6.3-bo'lim: Yechimlar seriyasi
Differensial tenglamalarning ketma-ket yechimlarini topishga kirishishdan oldin biz differentsial tenglamalarning ketma-ket yechimlarini qachon topishimiz mumkinligini aniqlashimiz kerak. Shunday qilib, keling, differentsial tenglamadan boshlaylik,
\[\begin{equation}p\left( x \right)y'' + q\left( x \right)y' + r\left( x \right)y = 0\label{eq:Eq1}\end{equation}\]Bu safar biz haqiqatan ham doimiy bo'lmagan koeffitsientlarni nazarda tutamiz. Shu paytgacha biz faqat doimiy koeffitsientlar bilan ishladik. Biroq, ketma-ket echimlar bilan biz endi doimiy bo'lmagan koeffitsientli differentsial tenglamalarga ega bo'lishimiz mumkin. Bundan tashqari, muammolarni biroz chiroyli qilish uchun biz faqat polinom koeffitsientlari bilan shug'ullanamiz.
Endi biz \(x=x_{0}\) ekanligini aytamiz oddiy nuqta ikkalasi ham taqdim etilgan bo'lsa
\[\frac{{q\left( x \right)}}{{p\left( x \right)}}\,\,\hspace{0.25in}\text{va}\hspace{0.25in}\frac{{r\left( x \right)}}{{p\left( x \right)}}\]bor analitik \(x=x_{0}\) da. Ya'ni, bu ikki miqdor \(x=x_{0}\) atrofida Teylor seriyasiga ega. Biz faqat polinom bo'lgan koeffitsientlar bilan shug'ullanamiz, shuning uchun bu shuni aytishga teng bo'ladi.
\[p\left( {{x_0}} \right) \ne 0\]muammolarning aksariyati uchun.
Agar nuqta oddiy nuqta bo'lmasa, biz uni a deb ataymiz yagona nuqta.
Differensial tenglamaning ketma-ket yechimini topishning asosiy g'oyasi biz yechimni quyidagi ko'rinishda darajali qator sifatida yozishimiz mumkin, deb taxmin qilishdir.
\[\begin{equation}y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x - {x_0}} \right)}^n}} \label{eq:eq2}\end{equation}\]va keyin \(a_{n}\) nima bo'lishi kerakligini aniqlashga harakat qiling. Agar \(x=x_{0}\) nuqtasi oddiy nuqta bo'lsa, biz buni qila olamiz. Biz odatda \(\eqref{eq:eq2}\) \(x=x_{0}\) atrofida ketma-ket yechim deb aytamiz.
Buning juda asosiy misolidan boshlaylik. Aslida, u shunchalik asosiy bo'ladiki, bizda doimiy koeffitsientlar bo'ladi. Bu bizga to'g'ri echimni olganimizni tekshirish imkonini beradi.
Yechim
E'tibor bering, bu holda \(p(x)=1\) va shuning uchun har bir nuqta oddiy nuqtadir. Biz shaklda yechim izlaymiz,
\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \]Biz buni differentsial tenglamaga kiritishimiz kerak, shuning uchun biz bir nechta hosilalarni topishimiz kerak.
\[y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} \]Quvvat seriyasini ko'rib chiqishdan eslang bo'limi Agar kerak bo'lsa, biz ularni \(n=0\) da ishga tushirishimiz mumkin bo'lgan quvvat seriyalarida, ammo ularni har doim bizda mavjud bo'lgan joydan boshlash yaxshidir. Agar ularni \(n=0\)-da boshlash osonroq bo'lgan bo'lsa, biz buni vaqti kelganda osongina tuzatamiz.
Shunday qilib, ularni bizning differentsial tenglamamizga kiriting. Buni qilish,
\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} + \sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0\]Keyingi qadam hamma narsani bitta seriyaga birlashtirishdir. Buni amalga oshirish uchun biz ikkala qatorni bir nuqtadan boshlashimiz va \(x\) ko'rsatkichi ikkala seriyada ham bir xil bo'lishini talab qiladi.
