3.4-bo'lim: Takrorlangan ildizlar
Ushbu bo'limda biz doimiy koeffitsient, chiziqli, bir hil ikkinchi tartibli differentsial tenglamalar uchun oxirgi holatni ko'rib chiqamiz. Bunday holda, biz yechimlarni xohlaymiz
\[ay'' + by' + cy = 0\]bu yerda xarakteristik tenglamaning yechimlari
\[a{r^2} + br + c = 0\]\({r_1} = {r_2} = r\) juft ildizlaridir.
Biroq, bu muammoga olib keladi. Eslatib o'tamiz, echimlar
\[{y_1}\left( t \right) = {{\bf{e}}^{{r_1}\,t}} = {{\bf{e}}^{r\,t}}\hspace{0.25in}{y_2}\left( t \right) = {{\bf{e}}^{{r_2}\,t}} = {{\bf{e}}^{r\,t}}\]Bular bir xil yechim va umumiy yechim yaratish uchun "etarlicha yoqimli" EMAS. Biz "etarlicha yaxshi" deb ta'riflaymiz, deb va'da beramiz oxir-oqibat! Shunday qilib, biz birinchi yechimdan foydalanishimiz mumkin, ammo bizga ikkinchi yechim kerak bo'ladi.
Ushbu ikkinchi yechimni topishdan oldin, keling, bir oz sayohat qilaylik. Yon safarning sababi oxir-oqibat aniq bo'ladi. Kvadrat formuladan biz xarakteristik tenglamaning ildizlari ekanligini bilamiz:
\[{r_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]Bunday holda, biz ikki ildizga ega bo'lganimiz uchun bizda bo'lishi kerak
\[{b^2} - 4ac = 0\]Bu biz ikki tomonlama ildiz olishning yagona yo'li va bu holda ildizlar bo'ladi
\[{r_{1,2}} = \frac{{ - b}}{{2a}}\]Shunday qilib, bizda bitta yechim bor
\[{y_1}\left( t \right) = {{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\]Ikkinchi yechimni topish uchun chiziqli bir jinsli differensial tenglamaning doimiy martalar yechimi ham yechim ekanligidan foydalanamiz. Agar bu rost bo'lsa balki bizga omad kulib boqadi va quyidagilar ham yechim bo'ladi
\[\begin{equation}{y_2}\left( t \right) = v\left( t \right){y_1}\left( t \right) = v\left( t \right){{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\label{eq:eq1}\end{equation}\]to'g'ri \(v(t)\) tanlovi bilan. Buni haqiqatan ham amalga oshirish mumkinligini aniqlash uchun keling, buni differensial tenglamaga qayta kiritamiz va biz nimaga erishamiz. Bizga birinchi navbatda bir nechta hosila kerak bo'ladi.
\[\begin{align*}{{y'}_2}\left( t \right) & = v'\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} - \frac{b}{{2a}}v\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\\ {{y''}_2}\left( t \right) & = v''\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} - \frac{b}{{2a}}v'\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} - \frac{b}{{2a}}v'\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} + \frac{{{b^2}}}{{4{a^2}}}v\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\\ & = v''\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} - \frac{b}{a}v'\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} + \frac{{{b^2}}}{{4{a^2}}}v\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\end{align*}\]Biz hosilalarni yozishni biroz soddalashtirish uchun \(\left( t \right)\) qismini \(v\)-ga tushirdik. Endi ularni differentsial tenglamaga kiriting.
\[a\left( {v''\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} - \frac{b}{a}v'\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} + \frac{{{b^2}}}{{4{a^2}}}v\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}} \right) + b\left( {v'\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}} - \frac{b}{{2a}}v\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}} \right) + c\left( {v\,{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}} \right) = 0\]Biz barcha atamalardan eksponentni hisobga olamiz, shuning uchun buni qilaylik. Shuningdek, biz \(v\) va uning hosilalarining barcha koeffitsientlarini to'playmiz.
