3.10-bo'lim: Parametrlarning o'zgarishi
Oxirgi bo'limda biz aniq echimni topish uchun aniqlanmagan koeffitsientlar usulini ko'rib chiqdik
\[\begin{equation}p\left( t \right)y'' + q\left( t \right)y' + r\left( t \right)y = g\left( t \right)\label{eq:eq1}\end{equation}\]Va biz ko'rdikki, u narsalarni faqat algebra muammosiga qisqartirgan bo'lsa-da, algebra juda tartibsiz bo'lib qolishi mumkin. Buning ustiga, aniqlanmagan koeffitsientlar faqat juda kichik funktsiyalar sinfi uchun ishlaydi.
Parametrlarni o'zgartirish usuli ancha umumiy usul bo'lib, undan ko'p hollarda foydalanish mumkin. Biroq, usulning ikkita kamchiliklari mavjud. Birinchidan, muammoni hal qilish uchun qo'shimcha echim mutlaqo talab qilinadi. Bu aniqlanmagan koeffitsientlar usulidan farqli o'laroq, qo'shimcha echimga ega bo'lish tavsiya etilgan, ammo talab qilinmagan. Ikkinchidan, biz ko'rib turganimizdek, usulni yakunlash uchun biz bir nechta integrallarni bajaramiz va biz integrallarni bajara olishimizga kafolat yo'q. Shunday qilib, ma'lum bir yechimni olish uchun formulani yozish har doim mumkin bo'lsa-da, agar integrallar juda qiyin bo'lsa yoki to'ldiruvchi yechimni topa olmasak, biz uni topa olmasligimiz mumkin.
Biz parametrlarni o'zgartirish formulasini olamiz. Biz \(\eqref{eq:eq1}\) ga qo'shimcha yechim ekanligini tan olishdan boshlaymiz.
\[{y_c}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)\]Shuni ham yodda tutingki, bu bir hil differentsial tenglamaning umumiy yechimi.
\[\begin{equation}p\left( t \right)y'' + q\left( t \right)y' + r\left( t \right)y = 0\label{eq:eq2}\end{equation}\]Shuni ham yodda tutingki, qo'shimcha echimni yozish uchun biz \(y_{1}(t)\) va \(y_{2}(t)\) asosiy echimlar to'plami ekanligini bilamiz.
Biz nima qilmoqchi bo'lsak, biz \(u_{1}(t)\) va \(u_{2}(t)\) juftligini topa olamizmi yoki yo'qligini ko'rishimiz kerak.
\[{Y_P}\left( t \right) = u_{1}\left( t \right){y_1}\left( t \right) + u_{2}\left( t \right){y_2}\left( t \right)\]\(\eqref{eq:eq1}\) uchun yechim bo'ladi. Bu erda ikkita noma'lum narsa bor va shuning uchun bizga oxir-oqibat ikkita tenglama kerak bo'ladi. Bitta tenglama oson. Bizning taklif qilayotgan yechimimiz differentsial tenglamani qondirishi kerak, shuning uchun biz taklif qilingan yechimimizni \(\eqref{eq:eq1}\) ga ulab birinchi tenglamani olamiz. Ikkinchi tenglama kelishi mumkin turli joylardan. Biz ikkinchi tenglamamizni ishimizni osonlashtiradigan taxmin qilish orqali olamiz. Tez orada bu haqda ko'proq gapiramiz.
Shunday qilib, boshlaylik. Agar biz taklif qilingan yechimimizni differentsial tenglamaga qo'shmoqchi bo'lsak, bizga ba'zi hosilalar kerak bo'ladi, shuning uchun ularni olamiz. Birinchi hosila
\[{Y'_P}\left( t \right) = {u'_1}{y_1} + {u_1}{y'_1} + {u'_2}{y_2} + {u_2}{y'_2}\]Mana bu taxmin. Shunchaki, birinchi hosila bilan ishlashni osonlashtirish uchun biz \(u_{1}(t)\) va \(u_{2}(t)\) nima bo'lishidan qat'i nazar, ular quyidagilarga javob beradi deb taxmin qilamiz.
\[\begin{equation}{u'_1}{y_1} + {u'_2}{y_2} = 0\label{eq:eq3}\end{equation}\]Endi buni amalga oshirish mumkinligiga ishonish uchun hech qanday sabab yo'q. Biroq, buning natijasi bo'lishini ko'ramiz. Biz shunchaki bu taxminni yo'lda muammolarga olib kelmasligi va birinchi hosilani osonroq qilish uchun umid qilamiz, shuning uchun bundan hayajonlanmang.