Biz buni har doim \(x\) ko'rsatkichini bir xil bo'lishidan boshlaymiz. Odatda ko'rsatkichni \(n\) deb olish yaxshidir. Ikkinchi qator allaqachon tegishli ko'rsatkichga ega va ko'rsatkichni \(n\) ga etkazish uchun birinchi qatorni 2 ga pastga siljitish kerak bo'ladi. Agar buni qanday qilishni eslamasangiz, birinchi sharhni tezda ko'rib chiqing bo'limi bu erda biz bir nechta muammolarni qildik.
Birinchi quvvat seriyasini o'zgartirish bizga beradi,
\[\sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n}} + \sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0\]E'tibor bering, siljish jarayonida biz ikkala seriyani ham bir joyda boshladik. Bu har doim ham sodir bo'lmaydi, lekin qachon sodir bo'lsa, biz buni qabul qilamiz. Endi biz ikkita seriyani qo'shishimiz mumkin. Bu beradi,
\[\sum\limits_{n = 0}^\infty {\left[ {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} + {a_n}} \right]{x^n}} = 0\]Endi eslab haqiqat quvvat seriyasini ko'rib chiqish bo'limidan biz bilamizki, agar bizda barcha \(x\) uchun nolga teng bo'lgan quvvat seriyasi bo'lsa (bu shunday bo'lsa), boshlash uchun barcha koeffitsientlar nolga teng bo'lishi kerak. Bu bizga quyidagilarni beradi,
\[\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} + {a_n} = 0,\hspace{0.25in}n = 0,1,2, \ldots \]Bu deyiladi takrorlanish munosabati va biz \(n\) qiymatlarini kiritganimizga e'tibor bering, ular uchun u to'g'ri bo'lishi kerak. Biz har doim takrorlanish munosabati to'g'ri bo'lgan \(n\) qiymatlarini kiritishni xohlaymiz, chunki ular bu holatda bo'lgani kabi har doim ham \(n\) = 0 dan boshlanmaydi.
Keling, birinchi navbatda nimaga erishganimizni eslaylik. Biz differentsial tenglamaning ketma-ket yechimini topmoqchi edik. Buni amalga oshirish uchun biz \(a_{n}\) qiymatlarini aniqlashimiz kerak edi. Biz buni qila oladigan darajaga yaqinlashdik. Takrorlanish munosabati ikki xil \(a_{n}\) ga ega, shuning uchun biz buni faqat \(a_{n}\) uchun hal qila olmaymiz va barcha \(n\) uchun ishlaydigan formulani olamiz. Biroq, biz bundan \(a_{n}\)-ning ikkitasidan tashqari barchasi nima ekanligini aniqlash uchun foydalanishimiz mumkin.
Buning uchun biz birinchi navbatda eng katta pastki indeksga ega \(a_{n}\) uchun takrorlanish munosabatini hal qilamiz. Buni qilish,
\[{a_{n + 2}} = - \frac{{{a_n}}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\hspace{0.25in}n = 0,1,2, \ldots \]Endi, biz faqat \(n\) qiymatini ulashni boshlashimiz va nima bo'lishini ko'rishimiz kerak,
| \(n=0\) | \({a_2} = \frac{{ - {a_0}}}{{\left( 2 \right)\left( 1 \right)}}\) | \(n=1\) | \({a_3} = \frac{{ - {a_1}}}{{\left( 3 \right)\left( 2 \right)}}\) |
| \(n=2\) | \(\begin{align*}{a_4} &= - \frac{{{a_2}}}{{\left( 4 \right)\left( 3 \right)}}\\ & = \frac{{{a_0}}}{{\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}}\end{align*}\) | \(n=3\) | \(\begin{align*}{a_5} &= - \frac{{{a_3}}}{{\left( 5 \right)\left( 4 \right)}}\\ & = \frac{{{a_1}}}{{\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( 2 \right)}}\end{align*}\) |
| \(n=4\) | \(\begin{align*}{a_6} & = - \frac{{{a_4}}}{{\left( 6 \right)\left( 5 \right)}}\\ & = \frac{{ - {a_0}}}{{\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}}\end{align*}\) | \(n=5\) | \(\begin{align*}{a_7} & = - \frac{{{a_5}}}{{\left( 7 \right)\left( 6 \right)}}\\ & = \frac{{ - {a_1}}}{{\left( 7 \right)\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( 2 \right)}}\end{align*}\) |
| \(\vdots \) | \(\vdots \) | ||
| \({a_{2k}} = \frac{{{{\left( { - 1} \right)}^k}{a_0}}}{{\left( {2k} \right)!}},\,\,\,k = 1,2, \ldots \) | \({a_{2k + 1}} = \frac{{{{\left( { - 1} \right)}^k}{a_1}}}{{\left( {2k + 1} \right)!}},\,\,\,k = 1,2, \ldots \) | ||
E'tibor bering, har bir qadamda biz har doim oldingi javobni qayta ulab turdik, shunda pastki indeks juft bo'lganda biz har doim \(a_{n}\) ni \(a_{0}\) ko'rinishida yozishimiz mumkin edi va koeffitsient toq bo'lsa, \(a_{n}\) ni \(a_{1}\) hisobida yozishimiz mumkin edi. Shuni ham yodda tutingki, bu holda biz juft koeffitsientli \(a_{n}\) va toq koeffitsientli \(a_{n}\) uchun umumiy formulani topa oldik. Buni qilish har doim ham mumkin bo'lmaydi.
