6.4-bo'lim: Eyler tenglamalari
Ushbu bo'limda biz echimlarni izlamoqchimiz
\[\begin{equation}a{x^2}y'' + bxy' + cy = 0\label{eq:eq1}\end{equation}\]\({x_0} = 0\) atrofida. Bunday turdagi differentsial tenglamalar deyiladi Eyler tenglamalari.
Oldingisidan eslang bo'limi nuqta oddiy nuqtadir, agar bo'laklar bo'lsa,
\[\frac{{bx}}{{a{x^2}}} = \frac{b}{{ax}}\hspace{0.25in}{\mbox{and }}\hspace{0.25in}\frac{c}{{a{x^2}}}\]\({x_0} = 0\) atrofida Teylor seriyasiga ega. Biroq, maxrajdagi \(x\) tufayli ularning hech biri \({x_0} = 0\) atrofida Teylor qatoriga ega bo'lmaydi va shuning uchun \({x_0} = 0\) yagona nuqtadir. Shunday qilib, oldingi bo'limdagi usul ishlamaydi, chunki u oddiy nuqtani talab qiladi.
Biroq, bu differentsial tenglamaning ketma-ket echimlari bo'lmagan echimlarini olish mumkin. Keling, \(x>0\) (buning sababi birinchi misolda ishlaganimizdan keyin ma'lum bo'ladi) va barcha echimlar shaklda, deb faraz qilishdan boshlaylik,
\[\begin{equation}y\left( x \right) = {x^r}\label{eq:eq2}\end{equation}\]Endi buni olish uchun differentsial tenglamaga ulang,
\[\begin{align*}a{x^2}\left( r \right)\left( {r - 1} \right){x^{r - 2}} + bx\left( r \right){x^{r - 1}} + c{x^r} & = 0\\ ar\left( {r - 1} \right){x^r} + b\left( r \right){x^r} + c{x^r} & = 0\\ \left( {ar\left( {r - 1} \right) + b\left( r \right) + c} \right){x^r} & = 0\end{align*}\]Endi biz \(x>0\) deb taxmin qildik va shuning uchun bu faqat nolga teng bo'ladi, agar:
\[\begin{equation}ar\left( {r - 1} \right) + b\left( r \right) + c = 0\label{eq:eq3}\end{equation}\]Shunday qilib, echimlar \(\eqref{eq:eq2}\) shaklida bo'ladi, agar \(r\) \(\eqref{eq:eq3}\) uchun yechim bo'lsa. Bu tenglama \(r\) da kvadratdir va shuning uchun bizda uchta holat ko'rib chiqiladi: haqiqiy, aniq ildizlar, qo'sh ildizlar va murakkab ildizlar.
Haqiqiy, aniq ildizlar
Bu holatda, albatta, ko'p narsa qilish kerak emas. Biz a hosil qiladigan ikkita yechimni olamiz asosiy echimlar to'plami (buni tekshirishni sizga qoldiramiz) va bizning umumiy yechimimiz shunday bo'ladi:
\[y\left( x \right) = {c_1}{x^{{r_1}}} + {c_2}{x^{{r_2}}}\]Yechim
Avval \(\eqref{eq:eq3}\) ning ildizlarini topishimiz kerak.
\[\begin{align*}2r\left( {r - 1} \right) + 3r - 15 & = 0\\ 2{r^2} + r - 15 = \left( {2r - 5} \right)\left( {r + 3} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_1} = \frac{5}{2},\,\,\,{r_2} = - 3\end{align*}\]Umumiy yechim shunday bo'ladi:
\[y\left( x \right) = {c_1}{x^{\frac{5}{2}}} + {c_2}{x^{ - 3}}\]Konstantalarni topish uchun biz ikkinchi tartibli differensial tenglamalar bobida bo'lgani kabi, biz farqlaymiz va boshlang'ich shartlarni kiritamiz.
\[y'\left( x \right) = \frac{5}{2}{c_1}{x^{\frac{3}{2}}} - 3{c_2}{x^{ - 4}}\] \[\left. \begin{aligned}& 0 = y\left( 1 \right) = {c_1} + {c_2}\\ & 1 = y'\left( 1 \right) = \frac{5}{2}{c_1} - 3{c_2}\end{aligned} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = \frac{2}{{11}},\,\,{c_2} = - \frac{2}{{11}}\]Haqiqiy yechim shundan iboratki,
\[y\left( x \right) = \frac{2}{{11}}{x^{\frac{5}{2}}} - \frac{2}{{11}}{x^{ - 3}}\]Ushbu misolning yechimi bilan biz endi nima uchun \(x>0\) talab qilinganligini bilib olamiz. Agar biz \(x=0\) ga ruxsat bergan bo‘lsak, ikkinchi a’zo nolga bo‘linadi va agar biz \(x<0\) ga ruxsat bergan bo‘lsak, birinchi a’zo salbiy sonlarning kvadrat ildizlarini beradi.