Ushbu faraz bilan birinchi hosila bo'ladi.
\[{Y'_P}\left( t \right) = {u_1}{y'_1} + {u_2}{y'_2}\]Ikkinchi hosila esa,
\[{Y''_P}\left( t \right) = {u'_1}{y'_1} + {u_1}{y''_1} + {u'_2}{y'_2} + {u_2}{y''_2}\]Eritma va uning hosilalarini \(\eqref{eq:eq1}\) ga ulang.
\[p\left( t \right)\left( {{u'_1}{y'_1} + {u_1}{y''_1} + {u'_2}{y'_2} + {u_2}{y''_2}} \right) + q\left( t \right)\left( {{u_1}{y'_1} + {u_2}{y'_2}} \right) + r\left( t \right)\left( {u{_1}{y_1} + u{_2}{y_2}} \right) = g\left( t \right)\]Bir oz qayta tartibga solish quyidagilarni beradi.
\[\begin{align*}& p\left( t \right)\left( {{u'_1}{y'_1} + {u'_2}{y'_2}} \right) + {u_1}\left( t \right)\left( {p\left( t \right){y''_1} + q\left( t \right){y'_1} + r\left( t \right){y_1}} \right) + \\ & \hspace{3.0in}{u_2}\left( t \right)\left( {p\left( t \right){y''_2} + q\left( t \right){y'_2} + r\left( t \right){y_2}} \right) = g\left( t \right)\end{align*}\]Endi \(y_{1}(t)\) ham, \(y_{2}(t)\) ham \(\eqref{eq:eq2}\) ning yechimlari va shuning uchun ikkinchi va uchinchi hadlar nolga teng. Buni tan olish va biroz tartibga solish bizga beradi,
\[p\left( t \right)\left( {{u'_1}{y'_1} + {u'_2}{y'_2}} \right) + {u_1}\left( t \right)\left( 0 \right) + {u_2}\left( t \right)\left( 0 \right) = g\left( t \right)\] \[\begin{equation}{u'_1}{y'_1} + {u'_2}{y'_2} = \frac{{g\left( t \right)}}{{p\left( t \right)}}\label{eq:eq4}\end{equation}\]Biz kerakli ikkita tenglamani deyarli oldik. Davom etishdan oldin biz orqaga qaytib, yana bir taxmin qilamiz. Oxirgi tenglama, \(\eqref{eq:eq4}\), aslida biz xohlagan tenglamadir, ammo biz uchun narsalarni soddalashtirish uchun biz funktsiyani nazarda tutamiz.
\(p(t) = 1\).
Boshqacha qilib aytganda, biz orqaga qaytib, differentsial tenglama bilan ishlashni boshlaymiz,
\[y'' + q\left( t \right)y' + r\left( t \right)y = g\left( t \right)\]Agar ikkinchi hosilaning koeffitsienti bitta bo'lmasa, uni bitta bo'lishi uchun ajrating. Biz oladigan formula buni nazarda tutadi! Buni qilgandan so'ng, biz noma'lum funktsiyalar uchun echmoqchi bo'lgan ikkita tenglama
\[\begin{equation}{u'_1}{y_1} + {u'_2}{y_2} = 0\label{eq:eq5}\end{equation}\] \[\begin{equation}{u'_1}{y'_1} + {u'_2}{y'_2} = g\left( t \right)\label{eq:eq6}\end{equation}\]E'tibor bering, ushbu tizimda biz ikkita echimni bilamiz va shuning uchun bu erda faqat ikkita noma'lum \({u'_1}\) va \({u'_2}\). Ushbu tizimni hal qilish aslida juda oddiy. Birinchidan, \(\eqref{eq:eq5}\)-ni \({u'_1}\) uchun hal qiling va uni \(\eqref{eq:eq6}\)-ga ulang va biroz soddalashtiring.
