7.4-bo'lim: Parametrlarning o'zgarishi
Endi biz differentsial tenglamaning muayyan yechimini aniqlashning ikkinchi usulini ko'rib chiqishimiz kerak. Biz birinchi marta qilganimizdek ko'rgan Parametrlarning o'zgarishi biz butun jarayonni ko'rib chiqamiz va ma'lum bir yechimni yaratish uchun ishlatilishi mumkin bo'lgan formulalar to'plamini olamiz.
Biroq, biz 2-ga qaraganimizda ko'rganimizdeknd tartib differensial tenglamalar bu usul baholash oson bo'lmagan integrallarga olib kelishi mumkin. Shunday qilib, bu usul har doim aniqlanmagan koeffitsientlardan farqli o'laroq, hech bo'lmaganda ma'lum bir yechim uchun formulani yozish uchun ishlatilishi mumkin bo'lsa-da, har doim ham haqiqiy echimni olish mumkin bo'lmaydi.
Shunday qilib, jarayonni boshlaylik. Biz differentsial tenglamadan boshlaymiz,
\[\begin{equation}{y^{\left( n \right)}} + {p_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {p_1}\left( t \right)y' + {p_0}\left( t \right)y = g\left( t \right)\label{eq:eq1}\end{equation}\]va bir hil differensial tenglama uchun \({y_1}\left( t \right),{y_2}\left( t \right), \ldots ,{y_n}\left( t \right)\) fundamental yechimlar to‘plamini topdik deb faraz qilaylik.
Bizda bir hil differensial tenglamaning asosiy yechimlari to'plami mavjud bo'lganligi sababli, biz endi to'ldiruvchi yechim ekanligini bilamiz:
\[y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)\]Parametrlarni o'zgartirish usuli \({u_1}\left( t \right),{u_2}\left( t \right), \ldots ,{u_n}\left( t \right)\) yangi funktsiyalar to'plamini topishga harakat qilishni o'z ichiga oladi, shunday qilib,
\[\begin{equation}Y\left( t \right) = {u_1}\left( t \right){y_1}\left( t \right) + {u_2}\left( t \right){y_2}\left( t \right) + \cdots + {u_n}\left( t \right){y_n}\left( t \right)\label{eq:eq2}\end{equation}\]bir jinsli bo'lmagan differensial tenglamaning yechimi bo'ladi. Bu mumkin yoki yo'qligini aniqlash va agar iloji bo'lsa, \({u_i}\left( t \right)\) ni topish uchun bizga (umid qilamanki) echishimiz mumkin bo'lgan noma'lum funktsiyalarni o'z ichiga olgan jami \(n\) tenglamalari kerak bo'ladi.
Tenglamalardan biri oson. \(\eqref{eq:eq2}\) taxmini dastlabki \(\eqref{eq:eq1}\) differensial tenglamasini qondirishi kerak. Shunday qilib, keling, ba'zi hosilalarni olishni boshlaylik va biz parametrlarning o'zgarishini birinchi marta ko'rib chiqqanimizdek, ishimizni va jarayonni soddalashtiradigan ba'zi taxminlarni qilamiz. bizga kerak bo'lgan qolgan tenglamalarni yarating.
