4.5-bo'lim: Ivpni laplas o'zgartirishlari bilan hal qilish
Endi differentsial tenglamalarga qaytish vaqti keldi. Biz oxirgi uchta bo'limni Laplas konvertatsiyasini va teskari Laplasni qanday olishni o'rgandik aylantiradi. Bu keyingi bo'limlar uchun bebaho ko'nikmalar bo'ladi, shuning uchun biz u erda nimani o'rganganimizni unutmang.
Differensial tenglamalarga o'tishdan oldin bizga yana bitta formula kerak bo'ladi. Biz hosilaning Laplas konvertatsiyasini qanday qabul qilishni bilishimiz kerak. Avval eslaylikki, \(f^{(n)}\) \(f\) funksiyasining \(n^{\mbox{th}}\) hosilasini bildiradi. Endi bizda quyidagi fakt bor.
Hosilaning Laplas konvertatsiyasi
Faraz qilaylik, \(f\), \(f'\), \(f''\),… \(f^{(n-1)}\) hammasi uzluksiz funksiyalar va \(f^{(n)}\) bo‘lak-bo‘lak uzluksiz funksiyalar. Keyin,
\[\mathcal{L}\left\{ {{f^{\left( n \right)}}} \right\} = {s^n}F\left( s \right) - {s^{n - 1}}f\left( 0 \right) - {s^{n - 2}}f'\left( 0 \right) - \cdots - s{f^{\left( {n - 2} \right)}}\left( 0 \right) - {f^{\left( {n - 1} \right)}}\left( 0 \right)\]Biz ikkinchi tartibli differensial tenglamalar bilan shug'ullanayotganimiz sababli, Laplasga ega bo'lish qulay bo'ladi birinchi ikkita hosilalarni o'zgartirish.
\[\begin{align*}\mathcal{L}\left\{ {y'} \right\} & = sY\left( s \right) - y\left( 0 \right)\\ \mathcal{L}\left\{ {y''} \right\} & = {s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)\end{align*}\]E'tibor bering, ushbu formulalarda ko'rinadigan ikkita funktsiyani baholash, \(y\left( 0 \right)\) va \(y'\left( 0 \right)\), ko'pincha biz IVP ning dastlabki holati uchun foydalanadigan narsadir. Bu shuni anglatadiki, agar biz IVP ni hal qilish uchun ushbu formulalardan foydalanmoqchi bo'lsak, bizga \(t = 0\) da dastlabki shartlar kerak bo'ladi.
Laplas konvertatsiyalari, ayniqsa, majburlash funktsiyasida Heaviside funktsiyalariga ega bo'lgan bir hil bo'lmagan differentsial tenglamalar uchun foydali bo'lsa-da, biz jarayonning qanday ishlashini ko'rsatish uchun bir nechta oddiy masalalardan boshlaymiz.
Yechim
Laplasdan foydalanishning birinchi qadami IVP ni yechish uchun transformatsiyalar differensial tenglamadagi har bir atamaning konvertatsiyasini olishdir.
\[\mathcal{L}\left\{ {y''} \right\} - 10\mathcal{L}\left\{ {y'} \right\} + 9\mathcal{L}\left\{ y \right\} = \mathcal{L}\left\{ {5t} \right\}\]Bizning tegishli formulalardan foydalanish Laplas konvertatsiyalari jadvali bizga quyidagilarni beradi.
\[{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 10\left( {sY\left( s \right) - y\left( 0 \right)} \right) + 9Y\left( s \right) = \frac{5}{{{s^2}}}\]Dastlabki shartlarni ulang va ularda \(Y(s)\) bo'lgan barcha shartlarni to'plang.
\[\left( {{s^2} - 10s + 9} \right)Y\left( s \right) + s - 12 = \frac{5}{{{s^2}}}\]\(Y(s)\) uchun yechish.
\[Y\left( s \right) = \frac{5}{{{s^2}\left( {s - 9} \right)\left( {s - 1} \right)}} + \frac{{12 - s}}{{\left( {s - 9} \right)\left( {s - 1} \right)}}\]Ayni paytda biz nima qilmoqchi ekanligimizni eslash qulay. IVP uchun \(y(t)\) yechimini topishga harakat qilyapmiz. Biz hozir topa olgan narsa bu yechim emas, balki uning Laplasidir aylantirish. Shunday qilib, yechim topish uchun biz qilishimiz kerak bo'lgan narsa teskari konvertatsiya qilishdir.
