Eslatib o'tamiz, \({x_0} = 0\) haqida ketma-ket yechimni faqat bu nuqta oddiy nuqta bo'lsa yoki boshqacha qilib aytganda, \({x_0} = 0\) da eng yuqori hosilaviy hadning koeffitsienti nolga teng bo'lmasagina topishimiz mumkin. Bu aniq, shuning uchun keling, yechimlar shaklidan, shuningdek, ushbu yechim uchun kerak bo'ladigan hosilalardan boshlaylik.

\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} \hspace{0.25in}y'''\left( x \right) = \sum\limits_{n = 3}^\infty {n\left( {n - 1} \right)\left( {n - 2} \right){a_n}{x^{n - 3}}} \]

Differensial tenglamaga ulanganda,

\[\sum\limits_{n = 3}^\infty {n\left( {n - 1} \right)\left( {n - 2} \right){a_n}{x^{n - 3}}} + {x^2}\sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} + x\sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0\]

Endi barcha koeffitsientlarni ketma-ketlikka o'tkazing va mos ravishda bajaring siljishlar Shunday qilib, barcha seriyalar \({x^n}\) jihatidan.

\[\begin{align*}\sum\limits_{n = 3}^\infty {n\left( {n - 1} \right)\left( {n - 2} \right){a_n}{x^{n - 3}}} + \sum\limits_{n = 1}^\infty {n{a_n}{x^{n + 1}}} + \sum\limits_{n = 0}^\infty {{a_n}{x^{n + 1}}} & = 0\\ \sum\limits_{n = 0}^\infty {\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 3}}{x^n}} + \sum\limits_{n = 2}^\infty {\left( {n - 1} \right){a_{n - 1}}{x^n}} + \sum\limits_{n = 1}^\infty {{a_{n - 1}}{x^n}} & = 0\end{align*}\]

Keyin, keling, ikkinchi seriyani \(n = 1\) da boshlashimiz mumkin, chunki bu atama nolga teng bo'ladi. Keling, buni qilaylik va keyin ikkinchi va uchinchi shartlarni birlashtirib olamiz,

\[\begin{align*}\sum\limits_{n = 0}^\infty {\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 3}}{x^n}} + \sum\limits_{n = 1}^\infty {\left[ {\left( {n - 1} \right) + 1} \right]{a_{n - 1}}{x^n}} & = 0\\ \sum\limits_{n = 0}^\infty {\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 3}}{x^n}} + \sum\limits_{n = 1}^\infty {n{a_{n - 1}}{x^n}} & = 0\end{align*}\]

Shunday qilib, biz yangi seriyada yanada soddalashtirishga yordam beradigan yaxshi soddalashtirishga ega bo'ldik. Yangi ikkinchi seriyani endi \(n = 0\) da boshlash mumkin va keyin birinchi seriya bilan birlashtirilishi mumkin,

\[\sum\limits_{n = 0}^\infty {\left[ {\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 3}} + n{a_{n - 1}}} \right]{x^n}} = 0\]

Endi biz oddiy takrorlanish munosabatini olish uchun koeffitsientlarni o'rnatishimiz mumkin.

\[\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 3}} + n{a_{n - 1}} = 0\hspace{0.25in}n = 0,1,2, \ldots \]

Qaytalanish munosabatini yechish,

\[{a_{n + 3}} = \frac{{ - n{a_{n - 1}}}}{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}\hspace{0.25in}n = 0,1,2, \ldots \]

Endi biz \(n\) qiymatlarini ulashni boshlashimiz kerak va bu yuqori tartibli differentsial tenglamaga o'tishda talab qilinadigan ish hajmining biroz sezilarli darajada oshishini ko'rishimiz mumkin bo'lgan asosiy sohalardan biri bo'ladi.

\( \displaystyle n = 0:\hspace{0.25in}{a_3} = 0\)

\( \displaystyle n = 1:\hspace{0.25in}{a_4} = \frac{{ - {a_0}}}{{\left( 2 \right)\left( 3 \right)\left( 4 \right)}}\)

\( \displaystyle n = 2:\hspace{0.25in}{a_5} = \frac{{ - 2{a_1}}}{{\left( 3 \right)\left( 4 \right)\left( 5 \right)}}\)

\( \displaystyle n = 3:\hspace{0.25in}{a_6} = \frac{{ - 3{a_2}}}{{\left( 4 \right)\left( 5 \right)\left( 6 \right)}}\)