\(\eqref{eq:eq2}\) ning birinchi hosilasi:
\[Y'\left( t \right) = {u_1}{y'_1} + {u_2}{y'_2} + \cdots + {u_n}{y'_n} + {u'_1}{y_1} + {u'_2}{y_2} + \cdots + {u'_n}{y_n}\]E'tibor bering, biz differentsiatsiya jarayonining natijalarini biroz o'zgartirdik va o'qishni biroz osonlashtirish uchun \(\left( t \right)\) qismini \(u\) va \(y\)-ga tushirdik. Endi, agar biz buni farqlashda davom etsak, u tezda noqulay bo'lib qoladi va shuning uchun bu erda narsalarni soddalashtirish uchun faraz qilaylik. Biz \({u_i}\left( t \right)\) dan keyin bo'lganimiz uchun, ehtimol, ulardagi hosilalarning juda katta bo'lishiga yo'l qo'ymaslikka harakat qilishimiz kerak. Shunday qilib, keling, taxmin qilaylik,
\[{u'_1}{y_1} + {u'_2}{y_2} + \cdots + {u'_n}{y_n} = 0\]Shu nuqtada tabiiy savol: buni qilish mantiqiymi? Javob: agar biz \({u_i}\left( t \right)\) uchun yechishimiz mumkin bo'lgan \(n\) tenglamalari tizimiga ega bo'lsak, ha, buni qilish mantiqiy bo'ladi. Albatta, boshqa javob shundaki, agar biz bu ish bo'lishini bilmaganimizda bu taxminni qilmagan bo'lardik. Biroq, bu javobni qabul qilish uchun to'g'ri taxminlar qilishimizga ishonishingiz kerak, shuning uchun birinchi javob hozirgi vaqtda eng yaxshisidir.
Ushbu nuqtada \(\eqref{eq:eq2}\) ning birinchi hosilasi,
\[Y'\left( t \right) = {u_1}{y'_1} + {u_2}{y'_2} + \cdots + {u_n}{y'_n}\]va shuning uchun biz ikkinchi hosilani olishimiz mumkin,
\[Y''\left( t \right) = {u_1}{y''_1} + {u_2}{y''_2} + \cdots + {u_n}{y''_n} + {u'_1}{y'_1} + {u'_2}{y'_2} + \cdots + {u'_n}{y'_n}\]Bu bizni soddalashtirishdan oldingi asl birinchi hosilaga juda o'xshaydi, shuning uchun yana soddalashtiramiz. Biz yana \({u_i}\left( t \right)\) da hosilalarni minimal darajada ushlab turishni xohlaymiz, shuning uchun bu safar faraz qilaylik,
\[{u'_1}{y'_1} + {u'_2}{y'_2} + \cdots + {u'_n}{y'_n} = 0\]va bu faraz bilan ikkinchi hosila bo'ladi,
\[Y''\left( t \right) = {u_1}{y''_1} + {u_2}{y''_2} + \cdots + {u_n}{y''_n}\]Umid qilamanki, siz bu erda naqsh rivojlanayotganini ko'rishni boshlaysiz. Agar biz ushbu jarayonni birinchi \(n - 1\) hosilalari uchun davom ettirsak, biz ushbu hosilalar uchun quyidagi formulaga erishamiz.
\[\begin{equation}{Y^{\left( k \right)}}\left( t \right) = {u_1}y_1^{\left( k \right)} + {u_2}y_2^{\left( k \right)} + \cdots + {u_n}y_n^{\left( k \right)}\hspace{0.25in}\hspace{0.25in}k = 1,2, \ldots ,n - 1\label{eq:eq3}\end{equation}\]Ushbu formulalarning har biriga kirish uchun biz shuni taxmin qilishimiz kerak edi,
\[\begin{equation}{u'_1}y_1^{\left( k \right)} + {u'_2}y_2^{\left( k \right)} + \cdots + {u'_n}y_n^{\left( k \right)} = 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}k = 0,1, \ldots n - 2\label{eq:eq4}\end{equation}\]va esda tutingki, 0th funktsiyaning hosilasi faqat funktsiyaning o'zi. Masalan, \(y_2^{\left( 0 \right)}\left( t \right) = {y_2}\left( t \right)\).
E'tibor bering, \(\eqref{eq:eq4}\)-dagi taxminlar to'plami haqiqatda bizga noma'lum funktsiyalarning hosilalari bo'yicha \(n - 1\) tenglamalarini beradi: \({u_1}\left( t \right),{u_2}\left( t \right), \ldots ,{u_n}\left( t \right)\).
