3.3-bo'lim: Murakkab ildizlar
Ushbu bo'limda biz differentsial tenglamaning echimlarini ko'rib chiqamiz
\[ay'' + by' + cy = 0\]xarakteristik tenglamaning qaysi ildizlarida,
\[a{r^2} + br + c = 0\]\({r_{1,2}} = \lambda \pm \mu \,i\) shaklidagi murakkab ildizlardir.
Endi eslaylikki, biz differensial tenglamaning barcha yechimlari shunday bo'ladi, deb faraz qilib, xarakteristik tenglamaga keldik.
\[y\left( t \right) = {{\bf{e}}^{rt}}\]Ikki ildizimizni yechimning umumiy shakliga ulab, differensial tenglamaning quyidagi yechimlarini beradi.
\[{y_1}\left( t \right) = {{\bf{e}}^{\left( {\lambda + \mu \,i} \right)\,\,t}}\hspace{0.25in}\text{va}\hspace{0.25in}{y_2}\left( t \right) = {{\bf{e}}^{\left( {\lambda - \mu \,i} \right)\,\,t}}\]Endi bu ikkita funktsiya "etarlicha yaxshi" (yana bu so'zlar bor ... biz ularni aniqlashga o'tamiz. oxir-oqibat) umumiy yechimni hosil qilish. Biroq, bizda muammo bor. Biz differensial tenglamamizda faqat haqiqiy sonlar bilan boshlaganimiz sababli, yechimimiz faqat haqiqiy sonlarni o'z ichiga olishini xohlaymiz. Yuqoridagi ikkita yechim murakkab va shuning uchun biz haqiqiy bo'lgan bir nechta ("etarli darajada yaxshi" ...) echimlarga murojaat qilmoqchimiz.
Buning uchun bizga Eyler formulasi kerak bo'ladi.
\[{{\bf{e}}^{i\theta }} = \cos \theta + i\sin \theta \]Bizga kerak bo'ladigan Eyler formulasining yaxshi varianti.
\[{{\bf{e}}^{ - i\theta }} = \cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right) = \cos \theta - i\sin \theta \]Endi bizning ikkita yechimimizni faqat haqiqiy ko'rsatkichlarga ega bo'lgan ko'rsatkichlarga va faqat xayoliy ko'rsatkichlarga ega bo'lgan ko'rsatkichlarga ajrating. Keyin ikkinchi eksponensialni qayta yozish uchun Eyler formulasidan yoki uning variantidan foydalaning.
\[\begin{align*}{y_1}\left( t \right) & = {{\bf{e}}^{\lambda \,t}}{{\bf{e}}^{i\mu \,\,t}} = {{\bf{e}}^{\lambda \,t}}\left( {\cos \left( {\mu t} \right) + i\sin \left( {\mu \,t} \right)} \right)\\ {y_2}\left( t \right) & = {{\bf{e}}^{\lambda \,t}}{{\bf{e}}^{ - i\mu \,\,t}} = {{\bf{e}}^{\lambda \,t}}\left( {\cos \left( {\mu t} \right) - i\sin \left( {\mu \,t} \right)} \right)\end{align*}\]Bu yechimlarning murakkab tabiatini yo'q qilmaydi, lekin bu ikkita echimni murakkab qismlarni yo'q qila oladigan shaklga qo'yadi.
dan eslang asoslar bo'limi agar ikkita yechim "etarlicha yaxshi" bo'lsa, har qanday yechim ikkita echimning kombinatsiyasi sifatida yozilishi mumkin. Boshqacha aytganda,
\[y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)\]ham yechim bo‘ladi.
Bundan foydalanib, agar biz ikkita echimni qo'shsak, biz erishamiz.
\[{y_1}\left( t \right) + {y_2}\left( t \right) = 2{{\bf{e}}^{\lambda \,t}}\cos \left( {\mu \,t} \right)\]Bu haqiqiy yechim va shunchaki 2 ni yo'q qilish uchun hamma narsani 2 ga bo'laylik. Bu biz qidirayotgan birinchi haqiqiy yechimni beradi.
\[u\left( t \right) = \frac{1}{2}{y_1}\left( t \right) + \frac{1}{2}{y_2}\left( t \right) = {{\bf{e}}^{\lambda \,t}}\cos \left( {\mu \,t} \right)\]E'tibor bering, bu faqat qabul qilish bilan tengdir
\[{c_1} = {c_2} = \frac{1}{2}\]Endi biz shunga o'xshash tarzda ikkinchi yechimga kelishimiz mumkin. Bu safar kelish uchun ikkita asl yechimni ayirib olaylik.