\( \displaystyle n = 4:\hspace{0.25in}{a_7} = \frac{{ - 4{a_3}}}{{\left( 5 \right)\left( 6 \right)\left( 7 \right)}} = 0\)

\( \displaystyle n = 5:\hspace{0.25in}{a_8} = \frac{{ - 5{a_4}}}{{\left( 6 \right)\left( 7 \right)\left( 8 \right)}} = \frac{{5{a_0}}}{{\left( 2 \right)\left( 3 \right)\left( 4 \right)\left( 6 \right)\left( 7 \right)\left( 8 \right)}}\)

\( \displaystyle n = 6:\hspace{0.25in}{a_9} = \frac{{ - 6{a_5}}}{{\left( 7 \right)\left( 8 \right)\left( 9 \right)}} = \frac{{\left( 2 \right)\left( 6 \right){a_1}}}{{\left( 3 \right)\left( 4 \right)\left( 5 \right)\left( 7 \right)\left( 8 \right)\left( 9 \right)}}\)

\( \displaystyle n = 7:\hspace{0.25in}{a_{10}} = \frac{{ - 7{a_6}}}{{\left( 8 \right)\left( 9 \right)\left( {10} \right)}} = \frac{{\left( 3 \right)\left( 7 \right){a_2}}}{{\left( 4 \right)\left( 5 \right)\left( 6 \right)\left( 8 \right)\left( 9 \right)\left( {10} \right)}}\)

\( \displaystyle n = 8:\hspace{0.25in}{a_{11}} = \frac{{ - 8{a_7}}}{{\left( 9 \right)\left( {10} \right)\left( {11} \right)}} = 0\)

\( \displaystyle n = 9:\hspace{0.25in}{a_{12}} = \frac{{ - 9{a_8}}}{{\left( {10} \right)\left( {11} \right)\left( {12} \right)}} = \frac{{ - \left( 5 \right)\left( 9 \right){a_0}}}{{\left( 2 \right)\left( 3 \right)\left( 4 \right)\left( 6 \right)\left( 7 \right)\left( 8 \right)\left( {10} \right)\left( {11} \right)\left( {12} \right)}}\)

\( \displaystyle n = 10:\hspace{0.25in}{a_{13}} = \frac{{ - 10{a_9}}}{{\left( {11} \right)\left( {12} \right)\left( {13} \right)}} = \frac{{ - \left( 2 \right)\left( 6 \right)\left( {10} \right){a_1}}}{{\left( 3 \right)\left( 4 \right)\left( 5 \right)\left( 7 \right)\left( 8 \right)\left( 9 \right)\left( {11} \right)\left( {12} \right)\left( {13} \right)}}\)

\( \displaystyle n = 11:\hspace{0.25in}{a_{14}} = \frac{{ - 11{a_{10}}}}{{\left( {12} \right)\left( {13} \right)\left( {14} \right)}} = \frac{{ - \left( 3 \right)\left( 7 \right)\left( {11} \right){a_2}}}{{\left( 4 \right)\left( 5 \right)\left( 6 \right)\left( 8 \right)\left( 9 \right)\left( {10} \right)\left( {12} \right)\left( {13} \right)\left( {14} \right)}}\)

Xo‘sh, endi biz \({a_{4k}}\), \({a_{4k + 1}}\), \({a_{4k + 2}}\) va \({a_{4k + 3}}\) tomonidan \(k = 0,1,2,3, \ldots \) uchun berilgan 4 ta kichik holatga koeffitsientlarni ajratishimiz mumkin, biz \({a_{4k}}\) uchun yarim batafsil ma’lumot beramiz, so‘ngra qolganlarini sizga qoldiramiz, chunki ular faqat bir-ikkita xulosalardir.

Avval e'tibor bering, barcha \({a_{4k}}\) atamalarida \({a_0}\) mavjud va ular bir-birini almashtirib turadi. Keyinchalik e'tibor bering, biz maxrajni faktorialga aylantirishimiz mumkin, aniqrog'i \(\left( {4k} \right)!\), agar biz yuqori va pastni hisoblagichda mavjud bo'lgan raqamlarga ko'paytirsak va bu bu raqamlarni kvadratlarga aylantiradi. Keyingi e'tibor bering, yuqoridagi mahsulot 1 dan boshlanadi va \(4k - 3\) ga yetguncha 4 ga ortadi. Shunday qilib, bularning barchasini hisobga olsak, biz

\[{a_{4k}} = \frac{{{{\left( { - 1} \right)}^k}{{\left( 1 \right)}^2}{{\left( 5 \right)}^2} \cdots {{\left( {4k - 3} \right)}^2}{a_0}}}{{\left( {4k} \right)!}}\hspace{0.25in}k = 1,2,3, \ldots \]

E'tibor bering, bu faqat \(k = 1\) dan boshlab ishlaydi, chunki biz \(k = 0\) ni ulash orqali ushbu formuladan \({a_0}\) ni olmaymiz.