\[\begin{equation}{u'_1} = - \frac{{{u'_2}{y_2}}}{{{y_1}}}\label{eq:eq7}\end{equation}\] \[\begin{align*}\left( { - \frac{{{u'_2}{y_2}}}{{{y_1}}}} \right){y'_1} + {u'_2}{y'_2} & = g\left( t \right)\\ {u'_2}\left( {{y'_2} - \frac{{{y_2}{y'_1}}}{{{y_1}}}} \right) & = g\left( t \right)\\ {u'_2}\left( {\frac{{{y_1}{y'_2} - {y_2}{y'_1}}}{{{y_1}}}} \right) & = g\left( t \right)\end{align*}\] \[\begin{equation}{u'_2} = \frac{{{y_1}g\left( t \right)}}{{{y_1}{y'_2} - {y_2}{y'_1}}}\label{eq:eq8}\end{equation}\]Shunday qilib, bizda \({u'_2}\) ifodasi mavjud. Buni \(\eqref{eq:eq7}\) ga ulash bizga \({u'_1}\) ifodasini beradi.
\[\begin{equation}{u'_1} = - \frac{{{y_2}g\left( t \right)}}{{{y_1}{y'_2} - {y_2}{y'_1}}}\label{eq:eq9}\end{equation}\]Keyin, keling, bunga e'tibor beraylik
\[W\left( {{y_1},{y_2}} \right) = {y_1}{y'_2} - {y_2}{y'_1} \ne 0\]Eslatib o'tamiz, \(y_{1}(t)\) va \(y_{2}(t)\) asosiy echimlar to'plamidir va shuning uchun biz Wronskian nolga teng bo'lmasligini bilamiz!
Va nihoyat, \(u_{1}(t)\) va \(u_{2}(t)\) nima ekanligini aniqlash uchun biz \(\eqref{eq:eq8}\) va \(\eqref{eq:eq9}\) ni birlashtirishimiz kerak. Buni qilish,
\[{u_1}\left( t \right) = - \int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{u_2}\left( t \right) = \int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\]Shunday qilib, agar biz ushbu integrallarni bajara olsak, differensial tenglamaning ma'lum bir yechimi bo'ladi
\[\begin{align*}{Y_P}\left( t \right) & = {y_1}{u_1} + {y_2}{u_2}\\ & = - {y_1}\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + {y_2}\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\end{align*}\]Shunday qilib, keling, bu erda aniqlagan narsalarni umumlashtiramiz.
Parametrlarning o'zgarishi
Differensial tenglamani ko'rib chiqing,
\[y'' + q\left( t \right)y' + r\left( t \right)y = g\left( t \right)\]Faraz qilaylik, \(y_{1}(t)\) va \(y_{2}(t)\) asosiy yechimlar to‘plamidir.
\[y'' + q\left( t \right)y' + r\left( t \right)y = 0\]Keyin bir hil bo'lmagan differensial tenglamaning maxsus yechimi:
\[{Y_P}\left( t \right) = - {y_1}\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + {y_2}\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\]Shaxsga va muammoga qarab, ba'zilar formulani eslab qolish va undan foydalanishni osonlashtiradi, boshqalari esa formulani olish uchun ishlatiladigan jarayonni osonlashtiradi. Ushbu bo'limdagi misollar formula yordamida amalga oshiriladi.
Bir-ikki misolni davom ettirishdan oldin integrallardan kelib chiqadigan integratsiya doimiylari bilan bog'liq masalalarni ko'rib chiqaylik. Integratsiya konstantalarini qo'yish quyidagilarni beradi.
\[\begin{align*}{Y_P}\left( t \right) & = - {y_1}\left( {\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + c} \right) + {y_2}\left( {\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + k} \right)\\ & = - {y_1}\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + {y_2}\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + \left( { - c{y_1} + k{y_2}} \right)\end{align*}\]Qavs ichidagi yakuniy miqdor bilan to'ldiruvchi yechimdan boshqa narsa emas c1 = -c va \(c\)2 = k va agar biz buni differentsial tenglamaga qo'shsak, u nolga soddalashadi, chunki bu bir hil differentsial tenglamaning yechimidir. Boshqacha qilib aytganda, bu atamalar muayyan yechimga hech narsa qo'shmaydi va shuning uchun biz davom etamiz va barcha misollarda \(c\) = 0 va \(k\) = 0 deb faraz qilamiz.