Buni amalga oshirishdan oldin, keling, uning hozirgi shaklida biz ikki marta qisman kasrlarni bajarishimiz kerakligini bilib olaylik. Biroq, agar biz ikkita atamani birlashtirsak, biz faqat bir marta qisman kasrlarni qilamiz. Faqat bu emas, balki qo'shma atamaning maxraji birinchi hadning maxraji bilan bir xil bo'ladi. Bu shuni anglatadiki, biz nima bo'lishidan qat'i nazar, bu maxraj bilan atamani qisman kasrga aylantiramiz, shuning uchun biz hisoblagichni biroz chalkashtirib, keyin bir marta qisman kasr qilishimiz mumkin.
Bu biz muammoni yanada chigallashtiradigan narsalardan biri, ammo bu jarayonda biz o'zimizni adolatli ish hajmini saqlab qolamiz!
Ikki atamani birlashtirganda,
\[Y\left( s \right) = \frac{{5 + 12{s^2} - {s^3}}}{{{s^2}\left( {s - 9} \right)\left( {s - 1} \right)}}\]Ushbu transformatsiya uchun qisman kasr parchalanishi:
\[Y\left( s \right) = \frac{A}{s} + \frac{B}{{{s^2}}} + \frac{C}{{s - 9}} + \frac{D}{{s - 1}}\]Numeratorlarni teng belgilash,
\[5 + 12{s^2} - {s^3} = As\left( {s - 9} \right)\left( {s - 1} \right) + B\left( {s - 9} \right)\left( {s - 1} \right) + C{s^2}\left( {s - 1} \right) + D{s^2}\left( {s - 9} \right)\]\(s\) ning tegishli qiymatlarini tanlash va doimiylarni echish,
\[\begin{align*} & s = 0 & 5 & = 9B & \Rightarrow \hspace{0.25in}B & = \frac{5}{9}\\ & s = 1 & 16 & = - 8D & \Rightarrow \hspace{0.25in}D & = - 2\\ & s = 9 & 248 & = 648C & \Rightarrow \hspace{0.25in}C & = \frac{{31}}{{81}}\\ & s = 2 & 45 & = - 14A + \frac{{4345}}{{81}} & \Rightarrow \hspace{0.25in}A & = \frac{{50}}{{81}}\end{align*}\]Konstantalarni ulash,
\[Y\left( s \right) = \frac{{\frac{{50}}{{81}}}}{s} + \frac{{\frac{5}{9}}}{{{s^2}}} + \frac{{\frac{{31}}{{81}}}}{{s - 9}} - \frac{2}{{s - 1}}\]Nihoyat, teskari konvertatsiya qilish bizga IVP yechimini beradi.
\[y\left( t \right) = \frac{{50}}{{81}} + \frac{5}{9}t + \frac{{31}}{{81}}{{\bf{e}}^{9t}} - 2{{\bf{e}}^t}\]Bu avvalgi bobdagi usullardan foydalangan holda tezroq hal qilinishi mumkin bo'lgan muammo uchun juda ko'p ish edi. Ammo bu muammoning maqsadi Laplasdan qanday foydalanishimizni ko'rsatish edi IVPni hal qilish uchun o'zgartiradi.