Bu erda yana bir narsaga e'tibor berish kerak. Biz ishlab chiqqan formulalar faqat \(k=1,2,\ldots\) uchun edi, ammo bu holda ular \(k=0\) uchun ham ishlaydi. Shunga qaramay, bu har doim ham ishlamaydigan narsa, lekin bu erda ishlaydi.
Biz \(a_{0}\) va \(a_{1}\) nima ekanligini bilmasligimizdan hayajonlanmang. Ko'rib turganingizdek, to'g'ri echimni olish uchun biz aslida muammoda bo'lishimiz kerak.
Endi bizda \(a_{n}\) uchun formulalar bor, keling, yechim topamiz. Biz qiladigan birinchi narsa, bir nechta \(a_{n}\) ulangan holda yechimni yozishdir.
\[\begin{align*}y\left( x \right) & = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \\ & = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \cdots + {a_{2k}}{x^{2k}} + {a_{2k + 1}}{x^{2k + 1}} + \cdots \\ & = {a_0} + {a_1}x - \frac{{{a_0}}}{{2!}}{x^2} - \frac{{{a_1}}}{{3!}}{x^3} + \cdots + \frac{{{{\left( { - 1} \right)}^k}{a_0}}}{{\left( {2k} \right)!}}{x^{2k}} + \frac{{{{\left( { - 1} \right)}^k}{a_1}}}{{\left( {2k + 1} \right)!}}{x^{2k + 1}} + \cdots \end{align*}\]Keyingi qadam, ulardagi bir xil koeffitsientga ega bo'lgan barcha shartlarni to'plash va keyin bu koeffitsientni hisobga olishdir.
\[\begin{align*}y\left( x \right) & = {a_0}\left\{ {1 - \frac{{{x^2}}}{{2!}} \cdots + \frac{{{{\left( { - 1} \right)}^k}{x^{2k}}}}{{\left( {2k} \right)!}} + \cdots } \right\} + {a_1}\left\{ {x - \frac{{{x^3}}}{{3!}} + \cdots + \frac{{{{\left( { - 1} \right)}^k}}}{{\left( {2k + 1} \right)!}}{x^{2k + 1}} + \cdots } \right\}\\ & = {a_0}\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}{x^{2k}}}}{{\left( {2k} \right)!}}} + {a_1}\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}{x^{2k + 1}}}}{{\left( {2k + 1} \right)!}}} \end{align*}\]Oxirgi bosqichda biz ikkala qismni kuch qatori sifatida yozish uchun umumiy formula nima ekanligini bilganimizdan ham foydalandik. Bu ham bizning yechimimiz. Biz tugatdik.
