Ushbu tenglamada o'zgaruvchilarni osongina ajratish mumkin. Buning uchun \(y^2\) ni chap tomonga, \(dx\) ni o'ng tomonga o'tkazamiz va har ikkala tomonni integrallaymiz:

\[\begin{align*} y^{-2}\,dy & = 6x\,dx\\ \int y^{-2}\,dy & = \int 6x\,dx\\ -\frac{1}{y} & = 3x^2 + c \end{align*}\]

E'tibor bering: integrallash doimiysini ikkala tomonga ham yozish mumkin edi, ammo ularni bitta tomonga o'tkazib, umumiy \(c\) o'zgarmasiga birlashtirish kifoyadir. Biz \(c\) doimiysini an'anaga ko'ra \(x\) turgan o'ng tomonda qoldiramiz.

Endi bizda oshkormas yechim mavjud: \(-\frac{1}{y} = 3x^2 + c\). Bu yerdan oshkor \(y(x)\) ni topishdan oldin, boshlang'ich shart yordamida \(c\) ning son qiymatini aniqlab olish ancha qulayroqdir. \(x = 1\) va \(y = 1/25\) ekanidan:

\[-\frac{1}{1/25} = 3(1)^2 + c \implies -25 = 3 + c \implies c = -28\]

\(c = -28\) qiymatni yechimga qo'yib, \(y\) ni topamiz:

\[-\frac{1}{y} = 3x^2 - 28 \implies \frac{1}{y} = 28 - 3x^2 \implies y(x) = \frac{1}{28 - 3x^2}\]

Bu tenglamaning oshkor yechimidir. Endi ushbu yechimning mavjudlik oralig'i ni topamiz.

Yechimning mavjudlik oralig'i uchun ikkita qat'iy talab bor: birinchidan, u hech qanday uzilishlarsiz uzluksiz oraliq bo'lishi kerak; ikkinchidan, u boshlang'ich shartdagi \(x = 1\) nuqtani o'z ichiga olishi shart.

Bizning yechimimiz kasr ko'rinishida bo'lgani sababli, maxraj nolga teng bo'lishiga yo'l qo'ymasligimiz kerak:

\[28 - 3x^2 = 0 \implies x^2 = \frac{28}{3} \implies x = \pm\sqrt{\frac{28}{3}} \approx \pm 3.05505\]

Ushbu ikki uzilish nuqtasi sonlar o'qini uchta mumkin bo'lgan oraliqqa ajratadi:

\[-\infty < x < -\sqrt{\frac{28}{3}},\hspace{0.25in} -\sqrt{\frac{28}{3}} < x < \sqrt{\frac{28}{3}},\hspace{0.25in} \sqrt{\frac{28}{3}} < x < \infty\]

Boshlang'ich shartdagi \(x = 1\) nuqta aynan ikkinchi oraliqda (\(-\sqrt{28/3} < 1 < \sqrt{28/3}\)) yotadi. Demak, ushbu Koshi masalasi yechimining mavjudlik oralig'i:

\[-\sqrt{\frac{28}{3}} < x < \sqrt{\frac{28}{3}}\]

Quyida yechimning grafigi keltirilgan:

Eslatma: agar boshlang'ich shart boshqacha berilganida, masalan \(y(-4) = -1/20\) bo'lganida, mavjudlik oralig'i birinchi oraliq \((-\infty, -\sqrt{28/3})\) bo'lar edi; agar \(y(6) = -1/80\) bo'lganida esa uchinchi oraliq \((\sqrt{28/3}, \infty)\) bo'lar edi.

