Vazifalar ketma-ketligi Последовательность задач

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Algoritmlash VAZIFALAR

Birinchi tartibli oddiy differentsial tenglama uchun Koshi masalasini taqribiy yechish 9.1-masala. Quyidagi birinchi tartibli differentsial tenglamaning x
Eyler usuli; Eylerning mukammallashgan usuli; Runge – Kutta usuli bilan hisoblang. Yechish.
Aniq integrallarni taqribiy hisoblashda Eyler va Koshi tenglamalari uchun quyidagi variantlarni qiymatlarini ikkalasida ham qo’llashingiz mumkin MUSTAQIL ISHLASH

7.39	9.39

Y		0,7	0,59	0,4	0,3	0,69

Aniq integrallarni taqribiy yechish usullari va ularning algoritm-dasturlari

Birinchi tartibli oddiy differentsial tenglama uchun
Koshi masalasini taqribiy yechish
9.1-masala. Quyidagi birinchi tartibli differentsial tenglamaning
x₀=1.8 y₀=2.6
boshlang’ich shartni qanoatlantiruvchi [1.8, 2.8] oraliqda yechimini h=0.1 qadami bilan, e=0.001 aniqlikda:

Eyler usuli;
Eylerning mukammallashgan usuli;
Runge – Kutta usuli bilan hisoblang.

Yechish.
1. Berilgan differentsial tenglamani Eyler usulida yyechamiz.
Buning uchun [1.8, 2.8] oraliqni

ya’ni, n=10 ta bo‘lakka ajratamiz. Bo‘linish nuqtalarini:
x_i=x_i-1+h, i=1,2,...,10
formulaga asosan topamiz.
x₁=x₀+h=1.8+0.1=1.9
x₂=x₁+h=1.9+0.1=2.0
shuningdek
x₃=2.1, x₄=2.2, x₅=2.3, x₆=2.4, x₇=2.5, x₈=2.6, x₉=2.7, x₁₀=2.8
Berilgan tenglamaning o‘ng tomonidagi
F(x;y)=x+cos(y/ )
funksiyaga asosan, Eyler qoidasi bilan quyidagi
y_i+1=y_i+ h f(x_i;y_i), i=1,2,...,10
formulaga asosan berilgan differentsial tenglama yechimining qiymatlarini quyidagicha topamiz.
y₁=y₀+hf (x_0,y₀)=y₀+h (x₀+cos(y₀/ ))=2.6+ 0.1(1.8+cos(26/ ))=2.6+0.1(18+0.3968)=2.81968
y₂=y₁+h f (x₁,y₁)=y₁+h(x₁+cos(y₁/ ))=2.819+ 0.1(1.9+cos(9.819/ ))=2.819+0.1(1.9+0.3968)=3.03948
SHuningdek, quyidagilarni topamiz:
y₃=3.261, y₄=3.4831, y₅=3.7045, y₆=3.926
y₇=4.1478, y₈=4.3701, y₉=4.5931, y₁₀=4.8173
Bu usul yordamida hisoblash quyidagicha dastur asosida berilgan.

Birinchi tartibli differentsial tenglama

Y¹=F(X,Y) uchun
Koshi masalasini Eyler usulida
taqribiy yechimini topish.
X(1)= 1.900 Y(1)= 2.8197
X(2)= 2.000 Y(2)= 3.0402
X(3)= 2.100 Y(3)= 3.2611
X(4)= 2.200 Y(4)= 3.4823
X(5)= 2.300 Y(5)= 3.7037
X(6)= 2.400 Y(6)= 3.9251
X(7)= 2.500 Y(7)= 4.1468