Boshqa muammoni hal qilishdan oldin, oldingi misolning echimini ko'rib chiqaylik. Birinchidan, biz shaklning ketma-ket yechimini xohlayotganimizni aytishdan boshladik,
\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \]va biz buni tushunmadik. Ikki xil quvvat seriyasini o'z ichiga olgan yechimga ega bo'ldik. Bundan tashqari, har bir yechimda noma'lum konstanta mavjud edi. Bu muammo emas. Aslida, bu biz sodir bo'lishini xohlagan narsadir. Bizdan ish Ikkinchi tartibli doimiy koeffitsientli differensial tenglamalar bilan biz bilamizki, oxirgi misoldagi differensial tenglamaning yechimi:
\[y\left( x \right) = {c_1}\cos \left( x \right) + {c_2}\sin \left( x \right)\]Ikkinchi tartibli differensial tenglamalar yechimlari ikkita alohida funksiyadan iborat bo‘lib, ularning har biri oldida noma’lum doimiysi bo‘lib, har qanday boshlang‘ich shartlarni qo‘llash orqali topiladi. Demak, oxirgi misoldagi yechimimizning shakli aynan biz olishni istagan narsadir. Quyidagi Teylor seriyasini ham eslang,
\[\cos \left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} \hspace{0.25in}\sin \left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} \]Bularni esga olsak, biz ketma-ket hal qilish usulidan olgan narsamiz aynan birinchi tamoyillardan olingan yechim ekanligini ko'ramiz, bundan mustasno, funktsiyalar haqiqiy funktsiyalarning o'rniga haqiqiy funktsiyalar uchun Teylor seriyasi edi.
Keling, doimiy bo'lmagan koeffitsientlarga misol keltiramiz, chunki bu erda ketma-ket echimlar eng foydalidir.
Yechim
Birinchi misolda bo'lgani kabi \(p(x)=1\) va shunga o'xshash bu differentsial tenglama uchun har bir nuqta oddiy nuqtadir. Endi biz buni xuddi birinchi misolda bo'lgani kabi boshlaymiz. Eritma shaklini yozamiz va hosilalarini olamiz.
\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \hspace{0.25in}y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} \]Differensial tenglamaga ulanganda,
\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} - x\sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0\]Birinchi misoldan farqli o'laroq, biz birinchi navbatda barcha koeffitsientlarni ketma-ketlikka ko'chirishimiz kerak.
\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} - \sum\limits_{n = 0}^\infty {{a_n}{x^{n + 1}}} = 0\]Endi ikkala seriyani \(x^{n}n\) nuqtai nazaridan olish uchun birinchi seriyani 2 ga pastga, ikkinchi qatorni esa 1 ga ko‘tarishimiz kerak bo‘ladi.
\[\sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n}} - \sum\limits_{n = 1}^\infty {{a_{n - 1}}{x^n}} = 0\]Keyinchalik, biz \(n\) ning bir xil qiymatidan boshlangan ikkita seriyani olishimiz kerak. Bu muammoni hal qilishning yagona yo'li \(n=0\) atamasini olib tashlashdir.
\[\begin{align*}\left( 2 \right)\left( 1 \right){a_2}{x^0} + \sum\limits_{n = 1}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n}} - \sum\limits_{n = 1}^\infty {{a_{n - 1}}{x^n}} & = 0\\ 2{a_2} + \sum\limits_{n = 1}^\infty {\left[ {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - {a_{n - 1}}} \right]{x^n}} & = 0\end{align*}\]Endi biz barcha koeffitsientlarni nolga tenglashtirishimiz kerak. Biroq, biz bu bilan ehtiyot bo'lishimiz kerak. \(n=0\) koeffitsienti seriyaning oldida va \(n=1,2,3,\dots\) hammasi qatorda. Shunday qilib, koeffitsientni nolga tenglashtirish,
\[\begin{align*} & n = 0 & 2{a_2} & = 0\\ & n = 1,2,3, \ldots & \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - {a_{n - 1}} & = 0\end{align*}\]Birinchi va takrorlanish munosabatini yechish,
\[\begin{align*} & n = 0 & {a_2} & = 0\\ & n = 1,2,3, \ldots & {a_{n + 2}} & = \frac{{{a_{n - 1}}}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\end{align*}\]Endi biz \(n\) qiymatlarini ulashni boshlashimiz kerak.