Misollar bilan davom etishdan oldin bitta so'nggi eslatma. Qo'shimcha yechimdagi ikkita yechimingizdan qaysi biri \(y_{1}(t)\) va qaysi biri \(y_{2}(t)\) ekanligi haqida tashvishlanmang. Bu muhim emas. Qaysi birini \(y_{1}(t)\) va qaysi birini \(y_{2}(t)\) deb tanlamasligingizdan qat’iy nazar, siz bir xil javob olasiz.
Keling, bir nechta misollar bilan ishlaylik.
Yechim
Birinchidan, parametrlarni o'zgartirish formulasi ikkinchi hosila oldida bir koeffitsientini talab qilganligi sababli, unutishdan oldin bu haqda g'amxo'rlik qilaylik. Biz aslida hal qiladigan differentsial tenglama
\[y'' + 9y = 3\tan \left( {3t} \right)\]Ushbu differensial tenglamaning to'ldiruvchi yechimi ekanligini tekshirishni sizga qoldiramiz
\[{y_c}\left( t \right) = {c_1}\cos \left( {3t} \right) + {c_2}\sin \left( {3t} \right)\]Demak, bizda bor
\[{y_1}\left( t \right) = \cos \left( {3t} \right)\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = \sin \left( {3t} \right)\]Bu ikki funktsiyaning Wronskian
\[W = \left| {\begin{array}{*{20}{c}}{\cos \left( {3t} \right)}&{\sin \left( {3t} \right)}\\{ - 3\sin \left( {3t} \right)}&{3\cos \left( {3t} \right)}\end{array}} \right| = 3{\cos ^2}\left( {3t} \right) + 3{\sin ^2}\left( {3t} \right) = 3\]Maxsus yechim shunday bo'ladi:
\[\begin{align*}{Y_P}\left( t \right) & = - \cos \left( {3t} \right)\int{{\frac{{3\sin \left( {3t} \right)\tan \left( {3t} \right)}}{3}\,dt}} + \sin \left( {3t} \right)\int{{\frac{{3\cos \left( {3t} \right)\tan \left( {3t} \right)}}{3}\,dt}}\\ & = - \cos \left( {3t} \right)\int{{\frac{{{{\sin }^2}\left( {3t} \right)}}{{\cos \left( {3t} \right)}}\,dt}} + \sin \left( {3t} \right)\int{{\sin \left( {3t} \right)\,dt}}\\ & = - \cos \left( {3t} \right)\int{{\frac{{1 - {{\cos }^2}\left( {3t} \right)}}{{\cos \left( {3t} \right)}}\,dt}} + \sin \left( {3t} \right)\int{{\sin \left( {3t} \right)\,dt}}\\ & = - \cos \left( {3t} \right)\int{{\sec \left( {3t} \right) - \cos \left( {3t} \right)\,dt}} + \sin \left( {3t} \right)\int{{\sin \left( {3t} \right)\,dt}}\\ & = - \frac{{\cos \left( {3t} \right)}}{3}\left( {\ln \left| {\sec \left( {3t} \right) + \tan \left( {3t} \right)} \right| - \sin \left( {3t} \right)} \right) + \frac{{\sin \left( {3t} \right)}}{3}\left( { - \cos \left( {3t} \right)} \right)\\ & = - \frac{{\cos \left( {3t} \right)}}{3}\ln \left| {\sec \left( {3t} \right) + \tan \left( {3t} \right)} \right|\end{align*}\]Umumiy yechim:
\[y\left( t \right) = {c_1}\cos \left( {3t} \right) + {c_2}\sin \left( {3t} \right) - \frac{{\cos \left( {3t} \right)}}{3}\ln \left| {\sec \left( {3t} \right) + \tan \left( {3t} \right)} \right|\]Yechim
Bizga birinchi navbatda ushbu differentsial tenglama uchun to'ldiruvchi yechim kerak. Qo'shimcha yechim ekanligini tekshirish uchun buni sizga qoldiramiz:
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}t{{\bf{e}}^t}\]Demak, bizda bor
\[{y_1}\left( t \right) = {{\bf{e}}^t}\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = t{{\bf{e}}^t}\]Bu ikki funktsiyaning Wronskian