Keyin qilishimiz kerak bo'lgan yagona narsa, biz bu jarayonni topish uchun boshlagan birinchi tenglamani yaratishni tugatamiz (i.e. \(\eqref{eq:eq2}\)-ni \(\eqref{eq:eq1}\)-ga ulash). Buni amalga oshirish uchun bizga taxminning yana bir hosilasi kerak bo'ladi. \(n\) ni olish uchun biz \(\eqref{eq:eq3}\) dan olishimiz mumkin bo'lgan \({\left( {n - 1} \right)^{{\mbox{st}}}}\) lotinini farqlashth hosila beradi,
\[{Y^{\left( n \right)}}\left( t \right) = {u_1}y_1^{\left( n \right)} + {u_2}y_2^{\left( n \right)} + \cdots + {u_n}y_n^{\left( n \right)} + {u'_1}y_1^{\left( {n - 1} \right)} + {u'_2}y_2^{\left( {n - 1} \right)} + \cdots + {u'_n}y_n^{\left( {n - 1} \right)}\]Bu safar biz buni soddalashtirish uchun hech qanday taxminlar qilmaymiz, aksincha buni \(\eqref{eq:eq3}\) da berilgan hosilalar bilan birga \(\eqref{eq:eq1}\) differensial tenglamasiga kiritamiz.
\[\begin{align*}{u_1}y_1^{\left( n \right)} + {u_2}y_2^{\left( n \right)} + \cdots + {u_n}y_n^{\left( n \right)} + {{u'}_1}y_1^{\left( {n - 1} \right)} + {{u'}_2}y_2^{\left( {n - 1} \right)} + \cdots + {{u'}_n}y_n^{\left( {n - 1} \right)} & + \\ {p_{n - 1}}\left( t \right)\left[ {{u_1}y_1^{\left( {n - 1} \right)} + {u_2}y_2^{\left( {n - 1} \right)} + \cdots + {u_n}y_n^{\left( {n - 1} \right)}} \right] & + \\ \vdots \hspace{1.1in} & \\ {p_1}\left( t \right)\left[ {{u_1}{{y'}_1} + {u_2}{{y'}_2} + \cdots + {u_n}{{y'}_n}} \right] & + \\ {p_0}\left( t \right)\left[ {{u_1}{y_1} + {u_2}{y_2} + \cdots + {u_n}{y_n}} \right] & = g\left( t \right)\end{align*}\]Keyin, buni olish uchun biroz o'zgartiring,
\[\begin{align*}{u_1}\left[ {y_1^{\left( n \right)} + {p_{n - 1}}\left( t \right)y_1^{\left( {n - 1} \right)} + \cdots + {p_1}\left( t \right){{y'}_1} + {p_0}\left( t \right){y_1}} \right] & + \\ {u_2}\left[ {y_2^{\left( n \right)} + {p_{n - 1}}\left( t \right)y_2^{\left( {n - 1} \right)} + \cdots + {p_1}\left( t \right){{y'}_2} + {p_0}\left( t \right){y_2}} \right] & + \\ \vdots \hspace{2.0in} & \\ {u_n}\left[ {y_n^{\left( n \right)} + {p_{n - 1}}\left( t \right)y_n^{\left( {n - 1} \right)} + \cdots + {p_1}\left( t \right){{y'}_n} + {p_0}\left( t \right){y_n}} \right] & + \\ {{u'}_1}y_1^{\left( {n - 1} \right)} + {{u'}_2}y_2^{\left( {n - 1} \right)} + \cdots + {{u'}_n}y_n^{\left( {n - 1} \right)} & = g\left( t \right)\end{align*}\]Eslatib o'tamiz, \({y_1}\left( t \right),{y_2}\left( t \right), \ldots ,{y_n}\left( t \right)\) bir hil differensial tenglamaning barcha yechimlari va shuning uchun \(\left[ {\,\,} \right]\) dagi barcha miqdorlar nolga teng va bu quyidagiga kamayadi:
\[{\kern 1pt} {u'_1}y_1^{\left( {n - 1} \right)} + {u'_2}y_2^{\left( {n - 1} \right)} + \cdots + {u'_n}y_n^{\left( {n - 1} \right)} = g\left( t \right)\]Shunday qilib, bu tenglama \(\eqref{eq:eq4}\) da berilganlar bilan bir qatorda bizga kerakli \(n\) tenglamalarini beradi. To'liqlik uchun ularning barchasini shu erda sanab o'tamiz.