O'zgaruvchilarni ajratamiz va har ikkala tomonni integrallaymiz:

\[\begin{align*} (2y - 4)\,dy & = (3x^2 + 4x - 4)\,dx\\ \int (2y - 4)\,dy & = \int (3x^2 + 4x - 4)\,dx\\ y^2 - 4y & = x^3 + 2x^2 - 4x + c \end{align*}\]

Oshkormas yechim hosil bo'ldi. Dastlab \(c\) ni topish uchun \(x = 1\) va \(y = 3\) shartni qo'yamiz:

\[(3)^2 - 4(3) = (1)^3 + 2(1)^2 - 4(1) + c \implies 9 - 12 = 1 + 2 - 4 + c \implies -3 = -1 + c \implies c = -2\]

Demak, oshkormas yechim quyidagicha:

\[y^2 - 4y = x^3 + 2x^2 - 4x - 2\]

Endi ushbu tenglikdan oshkor \(y(x)\) ni ajratishimiz kerak. Buning uchun barcha hadlarni chap tomonga o'tkazib, \(y\) ga nisbatan kvadrat tenglama sifatida qaraymiz:

\[y^2 - 4y - (x^3 + 2x^2 - 4x - 2) = 0\]

Kvadrat tenglamaning ildizlar formulasiga ko'ra (\(a = 1, b = -4, c = -(x^3 + 2x^2 - 4x - 2)\)):

\[\begin{align*} y(x) & = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-(x^3 + 2x^2 - 4x - 2))}}{2}\\ & = \frac{4 \pm \sqrt{16 + 4(x^3 + 2x^2 - 4x - 2)}}{2}\\ & = \frac{4 \pm 2\sqrt{4 + x^3 + 2x^2 - 4x - 2}}{2}\\ & = 2 \pm \sqrt{x^3 + 2x^2 - 4x + 2} \end{align*}\]

Bu yerda ikkita ishora (\(\pm\)) mavjud, ammo funksiya bitta qiymatli bo'lishi kerak. Ikkala ishoradan qaysi biri to'g'ri ekanini bilish uchun boshlang'ich shartni (\(x = 1\) da \(y = 3\)) qo'yib tekshiramiz:

\[y(1) = 2 \pm \sqrt{1 + 2 - 4 + 2} = 2 \pm \sqrt{1} = 2 \pm 1\]

\(2 + 1 = 3\) va \(2 - 1 = 1\). Bizga \(y = 3\) kerak bo'lgani sababli, musbat ("+") ishora to'g'ri keladi!

Shunday qilib, izlanayotgan oshkor yechim:

\[y(x) = 2 + \sqrt{x^3 + 2x^2 - 4x + 2}\]

Endi yechimning mavjudlik oralig'ini topaylik. Haqiqiy sonlar to'plamida qolishimiz uchun ildiz ostidagi ifoda nomanfiy bo'lishi shart:

\[x^3 + 2x^2 - 4x + 2 \ge 0\]

Ushbu uchinchi darajali ko'phad bitta haqiqiy ildizga ega: \(x \approx -3.36523\) (qolgan ikki ildizi kompleks sonlardir). Quyidagi grafikdan ko'rinib turibdiki, ko'phad \(x \ge -3.36523\) bo'lganda musbat bo'ladi:

Ushbu oraliq boshlang'ich shartdagi \(x = 1\) nuqtani o'z ichiga oladi. Demak, yechimning mavjudlik oralig'i: \(x \ge -3.36523\) (yoki differensial tenglamada hosila mavjud bo'lishi uchun ochiq oraliq: \(x > -3.36523\)) bo'ladi.

Quyida topilgan yechimning to'liq grafigi keltirilgan:

O'zgaruvchilarni ajratamiz:

\[y^{-3}\,dy = \frac{x}{\sqrt{1 + x^2}}\,dx = x(1 + x^2)^{-1/2}\,dx\]

Har ikkala tomonni integrallaymiz (o'ng tomonda \(u = 1 + x^2, du = 2x\,dx\) almashtirish bajariladi):

\[\int y^{-3}\,dy = \int x(1 + x^2)^{-1/2}\,dx \implies -\frac{1}{2y^2} = \sqrt{1 + x^2} + c\]

