5.7-bo'lim: Haqiqiy xususiy qiymatlar
Endi differentsial tenglamalar sistemasini echishni boshlash vaqti keldi. Bizda bor ko'rgan tizim uchun echimlar,
\[\vec x' = A\vec x\]shaklida bo'ladi
\[\vec x = \vec \eta {{\bf{e}}^{\lambda t}}\]Bu erda \(\lambda\) va \(\vec \eta \) \(A\) matritsasining xos qiymatlari va xos vektorlari. Biz \(2 \times 2\) tizimlari bilan ishlaymiz, shuning uchun biz ikkita yechimni qidiramiz, \({\vec x_1}\left( t \right)\) va \({\vec x_2}\left( t \right)\), bu erda matritsaning determinanti,
\[X = \left( {{{\vec x}_1}\,\,\,{{\vec x}_2}} \right)\]nolga teng.
Biz ikkita o'z qiymatlarimiz \({\lambda _{\,1}}\) va \({\lambda _{\,2}}\) haqiqiy va aniq bo'lgan holatni ko'rib chiqishdan boshlaymiz. Boshqacha qilib aytganda, ular haqiqiy, oddiy xos qiymatlar bo'ladi. Eslab qoling shuningdek, oddiy xos qiymatlar uchun xos vektorlar chiziqli mustaqildir. Bu shuni anglatadiki, biz ulardan oladigan echimlar ham chiziqli mustaqil bo'ladi. Agar echimlar chiziqli mustaqil bo'lsa, \(X\) matritsasi yagona bo'lmagan bo'lishi kerak va shuning uchun bu ikki yechim yechimlarning asosiy to'plami bo'ladi. Umumiy yechim bu holda bo'ladi,
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{{\lambda _1}t}}{\vec \eta ^{\left( 1 \right)}} + {c_2}{{\bf{e}}^{{\lambda _2}t}}{\vec \eta ^{\left( 2 \right)}}\]E'tibor bering, bizning har bir misolimiz ikkita misolga bo'linadi. Birinchi misol tizimni yechish va ikkinchi misol tizim uchun fazali portretni chizish bo'ladi. Fazali portretlar har doim ham differentsial tenglamalar kursida o'qitilmaydi va shuning uchun biz ularni yechim jarayonidan olib tashlaymiz, shunda siz ularni sinfingizda ko'rmagan bo'lsangiz, tizim uchun fazali portret misolini e'tiborsiz qoldirishingiz mumkin.
Yechim
Shunday qilib, biz qilishimiz kerak bo'lgan birinchi narsa - matritsa uchun xos qiymatlarni topish.
\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{1 - \lambda }&2\\3&{2 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} - 3\lambda - 4\\ & = \left( {\lambda + 1} \right)\left( {\lambda - 4} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\lambda _1} = - 1,\,\,{\lambda _2} = 4\end{align*}\]Endi bularning har biri uchun xos vektorlarni topamiz.
\({\lambda _{\,1}} = - 1\) :
Biz hal qilishimiz kerak,
Bu holda xos vektor,
\[\vec \eta = \left( {\begin{array}{*{20}{c}}{ - {\eta _2}}\\{{\eta _2}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 1 \right)}} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right),\,\,\,\,\,\,\,{\eta _2} = 1\]\({\lambda _{\,2}} = 4\) :
Biz hal qilishimiz kerak,
Bu holda xos vektor,
\[\vec \eta = \left( {\begin{array}{*{20}{c}}{\frac{2}{3}{\eta _2}}\\{{\eta _2}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 2 \right)}} = \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right),\,\,\,\,\,\,\,{\eta _2} = 3\]Keyin umumiy yechim shunday bo'ladi,
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) + {c_2}{{\bf{e}}^{4t}}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\]Endi biz doimiylarni topishimiz kerak. Buning uchun biz faqat dastlabki shartlarni qo'llashimiz kerak.
\[\left( {\begin{array}{*{20}{c}}0\\{ - 4}\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\]Biz hozir qilishimiz kerak bo'lgan yagona narsa - konstantalarni ko'paytirish va keyin biz konstantalar uchun hal qilishimiz mumkin bo'lgan ikkita tenglamani (har bir qator uchun bittadan) olamiz. Bu beradi,
\[\left. {\begin{array}{*{20}{c}}{ - {c_1} + 2{c_2} = 0}\\{{c_1} + 3{c_2} = - 4}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = - \frac{8}{5},\,\,\,{c_2} = - \frac{4}{5}\]Yechim shundan iboratki,
\[\vec x\left( t \right) = - \frac{8}{5}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) - \frac{4}{5}{{\bf{e}}^{4t}}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\]Keling, tizim uchun bosqich portretini ko'rib chiqaylik.