\[{y_1}\left( t \right) - {y_2}\left( t \right) = 2i\,{{\bf{e}}^{\lambda \,t}}\sin \left( {\mu \,t} \right)\]Tashqi tomondan, bu muammoni hal qilmaydi, chunki yechim hali ham murakkab. Biroq, ikkita konstanta, \(c_{1}\) va \(c_{2}\) kompleks sonlar bo'lishi mumkinligini bilib, biz buni \(2i\) ga bo'lish orqali haqiqiy echimga erishishimiz mumkin. Bu olish bilan tengdir
\[{c_1} = \frac{1}{{2i}}\hspace{0.25in}\text{va}\hspace{0.25in}{c_2} = - \frac{1}{{2i}}\]Bizning ikkinchi yechimimiz keyin bo'ladi
\[v\left( t \right) = \frac{1}{{2i}}{y_1}\left( t \right) - \frac{1}{{2i}}{y_2}\left( t \right) = {{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)\]Endi bizda differensial tenglamaning ikkita yechimi bor (ularning aslida yechim ekanligini tekshirishni sizga qoldiramiz).
\[u\left( t \right) = {{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right)\hspace{0.25in}\text{va}\hspace{0.25in}v\left( t \right) = {{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)\]Bundan tashqari, bu ikki yechim umumiy yechimni shakllantirish uchun "etarlicha yoqimli" ekanligi ma'lum bo'ldi.
Shunday qilib, agar xarakteristik tenglamaning ildizlari \({r_{1,2}} = \lambda \pm \mu \,i\) bo'lsa, differentsial tenglamaning umumiy yechimi bo'ladi.
\[y\left( t \right) = {c_1}{{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right) + {c_2}{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)\]Keling, bir nechta misollarni ko'rib chiqaylik.
Yechim
Bu differensial tenglamaning xarakteristik tenglamasi.
\[{r^2} - 4r + 9 = 0\]Bu tenglamaning ildizlari \({r_{1,2}} = 2 \pm \sqrt 5 \,i\). Differensial tenglamaning umumiy yechimi shunday bo'ladi.
\[y\left( t \right) = {c_1}{{\bf{e}}^{2t}}\cos \left( {\sqrt 5 t} \right) + {c_2}{{\bf{e}}^{2t}}\sin \left( {\sqrt 5 t} \right)\]Endi shuni ta'kidlaysizki, biz buni oxirgi bo'limda bo'lgani kabi darhol farq qilmaganmiz. Buning sababi oddiy. Farqlash unchalik qiyin bo'lmasa-da, u biroz chalkash bo'lishi mumkin. Shunday qilib, dastlab dastlabki shartlarni ko'rib chiqsak, birinchisidan shuni ko'rishimiz mumkinki, agar biz uni qo'llasak, biz quyidagilarni olamiz.
\[0 = y\left( 0 \right) = {c_1}\]Boshqacha qilib aytganda, birinchi shartni bajarish uchun birinchi muddat tugaydi. Bu uning hosilasi bilan birga eritmani hosil qiladi
\[\begin{align*}y\left( t \right) & = {c_2}{{\bf{e}}^{2t}}\sin \left( {\sqrt 5 t} \right)\\ y'\left( t \right) & = 2{c_2}{{\bf{e}}^{2t}}\sin \left( {\sqrt 5 t} \right) + \sqrt 5 {c_2}{{\bf{e}}^{2t}}\cos \left( {\sqrt 5 t} \right)\end{align*}\]Asl yechimni qilganimizdan ko'ra ancha yaxshi hosila. Endi, hosil qilish uchun ikkinchi boshlang'ich shartni qo'llang.
\[ - 8 = y'\left( 0 \right) = \sqrt 5 {c_2}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_2} = - \frac{8}{{\sqrt 5 }}\]Haqiqiy yechim shundan keyin.
\[y\left( t \right) = - \frac{8}{{\sqrt 5 }}{{\bf{e}}^{2t}}\sin \left( {\sqrt 5 t} \right)\]Yechim
Bu safar xarakterli tenglama.
\[{r^2} - 8r + 17 = 0\]Buning ildizlari \({r_{1,2}} = 4 \pm \,i\). Umumiy yechim, shuningdek, uning hosilasi
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{4t}}\cos \left( t \right) + {c_2}{{\bf{e}}^{4t}}\sin \left( t \right)\\ y'\left( t \right) & = 4{c_1}{{\bf{e}}^{4t}}\cos \left( t \right) - {c_1}{{\bf{e}}^{4t}}\sin \left( t \right) + 4{c_2}{{\bf{e}}^{4t}}\sin \left( t \right) + {c_2}{{\bf{e}}^{4t}}\cos \left( t \right)\end{align*}\]E'tibor bering, bu safar bizga boshidan lotin kerak bo'ladi, chunki bizda atamalardan birortasi tushib qolmaydi. Dastlabki shartlarni qo'llash quyidagi tizimni beradi.
\[\begin{align*} - 4 & = y\left( 0 \right) = {c_1}\\ - 1 & = y'\left( 0 \right) = 4{c_1} + {c_2}\end{align*}\]Ushbu tizimni hal qilish \({c_1} = - 4\) va \({c_2} = 15\) ni beradi. IVP uchun haqiqiy yechim shundan iborat.
\[y\left( t \right) = - 4{{\bf{e}}^{4t}}\cos \left( t \right) + 15{{\bf{e}}^{4t}}\sin \left( t \right)\]Yechim
Bu safar xarakterli tenglama.