\(x = 0\) va \(y = -1\) boshlang'ich shartni qo'llaymiz:

\[-\frac{1}{2(-1)^2} = \sqrt{1 + 0} + c \implies -\frac{1}{2} = 1 + c \implies c = -\frac{3}{2}\]

Hosil bo'lgan oshkormas yechim:

\[-\frac{1}{2y^2} = \sqrt{1 + x^2} - \frac{3}{2}\]

Endi \(y(x)\) ni topamiz:

\[\begin{align*} \frac{1}{y^2} & = 3 - 2\sqrt{1 + x^2}\\ y^2 & = \frac{1}{3 - 2\sqrt{1 + x^2}}\\ y(x) & = \pm \frac{1}{\sqrt{3 - 2\sqrt{1 + x^2}}} \end{align*}\]

Boshlang'ich shart \(y(0) = -1\) (manfiy son) bo'lganligi sababli, ildiz oldidagi manfiy ("-") ishora tanlanadi:

\[y(x) = -\frac{1}{\sqrt{3 - 2\sqrt{1 + x^2}}}\]

Mavjudlik oralig'i: Ildiz osti musbat bo'lishi va maxraj nolga teng bo'lmasligi shart:

\[3 - 2\sqrt{1 + x^2} > 0 \implies 3 > 2\sqrt{1 + x^2} \implies 9 > 4(1 + x^2) \implies \frac{9}{4} > 1 + x^2 \implies x^2 < \frac{5}{4}\]

Bundan ko'rinib turibdiki, mavjudlik oralig'i quyidagicha bo'ladi:

\[-\frac{\sqrt{5}}{2} < x < \frac{\sqrt{5}}{2}\]

Ushbu oraliq \(x = 0\) boshlang'ich nuqtani o'z ichiga oladi. Quyida yechimning grafigi keltirilgan:

O'zgaruvchilarni ajratamiz (har ikki tomonni \(e^y\) ga ko'paytiramiz):

\[e^y\,dy = (2x - 4)\,dx\]

Ikkala tomonni integrallaymiz:

\[\int e^y\,dy = \int (2x - 4)\,dx \implies e^y = x^2 - 4x + c\]

\(x = 5\) va \(y = 0\) boshlang'ich shartni qo'yamiz:

\[e^0 = (5)^2 - 4(5) + c \implies 1 = 25 - 20 + c \implies 1 = 5 + c \implies c = -4\]

Demak, oshkormas yechim: \(e^y = x^2 - 4x - 4\). Har ikki tomonni natural logarifmlasak, oshkor yechim hosil bo'ladi:

\[y(x) = \ln(x^2 - 4x - 4)\]

Mavjudlik oralig'i: Logarifm faqat qat'iy musbat sonlar uchun aniqlangan:

\[x^2 - 4x - 4 > 0\]

Kvadrat uchhadning ildizlari: \(x = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}\).

Tengsizlikning yechimi ikkita oraliqdan iborat: \((-\infty, 2 - 2\sqrt{2})\) va \((2 + 2\sqrt{2}, \infty)\). Boshlang'ich shartdagi \(x = 5\) qiymat ikkinchi oraliqda yotadi (chunki \(2 + 2\sqrt{2} \approx 2 + 2(1.414) = 4.828 < 5\)).

Demak, yechimning mavjudlik oralig'i:

\[2 + 2\sqrt{2} < x < \infty\]

Yechim grafigi quyidagicha:

O'zgaruvchilarni ajratib, integrallaymiz:

\[\frac{1}{r^2}\,dr = \frac{1}{\theta}\,d\theta \implies \int r^{-2}\,dr = \int \frac{1}{\theta}\,d\theta \implies -\frac{1}{r} = \ln|\theta| + c\]

\(\theta = 1\) bo'lganda \(r = 2\) shartdan:

\[-\frac{1}{2} = \ln|1| + c \implies -\frac{1}{2} = 0 + c \implies c = -\frac{1}{2}\]

Oshkormas yechim: \(-\frac{1}{r} = \ln|\theta| - \frac{1}{2} \implies \frac{1}{r} = \frac{1}{2} - \ln|\theta|\).