Yechim
Oxirgi misoldan bilamizki, bu tizim uchun xos qiymatlar va xos vektorlar:
\[\begin{align*}{\lambda _1} & = - 1 & \hspace{0.25in}{{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\\ {\lambda _2} & = 4 & \hspace{0.25in}{{\vec \eta }^{\left( 2 \right)}} & = \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\end{align*}\]Ma'lum bo'lishicha, bu bizga yo'nalish maydonini chizish uchun kerak bo'lgan barcha ma'lumotlar. Biroq, biz ishlarning to'g'ri ketayotganini ko'rishimiz uchun narsalarni o'z yechimimiz bilan bog'laymiz.
Biz ikkita xos vektor yo'nalishiga mos keladigan chiziqlarni chizishdan boshlaymiz. Bu beradi,
Endi, birinchi misoldan bizning umumiy yechimimiz
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) + {c_2}{{\bf{e}}^{4t}}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\]Agar bizda \({c_2} = 0\) bo'lsa, u holda yechim eksponensial marta vektor bo'ladi va eksponensial bajaradigan hamma narsa vektorning kattaligiga ta'sir qiladi va \(c_{1}\) doimiysi vektorning belgisiga ham, kattaligiga ham ta'sir qiladi. Boshqacha qilib aytganda, bu holda traektoriya \({\vec \eta ^{\left( 1 \right)}}\) vektoriga parallel bo'lgan to'g'ri chiziq bo'ladi. Shuni ham unutmangki, \(t\) ortishi bilan eksponensial kichikroq va kichikroq bo'ladi va shuning uchun traektoriya boshlang'ich tomon harakatlanadi. Agar \({c_1} > 0\) traektoriya II kvadrantda, \({c_1} < 0\) bo'lsa traektoriya bo'ladi. IV kvadrant.
Shunday qilib, yuqoridagi grafikdagi \({\vec \eta ^{\left( 1 \right)}}\) bilan belgilangan chiziq \({c_2} = 0\) ga mos keladigan traektoriyaning eskizi bo'ladi va bu traektoriya \(t\) ortishi bilan boshlang'ichga yaqinlashadi.
Agar biz narsalarni aylantirsak va \({c_1} = 0\) ga mos keladigan yechimga qarasak, biz \({\vec \eta ^{\left( 2 \right)}}\) ga parallel traektoriyaga ega bo'lamiz. Bundan tashqari, \(t\) oshgani sayin eksponensial ortib boradi va shuning uchun bu holda traektoriya endi \(t\) ortishi bilan boshlang'ichdan uzoqlashadi. Biz buni yuqoridagi grafikdagi chiziqlardagi strelkalar bilan belgilaymiz.
E'tibor bering, biz ushbu ma'lumotni yechimga bormasdan ham olishimiz mumkin edi. Biz qilishimiz kerak bo'lgan yagona narsa o'z qiymatlariga qarashdir. Manfiy bo'lgan xususiy qiymatlar \(t\) o'zining xos vektoriga parallel bo'lgan yo'nalishda ortishi bilan boshlang'ich tomon harakatlanadigan echimlarga mos keladi. Xuddi shunday, \(t\) o'zining xos vektoriga parallel bo'ladigan yo'nalishda ortishi bilan ijobiy bo'lgan xos qiymatlar boshlang'ichdan uzoqlashadi.
Agar ikkala konstanta ham yechimda bo'lsa, biz ushbu xatti-harakatlarning kombinatsiyasiga ega bo'lamiz. Katta manfiy \(t\) uchun yechim salbiy xos qiymatga ega bo'lgan qismga ustunlik qiladi, chunki bu hollarda ko'rsatkich katta va ijobiy bo'ladi. Katta manfiy \(t\) traektoriyalari \({\vec \eta ^{\left( 1 \right)}}\) ga parallel bo'ladi va bir xil yo'nalishda harakatlanadi.