| \(\displaystyle {a_3} = \frac{{{a_0}}}{{\left( 3 \right)\left( 2 \right)}}\) | \(\displaystyle {a_4} = \frac{{{a_1}}}{{\left( 4 \right)\left( 3 \right)}}\) | \(\displaystyle {a_5} = \frac{{{a_2}}}{{\left( 5 \right)\left( 4 \right)}} = 0\) |
| \(\begin{align*}{a_6} &= \frac{{{a_3}}}{{\left( 6 \right)\left( 5 \right)}}\\ & = \frac{{{a_0}}}{{\left( 6 \right)\left( 5 \right)\left( 3 \right)\left( 2 \right)}}\end{align*}\) | \(\begin{align*}{a_7} & = \frac{{{a_4}}}{{\left( 7 \right)\left( 6 \right)}}\\ & = \frac{{{a_1}}}{{\left( 7 \right)\left( 6 \right)\left( 4 \right)\left( 3 \right)}}\end{align*}\) | \(\displaystyle {a_8} = \frac{{{a_5}}}{{\left( 8 \right)\left( 7 \right)}} = 0\) |
| \( \vdots \) | \( \vdots \) | \( \vdots \) |
| \(\begin{align*}{a_{3k}} & = \frac{{{a_0}}}{{\left( 2 \right)\left( 3 \right)\left( 5 \right)\left( 6 \right) \cdots \left( {3k - 1} \right)\left( {3k} \right)}}\\ k & = 1,2,3, \ldots \end{align*}\) | \(\begin{align*}{a_{3k + 1}} & = \frac{{{a_1}}}{{\left( 3 \right)\left( 4 \right)\left( 6 \right)\left( 7 \right) \cdots \left( {3k} \right)\left( {3k + 1} \right)}}\\ k & = 1,2,3, \ldots \end{align*}\) | \(\begin{align*}{a_{3k + 2}} & = 0\\ k & = 0,1,2, \ldots \end{align*}\) |
Ushbu koeffitsientlarga e'tibor berish kerak bo'lgan bir nechta narsa bor. Birinchidan, har uchinchi koeffitsient nolga teng. Keyinchalik, bu erda formulalar biroz yoqimsiz va birinchi marta ko'rish unchalik oson emas. Nihoyat, bu formulalar birinchi misoldan farqli o'laroq \(k=0\) uchun ishlamaydi.
Endi yechim toping,
\[\begin{align*}y\left( x \right) & = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + {a_4}{x^4} + \cdots + {a_{3k}}{x^{3k}} + {a_{3k + 1}}{x^{3k + 1}} + \cdots \\ & = {a_0} + {a_1}x + \frac{{{a_0}}}{6}{x^3} + \frac{{{a_1}}}{{12}}{x^4} \cdots + \frac{{{a_0}{x^{3k}}}}{{\left( 2 \right)\left( 3 \right)\left( 5 \right)\left( 6 \right) \cdots \left( {3k - 1} \right)\left( {3k} \right)}} + \\ & \hspace{3.25in}\frac{{{a_1}{x^{3k + 1}}}}{{\left( 3 \right)\left( 4 \right)\left( 6 \right)\left( 7 \right) \cdots \left( {3k} \right)\left( {3k + 1} \right)}} + \cdots \end{align*}\]Shunga qaramay, bir xil koeffitsientni o'z ichiga olgan shartlarni to'plang, koeffitsientni ajratib oling va natijalarni yangi qator sifatida yozing,
\[y\left( x \right) = {a_0}\left\{ {1 + \sum\limits_{k = 1}^\infty {\frac{{{x^{3k}}}}{{\left( 2 \right)\left( 3 \right)\left( 5 \right)\left( 6 \right) \cdots \left( {3k - 1} \right)\left( {3k} \right)}}} } \right\} + {a_1}\left\{ {x + \sum\limits_{k = 1}^\infty {\frac{{{x^{3k + 1}}}}{{\left( 3 \right)\left( 4 \right)\left( 6 \right)\left( 7 \right) \cdots \left( {3k} \right)\left( {3k + 1} \right)}}} } \right\}\]Bu safar biz seriyamizni \(k=0\) da boshlay olmadik, chunki umumiy atama \(k=0\) uchun mos emas.
Endi biz \(x=0\) dan boshqa nuqtani ishlatadigan misol bilan ishlashimiz kerak. Aslida, keling, avvalgi misolni olaylik va \(x_{0}\) ning boshqa qiymati uchun uni qayta ishlaymiz. Shuningdek, ushbu misol uchun ko'rsatmalarni biroz o'zgartirishimiz kerak bo'ladi.
Yechim
Afsuski, biz uchun birinchi misoldan bu erda qayta foydalanish mumkin bo'lgan hech narsa yo'q. \({x_0} = - 2\) ga o'zgartirish muammoni butunlay o'zgartiradi. Bu holda bizning yechimimiz bo'ladi,
\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} \]Eritmaning hosilalari,
\[y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{{\left( {x + 2} \right)}^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} \]Bularni differentsial tenglamaga ulang.