Oshkor yechim:

\[r(\theta) = \frac{1}{\frac{1}{2} - \ln|\theta|}\]

Mavjudlik oralig'i: Bizda ikkita cheklov mavjud: birinchidan, \(\ln|\theta|\) logarifm mavjud bo'lishi uchun \(\theta \ne 0\); ikkinchidan, maxraj nolga teng bo'lmasligi kerak:

\[\frac{1}{2} - \ln|\theta| = 0 \implies \ln|\theta| = \frac{1}{2} \implies |\theta| = e^{1/2} = \sqrt{e} \implies \theta = \pm\sqrt{e}\]

Ushbu uchta nuqta (\(-\sqrt{e}\), \(0\), \(\sqrt{e}\)) sonlar o'qini 4 qismga ajratadi:

\[(-\infty, -\sqrt{e}),\quad (-\sqrt{e}, 0),\quad (0, \sqrt{e}),\quad (\sqrt{e}, \infty)\]

Boshlang'ich shartdagi \(\theta = 1\) nuqta (\(0 < 1 < \sqrt{e} \approx 1.6487\)) uchinchi oraliqda joylashgan. Demak, mavjudlik oralig'i:

\[0 < \theta < \sqrt{e}\]

Quyida yechimning grafigi keltirilgan:

Dastlab o'ng tomonni ko'paytuvchilarga ajratamiz: \(e^{y - t} = e^y e^{-t}\) va \(\sec(y) = \frac{1}{\cos y}\). Shunda:

\[\frac{dy}{dt} = \frac{e^y e^{-t}}{\cos y}(1 + t^2)\]

Endi \(y\) li ifodalarni chap tomonga, \(t\) li ifodalarni o'ng tomonga o'tkazamiz:

\[e^{-y}\cos y\,dy = e^{-t}(1 + t^2)\,dt\]

Har ikkala tomonni bo'laklab integrallash usuli bilan hisoblaymiz:

Chap tomon integrali: \(\int e^{-y}\cos y\,dy\). Ikki marta bo'laklab integrallasak (standart aylanma integral):

\[\int e^{-y}\cos y\,dy = \frac{e^{-y}}{2}(\sin y - \cos y)\]

O'ng tomon integrali: \(\int e^{-t}(1 + t^2)\,dt\). Ikki marta bo'laklab integrallasak:

\[\int e^{-t}(1 + t^2)\,dt = -e^{-t}(t^2 + 2t + 3) + c\]

Shunday qilib, umumiy yechim:

\[\frac{e^{-y}}{2}(\sin y - \cos y) = -e^{-t}(t^2 + 2t + 3) + c\]

\(t = 0\) va \(y = 0\) boshlang'ich shartni qo'yamiz (\(\sin 0 = 0, \cos 0 = 1, e^0 = 1\)):

\[\frac{1}{2}(0 - 1) = -(0 + 0 + 3) + c \implies -\frac{1}{2} = -3 + c \implies c = \frac{5}{2}\]

Natijada quyidagi oshkormas yechimga ega bo'lamiz:

\[\frac{e^{-y}}{2}(\sin y - \cos y) = -e^{-t}(t^2 + 2t + 3) + \frac{5}{2}\]

Ushbu tenglikdan \(y\) ni analitik tarzda oshkor ajratib olish (ya'ni \(y = \dots\) ko'rinishga keltirish) matematik jihatdan imkonsizdir. Shuning uchun yechim oshkormas ko'rinishda qoldiriladi. Shuningdek, oshkormas yechimlardan mavjudlik oralig'ini topish odatda juda murakkab bo'lgani bois, bunday masalalarda mavjudlik oralig'ini topish talab etilmaydi.