Katta musbat \(t\) uchun yechimlarda ijobiy xos qiymatga ega boʻlgan qism ustunlik qiladi. Bu holda traektoriyalar \({\vec \eta ^{\left( 2 \right)}}\) ga parallel bo'ladi va bir xil yo'nalishda harakatlanadi.
Umuman olganda, traektoriyalar \({\vec \eta ^{\left( 1 \right)}}\) “yaqinidan” boshlanadi, boshlang‘ich tomon harakatlanadi, so‘ngra kelib chiqishga yaqinlashganda ular \({\vec \eta ^{\left( 2 \right)}}\) tomon harakatlana boshlaydi va keyin bu vektor bo‘ylab yuqoriga ko‘tariladi. Ulardan ba'zilarini eskiz qilish keyingi bosqich portretini beradi. Mana ko'k rang bilan belgilangan xos vektorlarga mos keladigan traektoriyalar bilan buning eskizi.
Bu holda \(\left( {0,0} \right)\) muvozanat yechimi a deyiladi egar nuqtasi va beqaror. Bu holda beqaror, \(t\) ortishi bilan yechimlar undan uzoqlashishini bildiradi.
Shunday qilib, biz tizimni matritsa ko'rinishida hal qildik, lekin esda tutingki, biz matritsa ko'rinishidagi tizimlarsiz ish boshlaganmiz. Keling, dastlab matritsa shaklida bo'lmagan tizim misolini tez ko'rib chiqaylik.
Yechim
Avval buni matritsa shakliga aylantirishimiz kerak. Bu yetarlicha oson. Mana tizimning matritsa shakli.
\[\vec x' = \left( {\begin{array}{*{20}{c}}1&2\\3&2\end{array}} \right)\vec x,\hspace{0.25in}\vec x\left( 0 \right) = \left( {\begin{array}{*{20}{c}}0\\{ - 4}\end{array}} \right)\]Bu faqat birinchi misoldagi tizim va shuning uchun biz ushbu tizimning yechimini oldik. Mana.
\[\vec x\left( t \right) = - \frac{8}{5}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) - \frac{4}{5}{{\bf{e}}^{4t}}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\]Endi biz tizimning yechimini matritsa shaklida emasligini xohlayotganimiz sababli, bu erda bir qadam oldinga boraylik. Keling, konstantalar va eksponensiallarni vektorlarga ko'paytiramiz va keyin ikkita vektorni qo'shamiz.
\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{\frac{8}{5}{{\bf{e}}^{ - t}}}\\{ - \frac{8}{5}{{\bf{e}}^{ - t}}}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{\frac{8}{5}{{\bf{e}}^{4t}}}\\{\frac{{12}}{5}{{\bf{e}}^{4t}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{8}{5}{{\bf{e}}^{ - t}} - \frac{8}{5}{{\bf{e}}^{4t}}}\\{ - \frac{8}{5}{{\bf{e}}^{ - t}} - \frac{{12}}{5}{{\bf{e}}^{4t}}}\end{array}} \right)\]Endi eslang,
\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\end{array}} \right)\]Shunday qilib, tizimning yechimi shundan iboratki,
\[\begin{align*}{x_1}\left( t \right) & = \frac{8}{5}{{\bf{e}}^{ - t}} - \frac{8}{5}{{\bf{e}}^{4t}}\\ {x_2}\left( t \right) & = - \frac{8}{5}{{\bf{e}}^{ - t}} - \frac{{12}}{5}{{\bf{e}}^{4t}}\end{align*}\]Keling, boshqa misol bilan ishlaylik.
Yechim
Shunday qilib, biz qilishimiz kerak bo'lgan birinchi narsa - matritsa uchun xos qiymatlarni topish.
\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{ - 5 - \lambda }&1\\4&{ - 2 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} + 7\lambda + 6\\ & = \left( {\lambda + 1} \right)\left( {\lambda + 6} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\lambda _1} = - 1,\,\,{\lambda _2} = - 6\end{align*}\]Endi bularning har biri uchun xos vektorlarni topamiz.