\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - x\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} = 0\]Endi biz ushbu misol va oldingi misol o'rtasidagi birinchi haqiqiy farqga duch kelamiz. Bu holda biz \(x\) ni ikkinchi qatorga ko'paytirib bo'lmaydi, chunki seriya bilan birlashish uchun u \(x+2\) bo'lishi kerak. Shuning uchun, birinchi navbatda, ikkinchi qatorning koeffitsientini qatorga ko'paytirishdan oldin o'zgartirishimiz kerak bo'ladi.
\[\begin{align*}\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - \left( {x + 2 - 2} \right)\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} & = 0\\ \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - \left( {x + 2} \right)\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} + 2\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} & = 0\\ \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^{n + 1}}} + \sum\limits_{n = 0}^\infty {2{a_n}{{\left( {x + 2} \right)}^n}} & = 0\end{align*}\]Endi bizda ishlash uchun uchta seriya bor. Bunday muammolar ko'pincha sodir bo'ladi. Endi biz birinchi qatorni 2 ga pastga, ikkinchi qatorni esa 1 ga yuqoriga siljitishimiz kerak, buning uchun barcha seriyalardagi umumiy ko'rsatkichlar olinadi.
\[\sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{{\left( {x + 2} \right)}^n}} - \sum\limits_{n = 1}^\infty {{a_{n - 1}}{{\left( {x + 2} \right)}^n}} + \sum\limits_{n = 0}^\infty {2{a_n}{{\left( {x + 2} \right)}^n}} = 0\]Seriyalarni birlashtirish uchun biz birinchi va uchinchi seriyalardan \(n=0\) atamalarini olib tashlashimiz kerak.
\[\begin{align*}2{a_2} + 2{a_0} + \sum\limits_{n = 1}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{{\left( {x + 2} \right)}^n}} - \sum\limits_{n = 1}^\infty {{a_{n - 1}}{{\left( {x + 2} \right)}^n}} + \sum\limits_{n = 1}^\infty {2{a_n}{{\left( {x + 2} \right)}^n}} & = 0\\ 2{a_2} + 2{a_0} + \sum\limits_{n = 1}^\infty {\left[ {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - {a_{n - 1}} + 2{a_n}} \right]{{\left( {x + 2} \right)}^n}} & = 0\end{align*}\]Nolga teng koeffitsientlarni o'rnatish,
\[\begin{align*} & n = 0 & 2{a_2} + 2{a_0} & = 0\\ & n = 1,2,3, \ldots & \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - {a_{n - 1}} + 2{a_n} & = 0\end{align*}\]Endi biz ikkalasini ham hal qilishimiz kerak. Birinchi holda ikkita variant mavjud, biz \(a_{2}\) uchun hal qilishimiz mumkin yoki \(a_{0}\) uchun hal qilishimiz mumkin. Odatdagidan tashqari, men \(a_{0}\) uchun hal qilaman. Takrorlanish munosabatida biz oldingi misollarda bo'lgani kabi eng katta pastki belgisi bilan atama uchun hal qilamiz.
\[\begin{align*} & n = 0 & {a_2} & = - {a_0}\\ & n = 1,2,3, \ldots & {a_{n + 2}} & = \frac{{{a_{n - 1}} - 2{a_n}}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\end{align*}\]E'tibor bering, bu misolda biz oldingi misolda bo'lgani kabi har uchinchi davrni ham tashlab ketmaymiz.
Shu nuqtada, biz ushbu muammo bo'yicha ko'rsatmalar ham boshqacha ekanligini tan olamiz. Bu safar biz \(a_{n}\) uchun umumiy formulani olmaymiz, shuning uchun biz yechimning har bir qismi uchun birinchi ikki shartni olish bilan kifoyalanishimiz kerak. Bu ko'pincha ketma-ket echimlarga tegishli. \(a_{n}\) uchun umumiy formulalarni olish bu turdagi masalalarda qoida emas, istisno hisoblanadi.