X(1)= 2.600 Y(1)= 4.3688
X(9)= 2.700 Y(9)= 4.5914
X(10)= 2.800 Y(10)= 4.8150

2. Tenglama yechimini Eylerning ketma-ket yaqinlashish usulida hisoblaymiz. (9.6) formulada i=0 bo‘lganda
y₁⁽⁰⁾=y₀+hf (x₀;y₀)=y₀+k (x₀+cos(y₀/ ))=
=2.6+ 0.1(9.8+cos(9.6/ ))=2.6+0.1(9.8+0.3968)=2.81968
bo‘ladi. Bu Eyler usulidagi tenglama yechimining birinchi qiymati bo‘ladi.
Endi y₁⁽⁰⁾=2.81968 dan foydalanib (9.7) formulaga asosan i=1 bo‘lganda
y₁^(k)= y₀+ [f (x₀, y₀)+ f (x₁,y₁^(k-1))]
formulani k = 1,2,3,….. lar uchun ketma-ket
y₁⁽¹⁾, y₁⁽²⁾, y₁⁽³⁾, …, y₁^(k)
larni
y₁^(k-1) – y₁^(k)  < 0.001
shartni qanoatlantirguncha hisoblaymiz.
Demak,
k=1, y₁⁽¹⁾=y₀+ [f (x₀,y₀)+f (x₁,y₁⁽⁰⁾)]=2.6+0.05[2.1968+x₁+cos (y₁⁽⁰⁾/ )] =
=2.6+0.05[2.1968+1.9+0.36486] =2.7102
k=2, y₁⁽²⁾= 2.6+0.05[2.1968+x₁+cos(y₁⁽¹⁾/ )] =
=2.6+0.05[2.1968+1.9+ cos(9.7102/ )] =2.82239
k=3, y₁⁽³⁾= 2.6+0.05[2.1968+1.9+cos(9.83339/ )] =
=2.6+0.05[2.1968+1.9+ 0.303709] =2.82002
Endi xatolikni tekshiramiz.
y₁⁽²⁾ – y₁⁽³⁾  =2.8223 – 2.82202=0.0002< 0.001
Bundan 0.001 aniqlikdagi tenglama yechimining birinchi qiymati
y₁ = 2.82000  2.82
bo‘ladi.
Tenglama yechimi y₂ qiymatini topish uchun yuqoridagi qoidani takrorlaymiz.
i=1 uchun (9.6) formulaga asosan
y₂⁽⁰⁾= y₁ + hf (x₁, y₁) = 2.82 + 0.1(x₁+cos (y₁/ )) =
=2.82+0.1(9.9+cos(9.82/ ) = 3.04047
i=1 uchun (9.7) formulaga asosan
k=1, y₂⁽¹⁾= y₁+ [f (x₁, y₁)+ f (x₂, y₂⁽⁰⁾)] =
=2.82+0.05[2.20471+2+cos(3.0404/ )] =3.0407
k=2, y₂⁽²⁾=y₁+ [f (x₁,y₁)+f (x₂,y₂⁽²⁾)]= y₁+ [2.20471+2+ cos(u₂⁽¹⁾/ )]=
=2.82+0.05[2.20471+2+cos(3.0407/ )] =3.04071
k=3, y₂⁽³⁾= y₁+ [f (x₁, y₁)+ f (x₂, y₂⁽²⁾)] =
=2.82+0.05[2.20471+2+cos (3.0407/ )] =
=2.82+0.05[2.20471+2+cos (3.0407/ )] =3.0407
Endi xatolikni baholaymiz.
y₁⁽²⁾ – y₁⁽³⁾  =3.04071 – 3.04070=0.0001< 0.001
Bundan tenglama yechimining ikkinchi qiymati
y₂ = 3.0407
bo‘ladi.
Bu qoidani i=2,3,…,10 lar uchun ketma-ket davom ettirib tenglama yechimining qolgan qiymatlarini ham topamiz.
y₃=3.261, y₄=3.483, y₅=3.704, y₆=3.926
y₇=4.147, y₈=4.370, y₉=4.593, y₁₀=4.817
Bu usul yordamida hisoblash quyidagicha dastur asosida berilgan.

Aniq integrallarni taqribiy hisoblashda Eyler va Koshi tenglamalari uchun quyidagi variantlarni qiymatlarini ikkalasida ham qo’llashingiz mumkin
MUSTAQIL ISHLASH UCHUN TOPSHIRIQLAR
(7a-19 TJA 1-40 variantlar)
(7b-19 TJA 41-80variantlar)
(10-19 TJA 81-115 variantlar)
Quyidagi birinchi tartibli differentsial tenglamalar uchun Koshi masalasini ko'rsatilgan kesmada h=0,1 bo‘lganda:

Eyler usulida.
Eylerning ketma-ket yaqinlashish usulida.
Eylerning takomillashtirish usulida.

Taqribiy yechimini toping.