\({\lambda _{\,1}} = - 1\) :
Biz hal qilishimiz kerak,
Bu holda xos vektor,
\[\vec \eta = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{4{\eta _1}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 1 \right)}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right),\,\,\,\,\,\,\,{\eta _1} = 1\]\({\lambda _{\,2}} = - 6\) :
Biz hal qilishimiz kerak,
Bu holda xos vektor,
\[\vec \eta = \left( {\begin{array}{*{20}{c}}{ - {\eta _2}}\\{{\eta _2}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 2 \right)}} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right),\,\,\,\,\,\,\,{\eta _2} = 1\]Keyin umumiy yechim shunday bo'ladi,
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) + {c_2}{{\bf{e}}^{ - 6t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\]Endi biz doimiylarni topishimiz kerak. Buning uchun biz faqat dastlabki shartlarni qo'llashimiz kerak.
\[\left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\]Endi konstantalar uchun tizimni yeching.
\[\left. {\begin{array}{*{20}{c}}{{c_1} - {c_2} = 1}\\{4{c_1} + {c_2} = 2}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = \frac{3}{5},\,\,\,{c_2} = - \frac{2}{5}\]Yechim shundan iboratki,
\[\vec x\left( t \right) = \frac{3}{5}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) - \frac{2}{5}{{\bf{e}}^{ - 6t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\]Endi ushbu tizim uchun faza portretini topamiz.
Yechim
Oxirgi misoldan bilamizki, bu tizim uchun xos qiymatlar va xos vektorlar:
\[\begin{align*}{\lambda _1} & = - 1 & \hspace{0.25in}{{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\\ {\lambda _2} & = - 6 & \hspace{0.25in}{{\vec \eta }^{\left( 2 \right)}} & = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\end{align*}\]Bu birinchisidan bir oz farq qiladi. Biroq, xuddi shu tarzda boshlanadi. Biz birinchi navbatda xos vektorlarga mos keladigan traektoriyalarni chizamiz. E'tibor bering, ikkala o'z qiymatlari ham manfiy va shuning uchun \(t\) ortishi bilan ularning traektoriyalari boshlang'ich tomon siljiydi. Traektoriyalarni chizganimizda, \(t\) ortishi bilan ular olgan yo‘nalishni ko‘rsatish uchun strelkalar qo‘shamiz. Mana bu traektoriyalarning eskizi.
Endi bu erda birinchi bosqich portretidan ozgina farq paydo bo'ladi. \(t\) ortib borishi sababli barcha traektoriyalar boshlang'ich tomon harakatlanadi, chunki ikkala xos qiymat ham manfiy. Biz hal qilishimiz kerak bo'lgan masala - ular buni qanday qilishlari. Bu aslida birinchi qarashda ko'rinadiganidan osonroq.
Ikkinchi xos qiymat birinchisidan kattaroq. Katta va musbat \(t\) uchun bu xususiy qiymat uchun yechim birinchi xos qiymat uchun yechimdan kichikroq bo‘lishini anglatadi. Shuning uchun, \(t\) ortishi bilan traektoriya boshlang'ich tomon harakatlanadi va buni \({\vec \eta ^{\left( 1 \right)}}\) bilan parallel qiladi. Xuddi shunday, ikkinchi xususiy qiymat birinchisidan kattaroq bo'lganligi sababli, bu yechim katta va salbiy \(t\) uchun ustunlik qiladi. Shuning uchun, biz \(t\) ni kamaytirsak, traektoriya boshlang'ichdan uzoqlashadi va buni \({\vec \eta ^{\left( 2 \right)}}\) ga parallel qiladi.
Ba'zi traektoriyalarni qo'shish quyidagi eskizni beradi.
Bunday hollarda muvozanat yechimini \(\left( {0,0} \right)\) a deb ataymiz tugun va u asimptotik barqarordir. Agar \(t\) ortishi bilan barcha traektoriyalar unga qarab harakat qilsa, muvozanat yechimlari asimptotik barqaror hisoblanadi.
E'tibor bering, tugunlar ham beqaror bo'lishi mumkin. Oxirgi misolda, agar ikkala xos qiymat ham ijobiy bo'lganida, barcha traektoriyalar boshlang'ichdan uzoqlashgan bo'lar edi va bu holda muvozanat yechimi beqaror bo'lar edi.