IVP ni hal qilish uchun Laplas konvertatsiyalaridan foydalanishda bir nechta narsani ta'kidlash kerak. Birinchidan, Laplace yordamida transformatsiyalar differensial tenglamani algebra muammosiga qisqartiradi. Oxirgi misolda algebra oxirgi bobdagi to'g'ridan-to'g'ri yondashuvdan ko'ra murakkabroq edi. Biroq, keyingi muammolarda bu teskari bo'ladi. Algebra hali juda tartibsiz bo'lsa-da, ko'pincha to'g'ridan-to'g'ri yondashuvdan ko'ra osonroq bo'ladi.
Ikkinchidan, oxirgi bobdagi yondashuvdan farqli o'laroq, yechimni olish uchun avval umumiy yechim topish, uni farqlash, boshlang'ich shartlarni ulash va keyin konstantalarni hal qilish kerak emas edi. Laplas konvertatsiyalari bilan dastlabki shartlar birinchi bosqichda qo'llaniladi va oxirida biz umumiy yechim o'rniga haqiqiy yechimni olamiz.
Ko'pgina keyingi muammolarda Laplas O'zgartirishlar, agar biz oxirgi bobdagi to'g'ridan-to'g'ri yondashuvni bajarganimizdan ko'ra, muammolarni hal qilishni sezilarli darajada osonlashtiradi. Bundan tashqari, biz ko'rib turganimizdek, oxirgi bobdagi usullardan foydalangan holda amalga oshirib bo'lmaydigan ba'zi differensial tenglamalar mavjud va shuning uchun bunday hollarda Laplas o'zgarishlari bizning yagona yechimimiz bo'ladi.
Keling, yana bir oddiy muammoni ko'rib chiqaylik.
Yechim
Birinchi misoldagidek, avval differensial tenglamadagi barcha atamalarning Laplas konvertatsiyasini olaylik. Qabul qilish uchun biz boshlang'ich sharoitlarda vilka qilamiz,
\[\begin{align*}2\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right) + 3\left( {sY\left( s \right) - y\left( 0 \right)} \right) - 2Y\left( s \right) & = \frac{1}{{{{\left( {s + 2} \right)}^2}}}\\ \left( {2{s^2} + 3s - 2} \right)Y\left( s \right) + 4 & = \frac{1}{{{{\left( {s + 2} \right)}^2}}}\end{align*}\]Endi \(Y(s)\) ni hal qiling.
\[Y\left( s \right) = \frac{1}{{\left( {2s - 1} \right){{\left( {s + 2} \right)}^3}}} - \frac{4}{{\left( {2s - 1} \right)\left( {s + 2} \right)}}\]Endi, oxirgi misolda qilganimizdek, biz davom etamiz va ikkita atamani birlashtiramiz, chunki baribir birinchi maxrajni qisman kasrga ajratishimiz kerak bo'ladi, shuning uchun biz hisoblagichni biroz murakkabroq qilishimiz va faqat bitta qisman kasr qilishimiz mumkin. Bu beradi,
\[\begin{align*}Y\left( s \right) & = \frac{{1 - 4{{\left( {s + 2} \right)}^2}}}{{\left( {2s - 1} \right){{\left( {s + 2} \right)}^3}}}\\ & = \frac{{ - 4{s^2} - 16s - 15}}{{\left( {2s - 1} \right){{\left( {s + 2} \right)}^3}}}\end{align*}\]Bunda qisman fraksiya parchalanishi,
\[Y\left( s \right) = \frac{A}{{2s - 1}} + \frac{B}{{s + 2}} + \frac{C}{{{{\left( {s + 2} \right)}^2}}} + \frac{D}{{{{\left( {s + 2} \right)}^3}}}\]Numeratorning tengligini o'rnatish,
\[\begin{align*} - 4{s^2} - 16s - 15 & = A{\left( {s + 2} \right)^3} + B\left( {2s - 1} \right){\left( {s + 2} \right)^2} + C\left( {2s - 1} \right)\left( {s + 2} \right) + D\left( {2s - 1} \right)\\ & = \left( {A + 2B} \right){s^3} + \left( {6A + 7B + 2C} \right){s^2} + \left( {12A + 4B + 3C + 2D} \right)s\\ & \hspace{2.25in} + 8A - 4B - 2C - D\end{align*}\]Bunday holda, \(s\) qiymatlarini tanlashdan ko'ra, koeffitsientlarni tenglashtirish va natijada paydo bo'lgan tenglama tizimini echish osonroq bo'ladi. Shunday qilib, bu erda tizim va uning echimi.