Endi, \({a_{4k + 1}}\) uchun hosila deyarli bir xil va shuning uchun formula:

\[{a_{4k + 1}} = \frac{{{{\left( { - 1} \right)}^k}{{\left( 2 \right)}^2}{{\left( 6 \right)}^2} \cdots {{\left( {4k - 2} \right)}^2}{a_1}}}{{\left( {4k + 1} \right)!}}\hspace{0.25in}k = 1,2,3, \ldots \]

va yana bir bor e'tibor bering, bu \(k = 0\) uchun ishlamaydi

\({a_{4k + 2}}\) formulasi yana deyarli bir xil bo‘ldi, bundan tashqari, faktorial maxrajda paydo bo‘lishi uchun yuqori va pastni 2 ga ko‘paytirishimiz kerak bo‘ladi, shuning uchun bu erda formula:

\[{a_{4k + 2}} = \frac{{2{{\left( { - 1} \right)}^k}{{\left( 3 \right)}^2}{{\left( 7 \right)}^2} \cdots {{\left( {4k - 1} \right)}^2}{a_2}}}{{\left( {4k + 2} \right)!}}\hspace{0.25in}k = 1,2,3, \ldots \]

Bu \(k = 0\) uchun ishlamasligini yana bir bor payqadim.

Va nihoyat, bizda \(k = 0,1,2,3, \ldots \) uchun \({a_{4k + 3}} = 0\) mavjud

Endi bizda barcha koeffitsientlar bor, keling, yechimni topamiz,

\[\begin{align*}y\left( x \right) & = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \cdots + {a_{4k}}{x^{4k}} + {a_{4k + 1}}{x^{4k + 1}} + {a_{4k + 3}}{x^{4k + 3}} + {a_{4k + 3}}{x^{4k + 3}} + \cdots \\ & = {a_0} + {a_1}x + {a_2}{x^2} \cdots + \frac{{{{\left( { - 1} \right)}^k}{{\left( 1 \right)}^2}{{\left( 5 \right)}^2} \cdots {{\left( {4k - 3} \right)}^2}{a_0}\,}}{{\left( {4k} \right)!}}{x^{4k}} + \\ & \hspace{2.05in}\frac{{{{\left( { - 1} \right)}^k}{{\left( 2 \right)}^2}{{\left( 6 \right)}^2} \cdots {{\left( {4k - 2} \right)}^2}{a_1}}}{{\left( {4k + 1} \right)!}}{x^{4k + 1}} + \\ & \hspace{2.05in}\frac{{2{{\left( { - 1} \right)}^k}{{\left( 3 \right)}^2}{{\left( 7 \right)}^2} \cdots {{\left( {4k - 1} \right)}^2}{a_2}\,}}{{\left( {4k + 2} \right)!}}{x^{4k + 2}} + \cdots \end{align*}\]

Bir xil koeffitsientni o'z ichiga olgan atamalarni yig'ish (har bir holatda birinchisidan tashqari, chunki formula ular uchun ishlamaydi) va hamma narsani ketma-ketliklar to'plami sifatida yozish bizga yechimimizni beradi,

\[\begin{align*}y\left( x \right) & = {a_0}\left\{ {1 + \sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}{{\left( 1 \right)}^2}{{\left( 5 \right)}^2} \cdots {{\left( {4k - 3} \right)}^2}{x^{4k}}\,}}{{\left( {4k} \right)!}}} } \right\} + \\ & \hspace{0.5in}{a_1}\left\{ {x + \sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}{{\left( 2 \right)}^2}{{\left( 6 \right)}^2} \cdots {{\left( {4k - 2} \right)}^2}{x^{4k + 1}}}}{{\left( {4k + 1} \right)!}}} } \right\} + \\ & \hspace{0.5in}{a_2}\left\{ {{x^2} + \sum\limits_{k = 1}^\infty {\frac{{2{{\left( { - 1} \right)}^k}{{\left( 3 \right)}^2}{{\left( 7 \right)}^2} \cdots {{\left( {4k - 1} \right)}^2}{x^{4k + 2}}}}{{\left( {4k + 2} \right)!}}} } \right\}\end{align*}\]