Ikkita ildiz
Bu holat har safar qo'sh ildizlarga (yoki qo'sh o'ziga xos qiymatlarga) duch kelganimiz kabi muammoga olib keladi. Biz faqat bitta yechimni olamiz va ikkinchi yechim kerak bo'ladi. Bu holda ikkinchi yechim bo'lishini ko'rsatish mumkin,
\[{y_2}\left( x \right) = {x^r}\ln x\]va shuning uchun bu holatda umumiy yechim:
\[y\left( x \right) = {c_1}{x^r} + {c_2}{x^r}\ln x = {x^r}\left( {{c_1} + {c_2}\ln x} \right)\]Biz \(x>0\) ni talab qilish sababini yana ko'rishimiz mumkin. Agar shunday bo'lmasa, bizda bu logarifm bilan bog'liq har xil muammolar bo'lar edi.
Yechim
Avval \(\eqref{eq:eq3}\) ning ildizlari.
\[\begin{align*}r\left( {r - 1} \right) - 7r + 16 & = 0\\ {r^2} - 8r + 16 & = 0\\ {\left( {r - 4} \right)^2} & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}r = 4\end{align*}\]Shunday qilib, umumiy yechim shunday bo'ladi:
\[y\left( x \right) = {c_1}{x^4} + {c_2}{x^4}\ln x\]Murakkab ildizlar
Bunday holda, biz ildizlarimiz shaklda, deb taxmin qilamiz,
\[{r_{1,2}} = \lambda \pm \mu \,i\]Agar biz birinchi ildizni olsak, quyidagi yechimni olamiz.
\[{x^{\lambda + \mu \,i}}\]Bu muammo, chunki biz murakkab echimlarni xohlamaymiz, biz faqat haqiqiy echimlarni xohlaymiz. Buni eslash orqali bartaraf etishimiz mumkin,
\[{x^r} = {{\bf{e}}^{\ln {x^r}}} = {{\bf{e}}^{r\ln x}}\]Bunga ildizni ulash,
\[\begin{align*}{x^{\lambda + \mu \,i}} & = {{\bf{e}}^{\left( {\lambda + \mu \,i} \right)\ln x}}\\ & = {{\bf{e}}^{\lambda \ln x}}{{\bf{e}}^{\mu \,i\ln x}}\\ & = {{\bf{e}}^{\ln {x^\lambda }}}\left( {\cos \left( {\mu \ln x} \right) + i\sin \left( {\mu \ln x} \right)} \right)\\ & = {x^\lambda }\cos \left( {\mu \ln x} \right) + i{x^\lambda }\sin \left( {\mu \ln x} \right)\end{align*}\]E'tibor bering, biz foydalanishimiz kerak edi Eyler formulasi shuningdek, oxirgi bosqichga o'tish uchun. Endi, biz har safar qilganimizdek, shunga o'xshash echimlarni ko'rganmiz, biz haqiqiy qismni va xayoliy qismni olamiz va ulardan ikkita yechimimiz uchun foydalanishimiz mumkin.
Shunday qilib, murakkab ildizlar holatida umumiy yechim bo'ladi,
\[y\left( x \right) = {c_1}{x^\lambda }\cos \left( {\mu \ln x} \right) + {c_2}{x^\lambda }\sin \left( {\mu \ln x} \right) = {x^\lambda }\left( {{c_1}\cos \left( {\mu \ln x} \right) + {c_2}\sin \left( {\mu \ln x} \right)} \right)\]Yana bir bor, nima uchun biz \(x > 0\) ni talab qilishimiz kerakligini bilib olamiz.
Yechim
Avval har doimgidek \(\eqref{eq:eq3}\) ning ildizlarini oling.
\[\begin{align*}r\left( {r - 1} \right) + 3r + 4 & = 0\\ {r^2} + 2r + 4 & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_{1,2}} = - 1 \pm \sqrt 3 \,i\end{align*}\]Umumiy yechim shunday bo'ladi:
\[y\left( x \right) = {c_1}{x^{ - 1}}\cos \left( {\sqrt 3 \ln x} \right) + {c_2}{x^{ - 1}}\sin \left( {\sqrt 3 \ln x} \right)\]Endi \(x < 0\) bilan qanday kurashish kerakligi haqida gapirishimiz kerak, chunki bu ba'zida mumkin. Buni hal qilish uchun biz o'zgaruvchan transformatsiyadan foydalanishimiz kerak,
\[\eta = - x\]Bu holda \(x < 0\) dan beri biz \(\eta > 0\) ni olamiz. Endi aniqlang,
\[u\left( \eta \right) = y\left( x \right) = y\left( { - \eta } \right)\]Keyin zanjir qoidasidan foydalanib, biz buni ko'rishimiz mumkin,
\[u'\left( \eta \right) = - y'\left( x \right)\hspace{0.25in}\text{va}\hspace{0.25in}u''\left( \eta \right) = y''\left( x \right)\]Ushbu transformatsiya bilan differensial tenglama quyidagicha bo'ladi:
\[\begin{align*}a{\left( { - \eta } \right)^2}u'' + b\left( { - \eta } \right)\left( { - u'} \right) + cu & = 0\\ a{\eta ^2}u'' + b\eta u' + cu & = 0\end{align*}\]Boshqacha qilib aytganda, \(\eta>0\) dan boshlab biz ushbu differentsial tenglamaning yechimlarini olish uchun yuqoridagi ishdan foydalanishimiz mumkin. Shuningdek, biz \(x\)-ga o'zgaruvchini teskari o'zgartirish orqali qaytamiz.