\[\left. {\begin{aligned} & {s^3}: & A + 2B & = 0\\& {s^2}: & 6A + 7B + 2C & = - 4\\ & {s^1}: & 12A + 4B + 3C + 2D & = - 16\\& {s^0}: & 8A - 4B - 2C - D & = - 15\end{aligned}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned}A & = - \frac{{192}}{{125}} & B & = \frac{{96}}{{125}}\\ C & = - \frac{2}{{25}} & D & = - \frac{1}{5}\end{aligned}\]Biz ushbu koeffitsientlarning barchasida 125 ning umumiy maxrajini olamiz va ularni transformatsiyaga qayta ulashga kirishganimizda buni hisobga olamiz. Buni qilish,
\[Y\left( s \right) = \frac{1}{{125}}\left( {\frac{{ - 192}}{{2\left( {s - \frac{1}{2}} \right)}} + \frac{{96}}{{s + 2}} - \frac{{10}}{{{{\left( {s + 2} \right)}^2}}} - \frac{{25\frac{{2!}}{{2!}}}}{{{{\left( {s + 2} \right)}^3}}}} \right)\]E'tibor bering, biz birinchi hadning maxrajidan 2 ni ko'paytirishimiz va o'zgartirishlar jadvalimizdagi to'g'ri yozuvlarga mos kelishi uchun oxirgi hadning hisobini aniqlashimiz kerak edi.
Teskari konvertatsiyani qabul qilib, natija beradi,
\[y\left( t \right) = \frac{1}{{125}}\left( { - 96{{\bf{e}}^{\frac{t}{2}}} + 96{{\bf{e}}^{ - 2t}} - 10t{{\bf{e}}^{ - 2t}} - \frac{{25}}{2}{t^2}{{\bf{e}}^{ - 2t}}} \right)\]Yechim
Hamma narsaning Laplas konvertatsiyasini oling va boshlang'ich sharoitlarni ulang.
\[\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 6\left( {sY\left( s \right) - y\left( 0 \right)} \right) + 15Y\left( s \right) & = 2\frac{3}{{{s^2} + 9}}\\ \left( {{s^2} - 6s + 15} \right)Y\left( s \right) + s - 2 & = \frac{6}{{{s^2} + 9}}\end{align*}\]Endi \(Y(s)\) ni hal qiling va oldingi ikkita misolda qilganimizdek bitta atamaga birlashtiring.
\[Y\left( s \right) = \frac{{ - {s^3} + 2{s^2} - 9s + 24}}{{\left( {{s^2} + 9} \right)\left( {{s^2} - 6s + 15} \right)}}\]Endi bu borada qisman kasrlarni bajaring. Avval qisman kasr parchalanishini olaylik.
