yechimni x 0 hol uchun qidirayotganimiz uchun “x” ning oldidagi koeffitsiyentlarni “0”ga tenglashimiz lozim bo’ladi, ya’ni a₂=0, a₃=0, 12a₄–1=0 .

Bundan a₄= ; a₅=0; 30a₆-a₂=0; a₆=0 ;a₇=0 va 56a₈-a₄=0 x.k.

Shularni hisobga olgan holda yechimni quyidagicha yozish mumkin

u=1+ x⁴+ x⁸...