\[\begin{align*}{{u'}_1}{y_1} + {{u'}_2}{y_2} + \cdots + {{u'}_n}{y_n} & = 0\\ {{u'}_1}{{y'}_1} + {{u'}_2}{{y'}_2} + \cdots + {{u'}_n}{{y'}_n} & = 0\\ {{u'}_1}{{y''}_1} + {{u'}_2}{{y''}_2} + \cdots + {{u'}_n}{{y''}_n} & = 0\\ \vdots \hspace{0.85in} & \\ {{u'}_1}y_1^{\left( {n - 2} \right)} + {{u'}_2}y_2^{\left( {n - 2} \right)} + \cdots + {{u'}_n}y_n^{\left( {n - 2} \right)} & = 0\\ {{u'}_1}y_1^{\left( {n - 1} \right)} + {{u'}_2}y_2^{\left( {n - 1} \right)} + \cdots + {{u'}_n}y_n^{\left( {n - 1} \right)} & = g\left( t \right)\end{align*}\]Shunday qilib, bizda \(n\) tenglamalari bor, lekin shuni e'tiborga olingki, xuddi biz buni 2 uchun qilganimizda bo'lgani kabind Differensial tenglamalarni tartiblang. tizimdagi noma'lumlar \({u_1}\left( t \right),{u_2}\left( t \right), \ldots ,{u_n}\left( t \right)\) emas, balki ular \({u'_1}\left( t \right),{u'_2}\left( t \right), \ldots ,{u'_n}\left( t \right)\) hosilalaridir. Biroq, bu katta muammo emas. Agar biz ushbu tizimni hal qila olsak, biz o'zimiz kutayotgan funktsiyalarni olish uchun echimlarni birlashtira olamiz.
Shuni ham yodda tutingki, \({y_1}\left( t \right),{y_2}\left( t \right), \ldots ,{y_n}\left( t \right)\) ma'lum funktsiyalar sifatida qabul qilinadi va shuning uchun ular hosilalari (tizimda paydo bo'lgan) bilan birga tizimdagi barcha ma'lum miqdorlardir.
Endi biz ushbu tizimni qanday hal qilish haqida o'ylashimiz kerak. Agar juda ko'p tenglamalar bo'lmasa, agar xohlasak, uni to'g'ridan-to'g'ri hal qilishimiz mumkin. Biroq, katta \(n\) uchun (va bu erda kattalashtirish uchun ko'p narsa kerak bo'lmaydi) bu juda zerikarli va xatolarga moyil bo'lishi mumkin va bizda bo'lgani kabi umumiy \(n\) uchun umuman ishlamaydi.
Ushbu nuqtada foydalanish uchun eng yaxshi yechim usuli bu Kramer qoidasi. Biz ushbu kursda Kramer qoidasidan bir necha marta foydalanganmiz, ammo bu erda bizning maqsadlarimiz uchun eng yaxshi havola bu biz uni birinchi marta aniqlaganimizda foydalanganda. Yechimlarning asosiy to'plami qaytib 2nd materialga buyurtma berish.