\[Y\left( s \right) = \frac{{As + B}}{{{s^2} + 9}} + \frac{{Cs + D}}{{{s^2} - 6s + 15}}\]Endi hisoblagichlarni teng o'rnatish,
\[\begin{align*} - {s^3} + 2{s^2} - 9s + 24 & = \left( {As + B} \right)\left( {{s^2} - 6s + 15} \right) + \left( {Cs + D} \right)\left( {{s^2} + 9} \right)\\ & = \left( {A + C} \right){s^3} + \left( { - 6A + B + D} \right){s^2} + \left( {15A - 6B + 9C} \right)s + 15B + 9D\end{align*}\]Koeffitsientlarni teng o'rnatish va konstantalarni hal qilish,
\[\left. \begin{aligned}& {s^3}: & A + C & = - 1\\ & {s^2}: & - 6A + B + D & = 2\\ & {s^1}: & 15A - 6B + 9C & = - 9\\ & {s^0}: & 15B + 9D & = 24\end{aligned}\right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} A & = \frac{1}{{10}} & B & = \frac{1}{{10}}\\ C & = - \frac{{11}}{{10}} & D & = \frac{5}{2}\end{aligned}\]Endi ularni parchalanishga ulang, ikkinchi hadning maxrajidagi kvadratni to'ldiring va keyin teskari aylantirish jarayoni uchun hisoblagichlarni aniqlang.
\[\begin{align*}Y\left( s \right) & = \frac{1}{{10}}\left( {\frac{{s + 1}}{{{s^2} + 9}} + \frac{{ - 11s + 25}}{{{s^2} - 6s + 15}}} \right)\\ & = \frac{1}{{10}}\left( {\frac{{s + 1}}{{{s^2} + 9}} + \frac{{ - 11\left( {s - 3 + 3} \right) + 25}}{{{{\left( {s - 3} \right)}^2} + 6}}} \right)\\ & = \frac{1}{{10}}\left( {\frac{s}{{{s^2} + 9}} + \frac{{1\frac{3}{3}}}{{{s^2} + 9}} - \frac{{11\left( {s - 3} \right)}}{{{{\left( {s - 3} \right)}^2} + 6}} - \frac{{8\frac{{\sqrt 6 }}{{\sqrt 6 }}}}{{{{\left( {s - 3} \right)}^2} + 6}}} \right)\end{align*}\]Nihoyat, teskari konvertatsiya qiling.
\[y\left( t \right) = \frac{1}{{10}}\left( {\cos \left( {3t} \right) + \frac{1}{3}\sin \left( {3t} \right) - 11{{\bf{e}}^{3t}}\cos \left( {\sqrt 6 t} \right) - \frac{8}{{\sqrt 6 }}{{\bf{e}}^{3t}}\sin \left( {\sqrt 6 t} \right)} \right)\]Shu paytgacha biz faqat IVP-larni ko'rib chiqdik, unda boshlang'ich qiymatlari \(t = 0\) da bo'lgan. Buning sababi, biz Laplasni olishimiz uchun boshlang'ich qiymatlar shu nuqtada bo'lishi kerak hosilalarning transformatsiyasi. Bularning barchasi bilan bog'liq muammo shundaki, dunyoda \(t = 0\) dan boshqa joylarda boshlang'ich qiymatlarga ega bo'lgan IVPlar mavjud. Laplas Agar biz ushbu turdagi IVP-larda foydalana olmasak, transformatsiyalar unchalik foydali bo'lmaydi. Shunday qilib, biz ushbu turdagi muammolarni qanday hal qilishni ko'rish uchun dastlabki shartlar \(t = 0\) da bo'lmagan misolni ko'rib chiqishimiz kerak.
Yechim
Bu erda qilishimiz kerak bo'lgan birinchi narsa, dastlabki shartlar \(t = 0\) da emasligiga e'tibor berishdir. Biz hosilalarning Laplas konvertatsiyasini qabul qilishning yagona yo'li \(t = 0\) da dastlabki shartlarga ega bo'lishdir.