\[\eta = - x\]Keling, nima bo'lishini ko'rish uchun avvalo haqiqiy, aniq ishni olaylik.
\[\begin{align*}u\left( \eta \right) & = {c_1}{\eta ^{{r_1}}} + {c_2}{\eta ^{{r_2}}}\\ y\left( x \right) & = {c_1}{\left( { - x} \right)^{{r_1}}} + {c_2}{\left( { - x} \right)^{{r_2}}}\end{align*}\]Endi, agar xohlasak, qolgan hollarda buni qilishimiz mumkin edi, lekin buni amalga oshirishdan oldin, shuni ta'kidlaymizki, agar mutlaq qiymat ta'rifini eslasak,
\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.\]biz bu ishning ikkala yechimimizni bittaga birlashtirib, yechimni quyidagicha yozishimiz mumkin:
\[y\left( x \right) = {c_1}{\left| x \right|^{{r_1}}} + {c_2}{\left| x \right|^{{r_2}}},\hspace{0.25in}x \ne 0\]E'tibor bering, biz hali ham \(x = 0\) dan qochishimiz kerak, chunki biz hali ham nolga bo'linishimiz mumkin. Biroq, bu endi \(x = 0\) ni o'z ichiga olmagan har qanday interval uchun yechimdir.
Biz boshqa ikkita holat va \(x = 0\) ni o'z ichiga olmagan har qanday interval uchun quyidagi echimlar uchun ham shunday qilishimiz mumkin.
\[\begin{align*}y\left( x \right) & = {c_1}{\left| x \right|^r} + {c_2}{\left| x \right|^r}\ln \left| x \right|\\ y\left( x \right) & = {c_1}{\left| x \right|^\lambda }\cos \left( {\mu \ln \left| x \right|} \right) + {c_2}{\left| x \right|^\lambda }\sin \left( {\mu \ln \left| x \right|} \right)\end{align*}\]Yana bir misol bilan ishlashdan oldin yana bir umumlashma qilishimiz mumkin. Eyler tenglamasining umumiy shakli:
\[a{\left( {x - {x_0}} \right)^2}y'' + b\left( {x - {x_0}} \right)y' + cy = 0\]va biz \(x = {x_0}\) ni o'z ichiga olmagan har qanday oraliqda echimlarni so'rashimiz mumkin. Bu holda yechimlarni yaratish bo'yicha ish yuqoridagi barcha ishlar bilan bir xil va shuning uchun bu erda ko'rsatilmagan.
Ushbu umumiy holatda \(x = a\) ni o'z ichiga olmagan har qanday interval uchun echimlar:
\[\begin{align*}y\left( x \right) & = {c_1}{\left| {x - a} \right|^{{r_1}}} + {c_2}{\left| {x - a} \right|^{{r_2}}}\\ y\left( x \right) & = {\left| {x - a} \right|^r}\left( {{c_1} + {c_2}\ln \left| {x - a} \right|} \right)\\ y\left( x \right) & = {\left| {x - a} \right|^\lambda }\left( {{c_1}\cos \left( {\mu \ln \left| {x - a} \right|} \right) + {c_2}\sin \left( {\mu \ln \left| {x - a} \right|} \right)} \right)\end{align*}\]Ildizlar yechim bo'lgan joy
\[ar\left( {r - 1} \right) + b\left( r \right) + c = 0\]Yechim
4-misol \(x = - 6\) ni o‘z ichiga olmagan istalgan oraliqda quyidagi differensial tenglamaning yechimini toping.
\[3{\left( {x + 6} \right)^2}y'' + 25\left( {x + 6} \right)y' - 16y = 0\]Yechim
Shunday qilib, biz bu holda bir xil kvadratikdan ildizlarni olamiz.
\[\begin{align*}3r\left( {r - 1} \right) + 25r - 16 & = 0\\ 3{r^2} + 22r - 16 & = 0\\ \left( {3r - 2} \right)\left( {r + 8} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_1} = \frac{2}{3},\,\,{r_2} = - 8\end{align*}\]Umumiy yechim shunday bo'ladi:
\[y\left( x \right) = {c_1}{\left| {x + 6} \right|^{\frac{2}{3}}} + {c_2}{\left| {x + 6} \right|^{ - 8}}\]