Dastlabki to'rtta atamani olish uchun biz kerakli miqdordagi atamalarga ega bo'lgunimizcha shartlarni qo'shishni boshlaymiz. E'tibor bering, biz yechimning dastlabki ikki shartidan \(a_{0}\) va \(a_{1}\) bilan boshlaymiz, shuning uchun bizga \(a_{0}\) bilan yana uchta atama va \(a_{1}\) bilan yana uchta atama kerak bo'ladi.
\[n = 0\hspace{0.25in}{a_2} = - {a_0}\]Bizda ikkita \(a_{0}\) va bitta \(a_{1}\) bor.
\[n = 1\hspace{0.25in}{a_3} = \frac{{{a_0} - 2{a_1}}}{{\left( 3 \right)\left( 2 \right)}} = \frac{{{a_0}}}{6} - \frac{{{a_1}}}{3}\]Bizda uchta \(a_{0}\) va ikkita \(a_{1}\) bor.
\[n = 2\hspace{0.25in}{a_4} = \frac{{{a_1} - 2{a_2}}}{{\left( 4 \right)\left( 3 \right)}} = \frac{{{a_1} - 2\left( { - {a_0}} \right)}}{{\left( 4 \right)\left( 3 \right)}} = \frac{{{a_0}}}{6} + \frac{{{a_1}}}{{12}}\]Bizda to‘rtta \(a_{0}\) va uchta \(a_{1}\) bor. Bizda barcha kerakli \(a_{0}\) bor, lekin yana bitta \(a_{1}\) kerak. Shunday qilib, biz shunga o'xshash yana bir atama qilishimiz kerak.
\[n = 3\hspace{0.25in}{a_5} = \frac{{{a_2} - 2{a_3}}}{{\left( 5 \right)\left( 4 \right)}} = - \frac{{{a_0}}}{{20}} - \frac{1}{{10}}\left( {\frac{{{a_0}}}{6} - \frac{{{a_1}}}{3}} \right) = - \frac{{{a_0}}}{{15}} + \frac{{{a_1}}}{{30}}\]Bizda beshta \(a_{0}\) va to‘rtta \(a_{1}\) bor. Bizda barcha kerakli shartlar mavjud.
Endi biz qilishimiz kerak bo'lgan yagona narsa bizning yechimimizga ulanishdir.
\[\begin{align*}y\left( x \right) & = \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} \\ & = {a_0} + {a_1}\left( {x + 2} \right) + {a_2}{\left( {x + 2} \right)^2} + {a_3}{\left( {x + 2} \right)^3} + {a_4}{\left( {x + 2} \right)^4} + {a_5}{\left( {x + 2} \right)^5} + \cdots \\ & = {a_0} + {a_1}\left( {x + 2} \right) - {a_0}{\left( {x + 2} \right)^2} + \left( {\frac{{{a_0}}}{6} - \frac{{{a_1}}}{3}} \right){\left( {x + 2} \right)^3} + \\ & \hspace{0.25in}\left( {\frac{{{a_0}}}{6} + \frac{{{a_1}}}{{12}}} \right){\left( {x + 2} \right)^4} + \left( { - \frac{{{a_0}}}{{15}} + \frac{{{a_1}}}{{30}}} \right){\left( {x + 2} \right)^5} + \cdots \end{align*}\]Nihoyat, barcha shartlarni bir xil koeffitsient bilan to'plang va olish uchun koeffitsientni hisoblang,
\[\begin{align*}y\left( x \right) & = {a_0}\left\{ {1 - {{\left( {x + 2} \right)}^2} + \frac{1}{6}{{\left( {x + 2} \right)}^3} + \frac{1}{6}{{\left( {x + 2} \right)}^4} - \frac{1}{{15}}{{\left( {x + 2} \right)}^5} + \cdots } \right\} + \\ & \,\,\,{a_1}\left\{ {\left( {x + 2} \right) - \frac{1}{3}{{\left( {x + 2} \right)}^3} + \frac{1}{{12}}{{\left( {x + 2} \right)}^4} + \frac{1}{{30}}{{\left( {x + 2} \right)}^5} + \cdots } \right\}\end{align*}\]Bizga nisbatan bu muammoning yechimi. E'tibor bering, bu yechim avvalgi misoldagi yechimga o'xshamaydi. Bu bir xil differentsial tenglama, ammo \(x_{0}\) ni o'zgartirish yechimni butunlay o'zgartirdi.
Keling, bitta yakuniy muammoni hal qilaylik.