\[W = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^t}}&{t{{\bf{e}}^t}}\\{{{\bf{e}}^t}}&{{{\bf{e}}^t} + t{{\bf{e}}^t}}\end{array}} \right| = {{\bf{e}}^t}\left( {{{\bf{e}}^t} + t{{\bf{e}}^t}} \right) - {{\bf{e}}^t}\left( {t{{\bf{e}}^t}} \right) = {{\bf{e}}^{2t}}\]Maxsus yechim shunday bo'ladi:
\[\begin{align*}{Y_P}\left( t \right) & = - {{\bf{e}}^t}\int{{\frac{{t{{\bf{e}}^t}\,{{\bf{e}}^t}}}{{{{\bf{e}}^{2t}}\left( {{t^2} + 1} \right)}}\,dt}} + t{{\bf{e}}^t}\int{{\frac{{{{\bf{e}}^t}\,{{\bf{e}}^t}}}{{{{\bf{e}}^{2t}}\left( {{t^2} + 1} \right)}}\,dt}}\\ & = - {{\bf{e}}^t}\int{{\frac{t}{{{t^2} + 1}}\,dt}} + t{{\bf{e}}^t}\int{{\frac{1}{{{t^2} + 1}}\,dt}}\\ & = - \frac{1}{2}{{\bf{e}}^t}\ln \left( {1 + {t^2}} \right) + t{{\bf{e}}^t}{\tan ^{ - 1}}\left( t \right)\end{align*}\]Umumiy yechim:
\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}t{{\bf{e}}^t} - \frac{1}{2}{{\bf{e}}^t}\ln \left( {1 + {t^2}} \right) + t{{\bf{e}}^t}{\tan ^{ - 1}}\left( t \right)\]Bu usul, agar biz bog'langan bir jinsli differensial tenglama uchun asosiy yechimlar to'plamini bilsak, doimiy bo'lmagan koeffitsientli differensial tenglamalarda ham qo'llanilishi mumkin.
berilgan
\[{y_1}\left( t \right) = {{\bf{e}}^t}\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = t + 1\]bir jinsli differensial tenglama uchun asosiy yechimlar to‘plamini hosil qiling.
Yechim
Birinchi misolda bo'lgani kabi, biz avval \(t\) bilan ajratishimiz kerak.
\[y'' - \left( {1 + \frac{1}{t}} \right)y' + \frac{1}{t}y = t\]Yechimlarning asosiy to'plami uchun Wronskian
\[W = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^t}}&{t + 1}\\{{{\bf{e}}^t}}&1\end{array}} \right| = {{\bf{e}}^t} - {{\bf{e}}^t}\left( {t + 1} \right) = - t{{\bf{e}}^t}\]Maxsus yechim.
\[\begin{align*}{Y_P}\left( t \right) & = - {{\bf{e}}^t}\int{{\frac{{\left( {t + 1} \right)t}}{{ - t{{\bf{e}}^t}}}\,dt}} + \left( {t + 1} \right)\int{{\frac{{{{\bf{e}}^t}\left( t \right)}}{{ - t{{\bf{e}}^t}}}\,dt}}\\ & = {{\bf{e}}^t}\int{{\left( {t + 1} \right){{\bf{e}}^{ - t}}\,dt}} - \left( {t + 1} \right)\int{{\,dt}}\\ & = {{\bf{e}}^t}\left( { - {{\bf{e}}^{ - t}}\left( {t + 2} \right)} \right) - \left( {t + 1} \right)t\\ & = - {t^2} - 2t - 2\end{align*}\]Ushbu differentsial tenglamaning umumiy yechimi.
\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2} - 2t - 2\]Oldingi misolning yechimi haqida yana bir mavzuga murojaat qilishimiz kerak. Agar biz quyidagi amallarni bajarsak, yechimni biroz soddalashtirish mumkin.
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2} - 2t - 2\\ & = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2} - 2\left( {t + 1} \right)\\ & = {c_1}{{\bf{e}}^t} + \left( {{c_2} - 2} \right)\left( {t + 1} \right) - {t^2}\end{align*}\]Endi, \({c_2}\) noma'lum doimiy bo'lgani uchun undan 2 ni ayirish bu faktni o'zgartirmaydi. Shunday qilib, biz \({c_2} - 2\) ni \({c_2}\) deb yozishimiz va u bilan yakunlashimiz mumkin. Mana bu misol uchun yechimning soddalashtirilgan versiyasi.
\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2}\]Buni qilish har doim ham mumkin emas, lekin bu sodir bo'lganda, kelajakdagi ishni soddalashtirishingiz mumkin.