2.1-bo'lim: Chiziqli differensial tenglamalar (linear differential equations)
Birinchi tartibli differensial tenglamalarning biz ko'rib chiqadigan ilk muhim maxsus sinfi — bu birinchi tartibli chiziqli differensial tenglamalar dir. Boshqa ko'plab differensial tenglamalardan farqli o'laroq, bu holatda biz umumiy yechim uchun to'liq analitik formulani keltirib chiqara olamiz. Quyida ushbu umumiy yechim formulasi isbotlab beriladi. Biroq, tayyor formulani shunchaki yodlab olishni tavsiya etmaymiz. Formulani quruq yodlash o'rniga, uni keltirib chiqarish jarayoni va mantiqiy algoritmini tushunib olish ancha afzalroqdir. Aksariyat amaliy masalalarni tayyor formulaga qo'yishdan ko'ra, xuddi shu qadamlarni ketma-ket bajarish orqali yechish ancha oson va ishonchlidir.
Demak, birinchi tartibli chiziqli differensial tenglama qanday yechilishini ko'rib chiqaylik. Ushbu jarayonni o'rganayotganda asosiy maqsadimiz — \(y = y(t)\) ko'rinishidagi yechimni topish ekanini doimo yodda tuting.
Birinchi tartibli chiziqli differensial tenglamani yechish uchun biz ISHNI ALBATTA tenglamani quyidagi standart ko'rinishga keltirishdan boshlashimiz SHART. Agar tenglama ushbu ko'rinishda bo'lmasa, biz qo'llaydigan metod ish bermaydi:
\[\begin{equation}\frac{{dy}}{{dt}} + p\left( t \right)y = g\left( t \right) \label{eq:eq1} \end{equation}\]Bu yerda \(p(t)\) va \(g(t)\) funksiyalar uzluksiz funksiyalardir.
Endi faraz qilaylik, bizda \(\mu(t)\) deb nomlanuvchi maxsus funksiya mavjud bo'lib, u integrallovchi ko'paytuvchi (integrating factor) deyiladi. Hozircha bu funksiyaning qayerdan paydo bo'lgani haqida qayg'urmang. Umumiy yechim formulasini qurgach, \(\mu(t)\) ning aniq ifodasini birgalikda topamiz.
\(\mu(t)\) mavjud deb hisoblab, \(\eqref{eq:eq1}\) tenglamaning barcha hadlarini \(\mu(t)\) ga ko'paytiramiz:
\[\begin{equation}\mu \left( t \right)\frac{{dy}}{{dt}} + \mu \left( t \right)p\left( t \right)y = \mu \left( t \right)g\left( t \right) \label{eq:eq2} \end{equation}\]Mana endi \(\mu(t)\) funksiyasining ajoyib kuchi namoyon bo'ladi. Biz \(\mu(t)\) shunday tanlansinki, u quyidagi shartni qanoatlantirsin deb talab qilamiz:
\[\begin{equation}\mu \left( t \right)p\left( t \right) = \mu '\left( t \right) \label{eq:eq3} \end{equation}\]Ushbu \(\eqref{eq:eq3}\) munosabatni \(\eqref{eq:eq2}\) tenglamaga qo'ysak, quyidagi ifodaga ega bo'lamiz:
\[\begin{equation}\mu \left( t \right)\frac{{dy}}{{dt}} + \mu '\left( t \right)y = \mu \left( t \right)g\left( t \right) \label{eq:eq4} \end{equation}\]Ushbu bosqichda e'tibor bersangiz, \(\eqref{eq:eq4}\) tenglamaning chap tomoni ko'paytmaning hosilasi qoidasidan boshqa narsa emas:
\[\mu \left( t \right)\frac{{dy}}{{dt}} + \mu '\left( t \right)y = {\left( {\mu \left( t \right)y\left( t \right)} \right)^\prime }\]Shunday qilib, \(\eqref{eq:eq4}\) tenglamaning chap tomonini ushbu ko'paytmaning hosilasi bilan almashtirishimiz mumkin:
\[\begin{equation}{\left( {\mu \left( t \right)y\left( t \right)} \right)^\prime } = \mu \left( t \right)g\left( t \right) \label{eq:eq5} \end{equation}\]Endi eslang, biz \(y(t)\) funksiyani izlayapmiz. Buning uchun tenglamaning har ikkala tomonini \(t\) bo'yicha integrallasak kifoya:
\[\int{{{{\left( {\mu \left( t \right)y\left( t \right)} \right)}^\prime }\,dt}} = \int{{\mu \left( t \right)g\left( t \right)\,dt}}\] \[\begin{equation}\mu \left( t \right)y\left( t \right) + c = \int{{\mu \left( t \right)g\left( t \right)\,dt}} \label{eq:eq6} \end{equation}\]E'tibor bering: integrallash doimiysi \(c\) chap tomondagi integraldan hosil bo'ldi. Ushbu o'zgarmas sonni yozish nihoyatda muhimdir; agar u tushirib qoldirilsa, yechim xato chiqadi.