\[\begin{align*}{{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\left( {av'' + \left( { - b + b} \right)v' + \left( {\frac{{{b^2}}}{{4a}} - \frac{{{b^2}}}{{2a}} + c} \right)v} \right) & = 0\\ {{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\left( {av'' + \left( { - \frac{{{b^2}}}{{4a}} + c} \right)v} \right) & = 0\\ {{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\left( {av'' - \frac{1}{{4a}}\left( {{b^2} - 4ac} \right)v} \right) & = 0\end{align*}\]Endi, biz qo'sh ildiz bilan ishlayotganimiz uchun, ikkinchi atama nolga teng bo'lishini bilamiz. Bundan tashqari, ko'rsatkichlar hech qachon nolga teng bo'lmaydi. Shuning uchun \(\eqref{eq:eq1}\) differensial tenglamaning yechimi bo'ladi, agar \(v(t)\) quyidagi differensial tenglamani qanoatlantiradigan funksiya bo'lsa.
\[av'' = 0\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}v'' = 0\]Biz \(a\)-ni tashlab yuborishimiz mumkin, chunki u nolga teng bo'lmasligini bilamiz. Agar shunday bo'lganda edi, bizda ikkinchi tartibli differentsial tenglama bo'lmaydi! Shunday qilib, endi biz \(v(t)\) uchun ruxsat etilgan eng umumiy shaklni aniqlashimiz mumkin.
\[v' = \int{{v''\,dt}} = c\hspace{0.25in}v\left( t \right) = \int{{v'\,dt}} = ct + k\]Bu aslida bizga kerak bo'lgandan ko'ra murakkabroq va aslida biz bundan har ikkala konstantani ham olib tashlashimiz mumkin. Bu nima uchun ekanligini tushunish uchun keling, ikkinchi yechimni olish uchun davom etamiz. Keyin ikkita yechim mavjud
\[{y_1}\left( t \right) = {{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\hspace{0.25in}{y_2}\left( t \right) = \left( {ct + k} \right){{\bf{e}}^{ - \frac{{b\,t}}{{2a}}}}\]Oxir-oqibat bu ikki yechim umumiy yechimni shakllantirish uchun "etarlicha yoqimli" ekanligini ko'rsata olasiz. Umumiy yechim quyidagicha bo'ladi.
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}} + {c_2}\left( {ct + k} \right){{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}}\\ & = {c_1}{{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}} + \left( {{c_2}ct + {c_2}k} \right){{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}}\\ & = \left( {{c_1} + {c_2}k} \right){{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}} + {c_2}c\,t\,{{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}}\end{align*}\]E'tibor bering, biz narsalarni biroz o'zgartirdik. Endi \(c\), \(k\), \(c_{1}\) va \(c_{2}\) nomaʼlum konstantalardir, shuning uchun ularning har qanday birikmasi ham nomaʼlum konstantalar boʻladi. Xususan, \(c_{1}+c_{2}k\) va \(c_{2}c\) noma'lum konstantalardir, shuning uchun biz ularni quyidagicha qayta yozamiz.
\[y\left( t \right) = {c_1}{{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}} + {c_2}\,t\,{{\bf{e}}^{ - \,\frac{{b\,t}}{{2a}}}}\]Shunday qilib, agar biz \(v(t)\) uchun eng umumiy shaklga qaytsak, biz \(c=1\) va \(k=0\) ni olamiz va biz bir xil umumiy yechimga erishamiz.
Keling, xulosa qilaylik. Agar xarakteristik tenglamaning ildizlari \(r_{1} = r_{2} = r\) bo'lsa, u holda umumiy yechim shunday bo'ladi
\[y\left( t \right) = {c_1}{{\bf{e}}^{r\,t}} + {c_2}t{{\bf{e}}^{r\,t}}\]Endi bir nechta misollar bilan ishlaylik.