Tizimni yechish uchun Kramer qoidasidan foydalanganda, har bir \({u'_i}\) uchun natijada olingan yechim \(n\) x \(n\) matritsalarining ikkita determinantining koeffitsienti bo'ladi. Har bir yechimning maxraji ma'lum koeffitsientlar matritsasining determinanti bo'ladi,
\[\left| {\begin{array}{*{20}{c}}{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{{{y'}_1}}&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{y_1^{\left( {n - 1} \right)}}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right| = W\left( {{y_1},{y_2}, \ldots {y_n}} \right)\left( t \right)\]Biroq, bu yuqorida aytib o'tilganidek, faqat \({y_1}\left( t \right),{y_2}\left( t \right), \ldots ,{y_n}\left( t \right)\) ning Wronskianidir va biz bu echimlarning asosiy to'plamini tashkil qiladi deb taxmin qilganimiz sababli, Wronskian nolga teng bo'lmasligini ham bilamiz. Bu, o'z navbatida, yuqoridagi tizim haqiqatdan ham echilishi mumkinligini va biz yuqoridagi ko'kdan kelib chiqqan barcha taxminlar haqiqatda ishlaganligini ko'rsatadi.
\({u'_i}\) yechimining suratlari \(i\) bilan koeffitsientlar matritsasining determinanti bo'ladi.th ustun \(\left( {0,0,0, \ldots ,0,g\left( t \right)} \right)\) ustuniga almashtirildi. Masalan, birinchisining hisoblagichi \({u'_1}\):
\[\left| {\begin{array}{*{20}{c}}0&{{y_2}}& \cdots &{{y_n}}\\0&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{g\left( t \right)}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right|\]Endi, determinantlarning yaxshi xususiyatiga ko'ra, agar matritsaning ustunlaridan biriga biror narsani faktorlar qilsak, natijada paydo bo'lgan matritsaning determinanti yangi matritsaning determinantini koeffitsientga aylantiradi. Boshqacha qilib aytganda, agar biz ushbu matritsadan \(g\left( t \right)\) faktoriga erishsak,
\[\left| {\begin{array}{*{20}{c}}0&{{y_2}}& \cdots &{{y_n}}\\0&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{g\left( t \right)}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right| = g\left( t \right)\left| {\begin{array}{*{20}{c}}0&{{y_2}}& \cdots &{{y_n}}\\0&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\1&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right|\]Biz buni faqat birinchisi uchun qildik, lekin biz buni \(n\) yechimlarining har qandayida ham osonlik bilan bajarishimiz mumkin edi. Shunday qilib, \({W_i}\) biz \(i\) o'rnini bosgan holda oladigan determinantni ifodalasin.th Vronskian ustuni \(\left( {0,0,0, \ldots ,0,1} \right)\) ustuni va tizimning yechimi shunday yozilishi mumkin:
\[{u'_1} = \frac{{g\left( t \right){W_1}\left( t \right)}}{{W\left( t \right)}},\,\,\,\,\,\,{u'_2} = \frac{{g\left( t \right){W_2}\left( t \right)}}{{W\left( t \right)}},\,\,\,\,\,\, \cdots ,\,\,\,\,\,{u'_n} = \frac{{g\left( t \right){W_n}\left( t \right)}}{{W\left( t \right)}}\]Voy! Bu tizimni yaratish va hal qilish uchun juda ko'p kuch sarfladi, lekin biz deyarli u erdamiz. Qo'limizda mavjud bo'lgan tizim yechimi bilan biz endi ma'lum bo'lmagan funksiyalar, \({u_1}\left( t \right),{u_2}\left( t \right), \ldots ,{u_n}\left( t \right)\) nima ekanligini aniqlash uchun ushbu atamalarning har birini birlashtira olamiz.