\[4{r^2} + 24r + 37 = 0\]Buning ildizlari \({r_{1,2}} = - 3 \pm \frac{1}{2}\,i\). Umumiy yechim, shuningdek, uning hosilasi
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{ - 3t}}\cos \left( {\frac{t}{2}} \right) + {c_2}{{\bf{e}}^{ - 3t}}\sin \left( {\frac{t}{2}} \right)\\ y'\left( t \right) & = - 3{c_1}{{\bf{e}}^{ - 3t}}\cos \left( {\frac{t}{2}} \right) - \frac{{{c_1}}}{2}{{\bf{e}}^{ - 3t}}\sin \left( {\frac{t}{2}} \right) - 3{c_2}{{\bf{e}}^{ - 3t}}\sin \left( {\frac{t}{2}} \right) + \frac{{{c_2}}}{2}{{\bf{e}}^{ - 3t}}\cos \left( {\frac{t}{2}} \right)\end{align*}\]Dastlabki shartlarni qo'llash quyidagi tizimni beradi.
\[\begin{align*}1 & = y\left( \pi \right) = {c_1}{{\bf{e}}^{ - 3\pi }}\cos \left( {\frac{\pi }{2}} \right) + {c_2}{{\bf{e}}^{ - 3\pi }}\sin \left( {\frac{\pi }{2}} \right) = {c_2}{{\bf{e}}^{ - 3\pi }}\\ 0 & = y'\left( \pi \right) = - \frac{{{c_1}}}{2}{{\bf{e}}^{ - 3\pi }} - 3{c_2}{{\bf{e}}^{ - 3\pi }}\end{align*}\]\(t = \pi \)-ni eksponensialga ulashni unutmang! Bu o'quvchilarning ushbu muammolarni hal qilishda eng ko'p uchraydigan xatolaridan biridir. Bundan tashqari, bunday hollarda trig funksiyalarini imkon qadar ko'proq baholaganingizga ishonch hosil qiling. Bu sizning hayotingizni faqat soddalashtiradi. Ushbu tizimni hal qilish
\[{c_1} = - 6{{\bf{e}}^{3\pi }}\hspace{0.25in}{c_2} = {{\bf{e}}^{3\pi }}\]IVP uchun haqiqiy yechim shundan iborat.
\[\begin{align*}y\left( t \right) & = - 6{{\bf{e}}^{3\pi }}{{\bf{e}}^{ - 3t}}\cos \left( {\frac{t}{2}} \right) + {{\bf{e}}^{3\pi }}{{\bf{e}}^{ - 3t}}\sin \left( {\frac{t}{2}} \right)\\ y\left( t \right) & = - 6{{\bf{e}}^{ - 3\left( {t - \pi } \right)}}\cos \left( {\frac{t}{2}} \right) + {{\bf{e}}^{ - 3\left( {t - \pi } \right)}}\sin \left( {\frac{t}{2}} \right)\end{align*}\]Keyingi mavzuga o'tishdan oldin bitta so'nggi misol keltiraylik.
Yechim
Bu differentsial tenglamaning xarakteristik tenglamasi va uning ildizlari.
\[{r^2} + 16 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}r = \pm 4\,i\]Ushbu xarakterli polinom bilan ehtiyot bo'ling. Bu yerda talabalarning eng katta xatolaridan biri uni quyidagicha yozishdir:
\[{r^2} + 16r = 0\]
Muammo shundaki, agar differensial tenglamadagi ikkinchi hadda \(y'\) bo'lsa va bu aniq bo'lmasa, ikkinchi hadda \(r\) bo'ladi. Talabalar odatda \({r^2}\) dan boshlashadi va differensial tenglamadagi shartlar tugaguniga qadar vaqtlarni yozib olishadi. Bu ular noto'g'ri xarakterli polinomni yozishi mumkin va ko'pincha shuni anglatadiki, ehtiyot bo'ling.
OK, muammoga qaytish.
Ushbu differentsial tenglama va uning hosilasining umumiy yechimi.
\[\begin{align*}y\left( t \right) & = {c_1}\cos \left( {4t} \right) + {c_2}\sin \left( {4t} \right)\\ y'\left( t \right) & = - 4{c_1}\sin \left( {4t} \right) + 4{c_2}\cos \left( {4t} \right)\end{align*}\]Dastlabki sharoitlarda ulanish quyidagi tizimni beradi.
\[\begin{align*} - 10 & = y\left( {\frac{\pi }{2}} \right) = {c_1}\hspace{0.25in}{c_1} = - 10\\ 3 & = y'\left( {\frac{\pi }{2}} \right) = 4{c_2}\hspace{0.25in}{c_2} = \frac{3}{4}\end{align*}\]Shunday qilib, konstantalar ushbu tizim bilan to'g'ridan-to'g'ri tushadi va haqiqiy yechim.
\[y\left( t \right) = - 10\cos \left( {4t} \right) + \frac{3}{4}\sin \left( {4t} \right)\]