Fizika-matematika fakulteti

1-misol. Ushbu tenglama
y"xy'y 0
va
y(0) 1, y'(0) 0
boshlangich shartlar berilgan. Bularga mos integral tenglama tuzilsin. Yechish. Noma‟lum funksiyaning ikkinchi hosilasini quyidagicha
belgilaymiz.
y"u(x)
Bundan esa quyidagi kelib chiqadi:


x

1
^y'^{u(t)dt c .} 0
Berilgan shartlarga ko„ra x=0 bo„lganda u’=0 bo„ladi, demak c₁=0. Shuning uchun
x
^y'^u(t)dt. 0
U holda bu yerdan
x x x x
^{y }_^u(t)dt_^{dt c2 }^dt^{u(t)dt c2 .} 0 0 0 0
Endi x=0 bo„lganda y=1 bo„lgani sababli, so„nggi tenglikdan c₂=1 kelib chiqadi. Demak,


x x
^{y }^dt^{u(t)dt 1.} 0 0


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Koshi formulasiga asosan buni
x
^{y }^{(x t)u(t)dt 1} 0
ko„rinishda yozish mumkin.
_{Mana shu y}_,y_{,y lar uchun aniqlangan ifodalarni berilgan}
differensial tenglamaga qo„yamiz:
x x
^{y"xy'y u(x)x}^{u(t)dt }^{(xt)u(t)dt 10.} 0 0
Bu ifodadagi integrallarni birlashtirsak, ushbu


x
^{u(x) 1}^{(t 2x)u(t)dt} 0
integral tenglama hosil bo„ladi.


2-misol. Ushbu differensial tenglama y"y cos x
va
y(0) 0, y'(0) 1
Boshlang„ich shartlari berilgan. Bularga mos integral tenglama tuzilsin. Yechish. Odatdagicha
y"u(x)
deb belgilaymiz. Bundan
x

1
^y'^{u(t)dt c .} 0
Boshlangich shartlarga ko„ra x=0 bo„lganda y´=1, shu sababli, c₁=1 bo„ladi. Demak,
x
^y'1^u(t)dt. 0
Bundan yana bir marta integral olinsa,
x x x x
^{y }_^1^u(t)dt_^{dt c2 x}^dt^{u(t)dt c2.} 0 0 0 0
Boshlang„ich shartlarga ko„ra x=0 bo„lganda, u=0 bo„lishi kerak, shu sababli c₂=0 bo„ladi. Koshining yuqorida keltirilgan formulasiga muvofiq

13
x
^{y x }^{(x t)u(t)dt.} 0
Endi berilgan differensial tenglamaga y’, y” lar uchun aniqlangan ifodalarni qo„yamiz, u holda
x
^{y"y u(x)x}^{(xt)u(t)dt cos x.} 0
Bundan esa ushbu


x
^{u(x) x cos x }^{(t x)u(t)dt} 0
integral tenglama kelib chiqadi.


3-misol. Ushbu
y'''3y"6y'8y 0
differensial tenglama va
y(0) 1, y'(0) 1, y"(0) 1
boshlangich shartlar berilgan. Bularga mos integral tenglama tuzilsin. Yechish. Odatdagicha, ^{y'''(x)} deb belgilab olamiz va uning ikki
tomonini dx ga ko„paytirib, so„ngra integrallaymiz:


x

1
^y"^{u(t)dt c .} 0
Boshlangich shartlarga ko„ra bundan c₁=1 kelib chiqadi. U holda
x
^y"1^u(t)dt. 0
Bundan yana integral olinsa,
x x x x
^y'_^1^u(t)dt_^{dx c2 x c2 }^dx^u(t)dx 0 0 0 0
kelib chiqadi. Boshlangich shartlarga ko„ra c₂=1 bo„ladi. Demak,
x x
^{y'1x }^dx^u(t)dt 0 0
bo„lib, undan so„ngi marta integral olsak,
^{y }_^{1t }^dt^u(t)dt_^{dt c3 x x}2 ^^dt^dt^{u(t)dt c3} 0 0 0 0 0 0

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hosil bo„ladi. Boshlangich shartlarga asosan c₃ = 1 bo„ladi.
Endi Koshining (1.13) formulasiga muvofiq takroriy integrallarni oddiy integralga aylantirilsa va ^{y''', y", y', y} lar uchun aniqlangan ifodalarni berilgan differensial tenglamaga qo„yib ixchamlashtirilsa, quyidagi natija kelib chiqadi:


 

x
^{u(x) 12x4x}2 ^^{36(xt)4(xt)}2 ^u(t)dt. 0
Agar K(x,t) yadro t ga nisbatan n-tartibli ko„phad bo„lsa, ya‟ni


n
^{K(x,t) }^am^(x)tm^{, (1.16)} m0
u holda (1.14) integral tenglamani ketma-ket differensiallash orqali uni chiziqli differensial tenglama uchun Koshi masalasiga keltirish mumkin.


4-Misol. Ushbu
_y_2y_{y x}² _{, y(0)=1, y(0)=0} Koshi masalasiga mos keluvchi integral tenglamani tuzing.
Yechish. y(x) = u(x) almashtirish olib, boshlang„ich shartlarni e‟tiborga olib, quyidagilarni ketma-ket topamiz:
x x
^{y(x) y(0)}^{u(t)dt }^u(t)dt, 0 0

Bularni kelamiz:
yoki
x s x
^{y(x) y(0)}^ds^{u(t)dt 1}^{(x t)u(t)dt} 0 0 0
berilgan differensial tenglamaga qo„yib, quyidagi natijaga


_u(x) _2x _{u(t)dt }₁_x _{(x t)u(t)dt}_x² 0  0 


x
^{u(x) x}2 ^1^{(2x t)u(t)dt.} 0

5-Misol. Ushbu
x
^u(x)^{(2x t)u(t)dt x}2 ^(1.17) 0


15

integral tenglamani yeching.
Yechish. Tenglikning ikkala tomonini ikki marta ketma-ket differensiallab, quyidagilarni hosil qilamiz:
x
^{u(x)2u(x)}^{u(t)dt 2x, (1.18)} 0
_u_(x)2u_{(x)u(x) 2. (1.19)}
Endi (1.17) va (1.18) dan x 0 da ushbu
u(0) 0, u^(0) 0
boshlang„ich shartlarni hosil qilamiz. Odatdagi uslubda (1.19) differensial tenglamani yechsak, ushbu
u(x) 22e^x(1x) _(1.20) yechimga ega bo„lamiz.

6-Misol. Ushbu
x
^{u(x) sin x}^{sin( xt)u(t)dt} 0
integral tenglamani yeching.