Bu shuni anglatadiki, biz IVP ni dastlabki shartlar \(t = 0\) da bo'ladigan tarzda shakllantirishimiz kerak bo'ladi. Buni amalga oshirish juda oddiy, ammo biz uni ishlashi uchun o'zgaruvchini o'zgartirishimiz kerak bo'ladi. Biz aniqlaymiz
\[\eta = t - 3\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,t = \eta + 3\]Keling, asl differensial tenglamadan boshlaylik.
\[y''\left( t \right) + 4y'\left( t \right) = \cos \left( {t - 3} \right) + 4t\]E'tibor bering, biz bu erda narsalarni to'g'ri olishimizga ishonch hosil qilish uchun hosilalarga \(\left( t \right)\) qismini qo'yganmiz. Keyinchalik \(t\) ni almashtiramiz.
\[y''\left( {\eta + 3} \right) + 4y'\left( {\eta + 3} \right) = \cos \left( \eta \right) + 4\left( {\eta + 3} \right)\]Endi, hayotni biroz soddalashtirish uchun, keling, aniqlaymiz:
\[u\left( \eta \right) = y\left( {\eta + 3} \right)\]Keyin, zanjir qoidasiga ko'ra, biz birinchi hosila uchun quyidagilarni olamiz.
\[u'\left( \eta \right) = \frac{{du}}{{d\eta }} = \frac{{dy}}{{dt}}\frac{{dt}}{{d\eta }} = y'\left( {\eta + 3} \right)\]Shunga o'xshash dalil orqali biz ikkinchi hosila uchun quyidagini olamiz.
\[u''\left( \eta \right) = y''\left( {\eta + 3} \right)\]\(u\left( \eta \right)\) uchun dastlabki shartlar:
\[\begin{align*}u\left( 0 \right) & = y\left( {0 + 3} \right) = y\left( 3 \right) = 0\\ u'\left( 0 \right) & = y'\left( {0 + 3} \right) = y'\left( 3 \right) = 7\end{align*}\]Ushbu yangi o'zgaruvchilar ostida IVP keyin,
\[u'' + 4u' = \cos \left( \eta \right) + 4\eta + 12,\hspace{0.25in}u\left( 0 \right) = 0\,\,\,\,\,\,\,\,u'\left( 0 \right) = 7\]Bu biz barcha \(t\)-larni almashtirsak, Laplas konvertatsiyalaridan foydalanishimiz mumkin bo'lgan IVP. stol \(\eta \) bilan. Shunday qilib, ushbu yangi differensial tenglamaning Laplas konvertatsiyasini olib, yangi boshlang'ich shartlarni qo'shsak,
\[\begin{align*}{s^2}U\left( s \right) - su\left( 0 \right) - u'\left( 0 \right) + 4\left( {sU\left( s \right) - u\left( 0 \right)} \right) & = \frac{s}{{{s^2} + 1}} + \frac{4}{{{s^2}}} + \frac{{12}}{s}\\ \left( {{s^2} + 4s} \right)U\left( s \right) - 7 & = \frac{s}{{{s^2} + 1}} + \frac{{4 + 12s}}{{{s^2}}}\end{align*}\]\(U(s)\) ni hal qilish,
\[\begin{align*}\left( {{s^2} + 4s} \right)U\left( s \right) & = \frac{s}{{{s^2} + 1}} + \frac{{4 + 12s + 7{s^2}}}{{{s^2}}}\\ U\left( s \right) & = \frac{1}{{\left( {s + 4} \right)\left( {{s^2} + 1} \right)}} + \frac{{4 + 12s + 7{s^2}}}{{{s^3}\left( {s + 4} \right)}}\end{align*}\]E'tibor bering, oldingi misollardan farqli o'laroq, biz bu safar barcha atamalarni to'liq birlashtirmadik. Oldingi barcha misollarda biz shunday qildik, chunki atamalardan birining maxraji barcha atamalar uchun umumiy maxraj edi. Shuning uchun, birlashtirgandan so'ng, biz hisoblagichni biroz chalkashroq qildik va zarur bo'lgan qisman kasrlar sonini ikkitadan bittaga kamaytirdik. E'tibor bering, ushbu konvertatsiyadagi barcha atamalar maxrajda faqat \(s\) vakolatiga ega bo'lgan atamalar aynan shu sababga ko'ra birlashtirilgan.