Yakuniy qadam sifatida \(y(t)\) ni topish uchun algebraik amallarni bajaramiz:
\[\begin{equation}y\left( t \right) = \frac{{\int{{\mu \left( t \right)g\left( t \right)\,dt}} + c}}{{\mu \left( t \right)}} \label{eq:eq7} \end{equation}\]Endi integrallovchi ko'paytuvchi \(\mu(t)\) ning o'zini qanday topish kerakligini aniqlaylik. Biz \(\eqref{eq:eq3}\) shartdan boshlaymiz:
\[\mu \left( t \right)p\left( t \right) = \mu '\left( t \right) \implies \frac{{\mu '\left( t \right)}}{{\mu \left( t \right)}} = p\left( t \right)\]Chap tomon \((\ln \mu(t))'\) ga teng ekanligini bilamiz. Har ikki tomonni integrallab, eksponentaga ko'tarsak:
\[\begin{equation}\mu \left( t \right) = {{\bf{e}}^{\int{{p\left( t \right)\,dt}}}}\label{eq:eq10}\end{equation}\]Shunday qilib, \(\mu(t)\) uchun o'zgarmas ko'paytuvchini 1 deb olish kifoya qiladi (chunki u baribir oxirida umumiy \(c\) doimiysiga singib ketadi).
Yechish algoritmi (Bosqichma-bosqich qo'llanma)
Birinchi tartibli chiziqli differensial tenglamalarni yechish tartibi quyidagicha:
- Differensial tenglamani standart ko'rinishga keltiring: \(\frac{dy}{dt} + p(t)y = g(t)\).
- Integrallovchi ko'paytuvchi \(\mu(t)\) ni hisoblang: \(\mu(t) = e^{\int p(t) dt}\).
- Tenglamaning barcha hadlarini \(\mu(t)\) ga ko'paytiring va chap tomonni ko'paytmaning hosilasi sifatida yozing: \((\mu(t)y)' = \mu(t)g(t)\).
- Har ikkala tomonni integrallang va integrallash doimiysi \(c\) ni qo'shishni unutmang.
- Ikkala tomonni \(\mu(t)\) ga bo'lib, izlanayotgan \(y(t)\) umumiy yechimini toping.
Keling, bir nechta amaliy misollarni ko'rib chiqaylik.
Yechim
Avvalo, tenglamani standart ko'rinishga keltiramiz:
\[\frac{{dv}}{{dt}} + 0.196v = 9.8\]Bundan ko'rinib turibdiki, \(p(t) = 0.196\) va \(g(t) = 9.8\). Demak, integrallovchi ko'paytuvchi:
\[\mu \left( t \right) = {{\bf{e}}^{\int{{0.196dt}}}} = {{\bf{e}}^{0.196t}}\]Endi tenglamaning barcha hadlarini \(\mu(t) = e^{0.196t}\) ga ko'paytiramiz:
\[\begin{align*}{{\bf{e}}^{0.196t}}\frac{{dv}}{{dt}} + 0.196{{\bf{e}}^{0.196t}}v & = 9.8{{\bf{e}}^{0.196t}}\\ {\left( {{{\bf{e}}^{0.196t}}v} \right)^\prime } & = 9.8{{\bf{e}}^{0.196t}}\end{align*}\]Har ikkala tomonni \(t\) bo'yicha integrallaymiz:
\[\begin{align*}\int{{{{\left( {{{\bf{e}}^{0.196t}}v} \right)}^\prime }\,dt}} & = \int{{9.8{{\bf{e}}^{0.196t}}\,dt}}\\ {{\bf{e}}^{0.196t}}v & = \frac{9.8}{0.196}{{\bf{e}}^{0.196t}} + c = 50{{\bf{e}}^{0.196t}} + c\end{align*}\]Yakuniy bosqichda har ikkala tomonni \(e^{0.196t}\) ga bo'lamiz (yoki \(e^{-0.196t}\) ga ko'paytiramiz). Natijada umumiy yechim hosil bo'ladi:
\[v\left( t \right) = 50 + c{{\bf{e}}^{ - 0.196t}}\]Yechim
Dastlab tenglamani standart ko'rinishga keltirish uchun har bir hadni \(t\) ga bo'lamiz (chunki \(y'\) ning oldidagi koeffitsiyent 1 bo'lishi shart):
\[\frac{{dy}}{{dt}} + \frac{2}{t}y = t - 1 + \frac{1}{t}\]Bu yerda \(p(t) = \frac{2}{t}\). Integrallovchi ko'paytuvchini topamiz:
\[\mu \left( t \right) = {{\bf{e}}^{\int{{\frac{2}{t}dt}}}} = {{\bf{e}}^{2\ln \left| t \right|}} = {{\bf{e}}^{\ln \left( {{t^2}} \right)}} = {t^2}\]E'tibor bering: \(e^{2\ln t} = e^{\ln(t^2)} = t^2\). Endi tenglamani \(t^2\) ga ko'paytiramiz:
\[\begin{align*}{t^2}\frac{{dy}}{{dt}} + 2ty & = {t^3} - {t^2} + t\\ {\left( {{t^2}y} \right)^\prime } & = {t^3} - {t^2} + t\end{align*}\]Ikkala tomonni integrallaymiz:
\[\begin{align*}{t^2}y & = \int{{\left( {{t^3} - {t^2} + t} \right)\,dt}}\\ {t^2}y & = \frac{1}{4}{t^4} - \frac{1}{3}{t^3} + \frac{1}{2}{t^2} + c\end{align*}\]Har ikkala tomonni \(t^2\) ga bo'lib, umumiy yechimni topamiz:
\[y\left( t \right) = \frac{1}{4}{t^2} - \frac{1}{3}t + \frac{1}{2} + \frac{c}{{{t^2}}}\]Endi boshlang'ich shart \(y(1) = \frac{1}{2}\) dan foydalanib, \(c\) o'zgarmasini aniqlaymiz:
\[\frac{1}{2} = y\left( 1 \right) = \frac{1}{4} - \frac{1}{3} + \frac{1}{2} + c \implies c = \frac{1}{12}\]Demak, Koshi masalasining xususiy yechimi quyidagicha bo'ladi:
\[y\left( t \right) = \frac{1}{4}{t^2} - \frac{1}{3}t + \frac{1}{2} + \frac{1}{{12{t^2}}}\]Yechim
Tenglamaning ikkala tomonini \(t\) ga bo'lib, standart shaklga keltiramiz:
\[y' - \frac{2}{t}y = {t^4}\sin \left( {2t} \right) - {t^2} + 4{t^3}\]Integrallovchi ko'paytuvchi:
\[\mu \left( t \right) = {{\bf{e}}^{\int{{ - \frac{2}{t}dt}}}} = {{\bf{e}}^{ - 2\ln t}} = {{\bf{e}}^{\ln \left( {{t^{ - 2}}} \right)}} = {t^{ - 2}} = \frac{1}{{{t^2}}}\]Tenglamani \(\mu(t) = t^{-2}\) ga ko'paytiramiz:
\[{\left( {{t^{ - 2}}y} \right)^\prime } = {t^2}\sin \left( {2t} \right) - 1 + 4t\]Ikkala tomonni integrallaymiz (birinchi had \(\int t^2 \sin(2t) dt\) bo'laklab integrallash usuli bilan hisoblanadi):
\[{t^{ - 2}}y = - \frac{1}{2}{t^2}\cos \left( {2t} \right) + \frac{1}{2}t\sin \left( {2t} \right) + \frac{1}{4}\cos \left( {2t} \right) - t + 2{t^2} + c\]Ikkala tomonni \(t^2\) ga ko'paytirib, \(y(t)\) ni topamiz va \(y(\pi) = \frac{3}{2}\pi^4\) boshlang'ich shartni qo'yib, \(c = 0\) ekanini aniqlaymiz.
Yakuniy yechim:
\[y\left( t \right) = - \frac{1}{2}{t^4}\cos \left( {2t} \right) + \frac{1}{2}{t^3}\sin \left( {2t} \right) + \frac{1}{4}{t^2}\cos \left( {2t} \right) - {t^3} + 2{t^4}\]Ushbu yechimning mavjudlik oralig'i: \(0 < t < \infty\).