Yechim
Biz nihoyat ikkinchi hosila uchun doimiy koeffitsientga ega bo'lmagan differentsial tenglamaga egamiz.
\[p\left( x \right) = {x^2} + 1\hspace{0.25in}p\left( 0 \right) = 1 \ne 0\]Shunday qilib, \({x_0} = 0\) bu differentsial tenglama uchun oddiy nuqtadir. Bizga birinchi navbatda yechim va uning hosilalari kerak,
\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \hspace{0.25in}y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} \]Bularni differentsial tenglamaga ulang.
\[\left( {{x^2} + 1} \right)\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} - 4x\sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} + 6\sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0\]Endi birinchi hadni ikkiga bo'ling, shunda biz koeffitsientni qatorga ko'paytiramiz va ikkinchi va uchinchi qatorlarning koeffitsientlarini ham ko'paytiramiz.
\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^n}} + \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} - \sum\limits_{n = 1}^\infty {4n{a_n}{x^n}} + \sum\limits_{n = 0}^\infty {6{a_n}{x^n}} = 0\]Barcha ketma-ketlikdagi barcha ko'rsatkichlar bir xil bo'lishi uchun biz ikkinchi qatorni ikkiga pastga siljitishimiz kerak bo'ladi.
\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^n}} + \sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n}} - \sum\limits_{n = 1}^\infty {4n{a_n}{x^n}} + \sum\limits_{n = 0}^\infty {6{a_n}{x^n}} = 0\]Shu nuqtada biz \(n=2\) da boshlangan barcha seriyalarni olish uchun ba'zi shartlarni olib tashlashimiz mumkin, ammo bu aslida kerak bo'lgandan ko'ra ko'proq ish. Buning o'rniga shuni ta'kidlaymizki, agar biz xohlasak, \(n=0\) da uchinchi seriyani boshlashimiz mumkin, chunki bu atama nolga teng. Xuddi shunday, birinchi seriyadagi atamalar \(n=1\) va \(n=0\) uchun nolga teng va shuning uchun biz ushbu seriyani \(n=0\) da boshlashimiz mumkin. Agar biz buni qilsak, barcha seriyalar endi \(n=0\) da boshlanadi va biz ularni har qanday seriyadan shartlarni ajratmasdan qo'shishimiz mumkin.
\[\begin{align*}\sum\limits_{n = 0}^\infty {\left[ {n\left( {n - 1} \right){a_n} + \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - 4n{a_n} + 6{a_n}} \right]{x^n}} & = 0\\ \sum\limits_{n = 0}^\infty {\left[ {\left( {{n^2} - 5n + 6} \right){a_n} + \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}} \right]{x^n}} & = 0\\ \sum\limits_{n = 0}^\infty {\left[ {\left( {n - 2} \right)\left( {n - 3} \right){a_n} + \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}} \right]{x^n}} & = 0\end{align*}\]Endi koeffitsientlarni nolga tenglashtiring.
\[\left( {n - 2} \right)\left( {n - 3} \right){a_n} + \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} = 0,\hspace{0.25in}n = 0,1,2, \ldots \]Buni hal qilish,
\[{a_{n + 2}} = - \frac{{\left( {n - 2} \right)\left( {n - 3} \right){a_n}}}{{\left( {n + 2} \right)\left( {n + 1} \right)}},\hspace{0.25in}n = 0,1,2, \ldots \]Endi biz \(n\) qiymatlarini kiritamiz.
\[n = 0:\hspace{0.25in}{a_2} = - 3{a_0}\] \[n = 1:\hspace{0.25in}{a_3} = - \frac{1}{3}{a_1}\] \[n = 2:\hspace{0.25in}{a_4} = - \frac{0}{{12}}{a_2} = 0\] \[n = 3:\hspace{0.25in}{a_5} = - \frac{0}{{20}}{a_3} = 0\]Endi, shu nuqtadan boshlab barcha koeffitsientlar nolga teng. Bunday holda, eritmadagi ikkala seriya ham tugaydi. Bu har doim ham sodir bo'lmaydi va ko'pincha ulardan faqat bittasi tugaydi.
Bu holatda yechim,
\[y\left( x \right) = {a_0}\left\{ {1 - 3{x^2}} \right\} + {a_1}\left\{ {x - \frac{1}{3}{x^3}} \right\}\]