Yechim
Xarakteristik tenglama va uning ildizlari.
\[{r^2} - 4r + 4 = {\left( {r - 2} \right)^2} = 0\hspace{0.25in}{r_{1,2}} = 2\]Umumiy yechim va uning hosilasi
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{2t}} + {c_2}t{{\bf{e}}^{2t}}\\ y'\left( t \right) & = 2{c_1}{{\bf{e}}^{2t}} + {c_2}{{\bf{e}}^{2t}} + 2{c_2}t{{\bf{e}}^{2t}}\end{align*}\]Ikkinchi muddatda mahsulot qoidasini unutmang! Dastlabki sharoitlarda ulanish quyidagi tizimni beradi.
\[\begin{align*}12 & = y\left( 0 \right) = {c_1}\\ - 3 & = y'\left( 0 \right) = 2{c_1} + {c_2}\end{align*}\]Ushbu tizim \(c_{1} = 12\) va \(c_{2} = -27\) ni olish uchun osonlik bilan hal qilinadi. IVP uchun haqiqiy yechim shundan iborat.
\[y\left( t \right) = 12{{\bf{e}}^{2t}} - 27t{{\bf{e}}^{2t}}\]Yechim
Xarakteristik tenglama va uning ildizlari.
\[16{r^2} - 40r + 25 = {\left( {4r - 5} \right)^2} = 0\hspace{0.25in}{r_{1,2}} = \frac{5}{4}\]Umumiy yechim va uning hosilasi
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{\frac{{5\,t}}{4}}} + {c_2}t{{\bf{e}}^{\frac{{5\,t}}{4}}}\\ y'\left( t \right) & = \frac{5}{4}{c_1}{{\bf{e}}^{\frac{{5\,t}}{4}}} + {c_2}{{\bf{e}}^{\frac{{5\,t}}{4}}} + \frac{5}{4}{c_2}t{{\bf{e}}^{\frac{{5\,t}}{4}}}\end{align*}\]Ikkinchi muddatda mahsulot qoidasini unutmang! Dastlabki sharoitlarda ulanish quyidagi tizimni beradi.
\[\begin{align*}3 = y\left( 0 \right) & = {c_1}\\ - \frac{9}{4} = y'\left( 0 \right) & = \frac{5}{4}{c_1} + {c_2}\end{align*}\]Ushbu tizim \(c_{1} = 3\) va \(c_{2} = -6\) ni olish uchun osonlikcha hal qilinadi. IVP uchun haqiqiy yechim shundan iborat.
\[y\left( t \right) = 3{{\bf{e}}^{\frac{{5\,t}}{4}}} - 6t{{\bf{e}}^{\frac{{5\,t}}{4}}}\]Yechim
Xarakteristik tenglama va uning ildizlari.
\[{r^2} + 14r + 49 = {\left( {r + 7} \right)^2} = 0\hspace{0.25in}{r_{1,2}} = - 7\]Umumiy yechim va uning hosilasi
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{ - 7t}} + {c_2}t{{\bf{e}}^{ - 7t}}\\ y'\left( t \right) & = - 7{c_1}{{\bf{e}}^{ - 7t}} + {c_2}{{\bf{e}}^{ - 7t}} - 7{c_2}t{{\bf{e}}^{ - 7t}}\end{align*}\]Dastlabki shartlarni ulash quyidagi tenglamalar sistemasini beradi.
\[\begin{align*} - 1 & = y\left( { - 4} \right) = {c_1}{{\bf{e}}^{28}} - 4{c_2}{{\bf{e}}^{28}}\\ 5 & = y'\left( { - 4} \right) = - 7{c_1}{{\bf{e}}^{28}} + {c_2}{{\bf{e}}^{28}} + 28{c_2}{{\bf{e}}^{28}} = - 7{c_1}{{\bf{e}}^{28}} + 29{c_2}{{\bf{e}}^{28}}\end{align*}\]Bu sistemani yechish quyidagi konstantalarni beradi.
\[{c_1} = - 9{{\bf{e}}^{ - 28}}\hspace{0.25in}{c_2} = - 2{{\bf{e}}^{ - 28}}\]IVP uchun haqiqiy yechim shundan iborat.
\[\begin{align*}y\left( t \right) & = - 9{{\bf{e}}^{ - 28}}{{\bf{e}}^{ - 7t}} - 2t{{\bf{e}}^{ - 28}}{{\bf{e}}^{ - 7t}}\\ y\left( t \right) & = - 9{{\bf{e}}^{ - 7\left( {t + 4} \right)}} - 2t{{\bf{e}}^{ - 7\left( {t + 4} \right)}}\end{align*}\]