\[{u_1} = \int{{\frac{{g\left( t \right){W_1}\left( t \right)}}{{W\left( t \right)}}\,dt}},\,\,\,\,\,\,{u_2} = \int{{\frac{{g\left( t \right){W_2}\left( t \right)}}{{W\left( t \right)}}\,dt}},\,\,\,\,\,\, \cdots ,\,\,\,\,\,{u_n} = \int{{\frac{{g\left( t \right){W_n}\left( t \right)}}{{W\left( t \right)}}\,dt}}\]Nihoyat, \(\eqref{eq:eq1}\) ga ma'lum bir yechim quyidagicha beriladi:
\[Y\left( t \right) = {y_1}\left( t \right)\int{{\frac{{g\left( t \right){W_1}\left( t \right)}}{{W\left( t \right)}}\,dt}} + {y_2}\left( t \right)\int{{\frac{{g\left( t \right){W_2}\left( t \right)}}{{W\left( t \right)}}\,dt}} + \cdots + {y_n}\left( t \right)\int{{\frac{{g\left( t \right){W_n}\left( t \right)}}{{W\left( t \right)}}\,dt}}\]Shuni ham ta'kidlash kerakki, hosila olish jarayonida biz \({y^{\left( n \right)}}\) atamasi koeffitsienti bitta deb taxmin qilganmiz va bu yuqoridagi formulaga kiritilgan. Agar ushbu atamaning koeffitsienti bitta bo'lmasa, biz ushbu formuladan foydalanishdan oldin ishonch hosil qilishimiz va uni ajratishimiz kerak.
Bu erda misol bilan ishlashdan oldin shuni ta'kidlashimiz kerakki, biz ushbu formulani yozib olishimiz mumkin bo'lsa-da, aslida bu integrallarni hisoblash imkonsiz bo'lishi mumkin.
Yaxshi, keling, tezkor misolni ko'rib chiqaylik.
Yechim
Xarakteristik tenglama,
\[{r^3} - 2{r^2} - 21r - 18 = \left( {r + 3} \right)\left( {r + 1} \right)\left( {r - 6} \right) = 0\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_{\,1}} = - 3,\,\,\,{r_{\,2}} = - 1,\,\,\,{r_{\,3}} = 6\]Shunday qilib, bizda uchta haqiqiy ildiz bor va shuning uchun qo'shimcha echim:
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 3t}} + {c_2}{{\bf{e}}^{ - t}} + {c_3}{{\bf{e}}^{6t}}\]OK, endi bizda hisoblash uchun bir nechta determinantlar bor. Quyidagi aniqlovchi hisob-kitoblarni tekshirishni sizga qoldiramiz.
\[\begin{align*} W & = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^{ - 3t}}}&{{{\bf{e}}^{ - t}}}&{{{\bf{e}}^{6t}}}\\{ - 3{{\bf{e}}^{ - 3t}}}&{ - {{\bf{e}}^{ - t}}}&{6{{\bf{e}}^{6t}}}\\{9{{\bf{e}}^{ - 3t}}}&{{{\bf{e}}^{ - t}}}&{36{{\bf{e}}^{6t}}}\end{array}} \right| = 126{{\bf{e}}^{2t}} & \hspace{0.25in}{W_1} & = \left| {\begin{array}{*{20}{c}}0&{{{\bf{e}}^{ - t}}}&{{{\bf{e}}^{6t}}}\\0&{ - {{\bf{e}}^{ - t}}}&{6{{\bf{e}}^{6t}}}\\1&{{{\bf{e}}^{ - t}}}&{36{{\bf{e}}^{6t}}}\end{array}} \right| = 7{{\bf{e}}^{5t}}\\ {W_2} & = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^{ - 3t}}}&0&{{{\bf{e}}^{6t}}}\\{ - 3{{\bf{e}}^{ - 3t}}}&0&{6{{\bf{e}}^{6t}}}\\{9{{\bf{e}}^{ - 3t}}}&1&{36{{\bf{e}}^{6t}}}\end{array}} \right| = - 9{{\bf{e}}^{3t}} & \hspace{0.25in} {W_3} & = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^{ - 3t}}}&{{{\bf{e}}^{ - t}}}&0\\{ - 3{{\bf{e}}^{ - 3t}}}&{ - {{\bf{e}}^{ - t}}}&0\\{9{{\bf{e}}^{ - 3t}}}&{{{\bf{e}}^{ - t}}}&1\end{array}} \right| = 2{{\bf{e}}^{ - 4t}}\end{align*}\]Endi \(g\left( t \right) = 3 + 4{{\bf{e}}^{ - t}}\) ni hisobga olsak, biz \({u_i}\) ning har birini hisoblashimiz mumkin. Mana bu integrallar.