Biroq, bu o'zgartirishda, agar qolgan ikkala atamani bitta atamaga birlashtirsak, bizda qisman qismli kasr muammosi qoladi. Shuning uchun, bu holda, qisman kasrlarni ikki marta bajarish osonroq bo'ladi. Biz ushbu bo'limda bir nechta qisman kasr masalalarini va oldingi ikki bo'limda ko'plab qisman kasr masalalarini bajardik, shuning uchun biz qisman kasrlar tafsilotlarini tekshirish uchun sizga qoldiramiz. O'zgartirishimizdagi har bir atamani qisman qismlarga ajratish bizga quyidagilarni beradi.
\[\begin{align*}\frac{1}{{\left( {s + 4} \right)\left( {{s^2} + 1} \right)}} & = \frac{{\frac{1}{{17}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s + 4}}{{{s^2} + 1}}} \right)\\ \frac{{4 + 12s + 7{s^2}}}{{{s^3}\left( {s + 4} \right)}} & = \frac{1}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \frac{{\frac{{17}}{{16}}}}{s} - \,\frac{{\frac{{17}}{{16}}}}{{s + 4}}\end{align*}\]Ularni o'zgartirishimizga kiritish va o'xshash atamalarni birlashtirish bizga beradi
\[\begin{align*}U\left( s \right) & = \frac{1}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \frac{{\frac{{17}}{{16}}}}{s} - \frac{{\frac{{273}}{{272}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s + 4}}{{{s^2} + 1}}} \right)\\ & = \frac{{1\frac{{2!}}{{2!}}}}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \,\frac{{\frac{{17}}{{16}}}}{s} - \frac{{\frac{{273}}{{272}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s}}{{{s^2} + 1}} + \frac{4}{{{s^2} + 1}}} \right)\end{align*}\]Endi teskari konvertatsiya qilish bizning yangi IVP ga yechim beradi. \(t\) o'rniga \(\eta \) dan foydalanishni unutmang!
\[u\left( \eta \right) = \frac{1}{2}{\eta ^2} + \frac{{11}}{4}\eta + \frac{{17}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\eta }} + \frac{1}{{17}}\left( {4\sin \left( \eta \right) - \cos \left( \eta \right)} \right)\]Bu, albatta, biz izlayotgan yechim emas. Biz \(y(t)\) ortidan turibmiz. Biroq, biz buni payqash orqali olishimiz mumkin
\[y\left( t \right) = y\left( {\eta + 3} \right) = u\left( \eta \right) = u\left( {t - 3} \right)\]Shunday qilib, asl IVP ning yechimi,
\[\begin{align*}y\left( t \right) & = \frac{1}{2}{\left( {t - 3} \right)^2} + \frac{{11}}{4}\left( {t - 3} \right) + \frac{{17}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\left( {t - 3} \right)}} + \frac{1}{{17}}\left( {4\sin \left( {t - 3} \right) - \cos \left( {t - 3} \right)} \right)\\ y\left( t \right) & = \frac{1}{2}{t^2} - \frac{1}{4}t - \frac{{43}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\left( {t - 3} \right)}} + \frac{1}{{17}}\left( {4\sin \left( {t - 3} \right) - \cos \left( {t - 3} \right)} \right)\end{align*}\]Shunday qilib, biz endi \(t = 0\) da boshlang'ich shartlari bo'lmagan IVP-ni amalga oshirishimiz mumkin. Bundan tashqari, oxirgi misolda biz ushbu misoldan oldin qilganimizdek, barcha atamalarni bitta qisman kasr muammosiga birlashtirish har doim ham yaxshi emasligini ko'rdik.
Agar biz oldingi bobdagi usullardan foydalanganimizda, bu bo'limda ishlagan misollar ham xuddi shunday oson bo'lar edi. Ular bu yerda Laplas yordamida ishlagan texnika va usulni tasvirlash uchun o'zgartiradi.