Keyingiga o'tishdan oldin yana bir misol qilishimiz kerak. Biz tizimlar haqida birinchi marta gapira boshlaganimizda, biz yuqori tartibli differensial tenglamani tizimga aylantirishimiz mumkinligi aytilgan edi. Biz tizimlar yordamida yuqori tartibli differensial tenglamalarni qanday yechishni ko'rishimiz uchun shunga o'xshash misol qilishimiz kerak.
Yechim
Shunday ekan, avvalo buni tizimga aylantirishimiz kerak. Mana o'zgaruvchilarning o'zgarishi,
\[\begin{align*}{x_1} & = y & \hspace{0.25in}{{x'}_1} & = y' = {x_2}\\ {x_2} & = y' & \hspace{0.25in}{{x'}_2} & = y'' = \frac{3}{2}y - \frac{5}{2}y' = \frac{3}{2}{x_1} - \frac{5}{2}{x_2}\end{align*}\]Keyin tizim,
\[\vec x' = \left( {\begin{array}{*{20}{c}}0&1\\{\frac{3}{2}}&{ - \frac{5}{2}}\end{array}} \right)\vec x\hspace{0.25in}\vec x\left( 0 \right) = \left( {\begin{array}{*{20}{c}}{ - 4}\\9\end{array}} \right)\]qayerda,
\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{y\left( t \right)}\\{y'\left( t \right)}\end{array}} \right)\]Endi matritsa uchun xos qiymatlarni topishimiz kerak.
\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{ - \lambda }&1\\{\frac{3}{2}}&{ - \frac{5}{2} - \lambda }\end{array}} \right|\\ & = {\lambda ^2} + \frac{5}{2}\lambda - \frac{3}{2}\\ & = \frac{1}{2}\left( {\lambda + 3} \right)\left( {2\lambda - 1} \right)\hspace{0.25in}{\lambda _1} = - 3,\,\,\,{\lambda _2} = \frac{1}{2}\end{align*}\]Endi xos vektorlarni topamiz.
\({\lambda _{\,1}} = - 3\) :
Biz hal qilishimiz kerak,
Bu holda xos vektor,
\[\vec \eta = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{ - 3{\eta _1}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 1 \right)}} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right),\,\,\,\,\,\,\,{\eta _1} = 1\]\({\lambda _{\,2}} = \frac{1}{2}\):
Biz hal qilishimiz kerak,
Bu holda xos vektor,
\[\vec \eta = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{\frac{1}{2}{\eta _1}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 2 \right)}} = \left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right),\,\,\,\,\,\,\,{\eta _1} = 2\]Umumiy yechim shunday bo'ladi:
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - 3t}}\left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right) + {c_2}{{\bf{e}}^{\frac{t}{2}}}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\]Dastlabki shartni qo'llang.
\[\left( {\begin{array}{*{20}{c}}{ - 4}\\9\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\]Bu doimiylar uchun yechishimiz mumkin bo'lgan tenglamalar sistemasini beradi.
\[\left. {\begin{array}{*{20}{c}}{{c_1} + 2{c_2} = - 4}\\{ - 3{c_1} + {c_2} = 9}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = - \frac{{22}}{7},\,\,\,{c_2} = - \frac{3}{7}\]Tizimning haqiqiy yechimi shundan iboratki,
\[\vec x\left( t \right) = - \frac{{22}}{7}{{\bf{e}}^{ - 3t}}\left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right) - \frac{3}{7}{{\bf{e}}^{\frac{t}{2}}}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\]Endi shuni eslasak,
\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{y\left( t \right)}\\{y'\left( t \right)}\end{array}} \right)\]biz dastlabki differensial tenglamaning yechimi matritsa sistemasi yechimining faqat yuqori qatori ekanligini ko'rishimiz mumkin. Dastlabki differensial tenglamaning yechimi,
\[y\left( t \right) = - \frac{{22}}{7}{{\bf{e}}^{ - 3t}} - \frac{6}{7}{{\bf{e}}^{\frac{t}{2}}}\]E'tibor bering, chek sifatida, bu holda, pastki qator yuqori qatorning hosilasi bo'lishi kerak.