\[{u_1} = \int{{\frac{{\left( {3 + 4{{\bf{e}}^{ - t}}} \right)\left( {7{{\bf{e}}^{5t}}} \right)}}{{126{{\bf{e}}^{2t}}}}\,dt}} = \frac{1}{{18}}\int{{3{{\bf{e}}^{3t}} + 4{{\bf{e}}^{2t}}\,dt}} = \frac{1}{{18}}\left( {{{\bf{e}}^{3t}} + 2{{\bf{e}}^{2t}}} \right)\] \[{u_2} = \int{{\frac{{\left( {3 + 4{{\bf{e}}^{ - t}}} \right)\left( { - 9{{\bf{e}}^{3t}}} \right)}}{{126{{\bf{e}}^{2t}}}}\,dt}} = - \frac{1}{{14}}\int{{3{{\bf{e}}^t} + 4\,dt}} = - \frac{1}{{14}}\left( {3{{\bf{e}}^t} + 4t} \right)\] \[{u_3} = \int{{\frac{{\left( {3 + 4{{\bf{e}}^{ - t}}} \right)\left( {2{{\bf{e}}^{ - 4t}}} \right)}}{{126{{\bf{e}}^{2t}}}}\,dt}} = \frac{1}{{63}}\int{{3{{\bf{e}}^{ - 6t}} + 4{{\bf{e}}^{ - 7t}}\,dt}} = \frac{1}{{63}}\left( { - \frac{1}{2}{{\bf{e}}^{ - 6t}} - \frac{4}{7}{{\bf{e}}^{ - 7t}}} \right)\]E'tibor bering, biz ularning har biriga integratsiya konstantalarini kiritmadik, chunki ularni qo'shish biz kabi qo'shimcha echimga singib ketadigan atamani kiritgan bo'lar edi. ko'rgan biz 2 bilan muomala qilganimizdand tartibli differentsial tenglamalar.
Nihoyat, bu differensial tenglama uchun maxsus yechim,
\[\begin{align*}{Y_P} & = {u_1}{y_1} + {u_2}{y_2} + {u_3}{y_3}\\ & = \frac{1}{{18}}\left( {{{\bf{e}}^{3t}} + 2{{\bf{e}}^{2t}}} \right){{\bf{e}}^{ - 3t}} - \frac{1}{{14}}\left( {3{{\bf{e}}^t} + 4t} \right){{\bf{e}}^{ - t}} + \frac{1}{{63}}\left( { - \frac{1}{2}{{\bf{e}}^{ - 6t}} - \frac{4}{7}{{\bf{e}}^{ - 7t}}} \right){{\bf{e}}^{6t}}\\ & = - \frac{1}{6} + \frac{5}{{49}}{{\bf{e}}^{ - t}} - \frac{2}{7}t{{\bf{e}}^{ - t}}\end{align*}\]Umumiy yechim shunday bo'ladi:
\[y\left( t \right) = {c_1}{{\bf{e}}^{ - 3t}} + {c_2}{{\bf{e}}^{ - t}} + {c_3}{{\bf{e}}^{6t}} - \frac{1}{6} + \frac{5}{{49}}{{\bf{e}}^{ - t}} - \frac{2}{7}t{{\bf{e}}^{ - t}}\]Jarayonni ko'proq tasvirlash uchun biz ushbu bo'limda faqat bitta misol keltiramiz, shuning uchun biz ushbu bo'limni yopamiz.