1		2	3		4		5
y'=x/(x+y) y(0)=1, Xϵ [0,1]	y'-2y=3e^x y(0,3)=1,415 Xϵ [0,1;0,5]			y'=x+y²y(0)=0, Xϵ [0;0,3]		y'=y²-x² y(1)=1, Xϵ[1;2]		y'=x²+y² y(0)=0.27 Xϵ [0;1]
6	7			8		9		10
y'+xy(9-y²)=0 y(0)=0.5 Xϵ [0;1]	y'=x²-xy+y² y(0)=0.1 Xϵ [0;1]			y'=(2y-x)/y y(1)=2 Xϵ [1;2]		y'=x²+xy+y²+1 y(0)=0 Xϵ [0;1]		y'+y=x³ y(1)=-1 Xϵ [1;2]
11	12			13		14		15
y'=xy+e^y y(0)=0 Xϵ [0;0.1]	y'=2xy+x² y(0)=0 Xϵ [0;0.5]			y'=x+ y(0)=1, [0;1]		y'=e^x-y² y(0)=0 Xϵ [0;0.4]		y'=2x+cosy y(0)=0 Xϵ [0;0.1]
16	17			18		19		20
y'=x³+y² y(0)=0.5 Xϵ [0;0.5]	y'=xy³-y y(0)=1 Xϵ [0;1]			y'=y²e^x-2y y(0)=1 Xϵ [0;1]		y'= y(1)=0, Xϵ [1;2]		y'= y(1)=1, Xϵ [1;2]
21	22			23		24		25
y'=e^xcosy/x y(1)=1 , Xϵ [1;2]	y'=e^xsiny/x y(1)=1 Xϵ [1;2]			y'cosx-ysinx=2x y(0)=0 Xϵ [0;1]		y’=ytgx- y(0)=0 , Xϵ [0;1]		y'+ycosx=cosx y(0)=0 Xϵ [0;1]
26	27			28		29		30
y’= y(0)=0, Xϵ [0;1]	y'=(9+ )²y(1)=1, Xϵ [1;2]			xy'- -x=0 y(1)=1/2, Xϵ [1;2]		y'= (9+lny-lnx) y(1)=e , Xϵ [1;2]		y³xdx=(x²y+2)dy y(0.348)=2 Xϵ [0;1]
31	32			33		34		35
y'=x/(x+y) y(0)=3, Xϵ [0,1]	y'-2y=3e^x y(0,3)=1,4 Xϵ [0,1;0,5]			y'=x+y² y(1)=0, Xϵ [0;0,3]		y'=y²-x² y(1)=0, Xϵ [1;2]		y'=x²+y² y(0)=2 Xϵ [0;1]
36	37			38		39		40
y'+xy(9-y²)=0 y(0)=5 Xϵ [0;1]	y'=x²-xy+y² y(0)=1 Xϵ [0;1]			y'=(2y-x)*y y(0)=2 Xϵ [1;2]		y'=x²+xy+y²+1 y(0)=5 Xϵ [0;1]		y'+y=x³ y(1)=-2 Xϵ [1;2]
41	42			43		44		45
y'=xy+ey y(0)=0 [0;0.1]	y'=2xy+x² y(0)=0 [0;0.5]			y'=x+ y(0)=1, [0;1]		y'=e^x-y2 y(0)=0 [0;0.4]		y'=2x+cosy y(0)=0 [0;0.1]
46	47			48		49		50
y'=x³+y² y(0)=0.5 Xϵ [0;0.5]	y'=xy³-y y(0)=1 Xϵ [0;1]			y'=y²ex-2y y(0)=1 Xϵ [0;1]		y'= y(1)=0, [1;2]		y'= y(1)=1, [1;2]
51	52			53		54		55
y'=e^xcosy/x y(1)=1 , Xϵ [1;2]	y'=e^xsiny/x y(1)=1 Xϵ [1;2]			y'cosx-ysinx=2x y(0)=0 Xϵ [0;1]		y’=ytgx- y(0)=0 , Xϵ [0;1]		y'+ycosx=cosx y(0)=0 Xϵ [0;1]
56	57			58		59		60
y’= y(0)=0, Xϵ [0;1]	y'=(9+ )2 y(1)=1, Xϵ [1;2]			xy'- -x=0 y(1)=1/2, Xϵ [1;2]		y'= (9+lny-lnx) y(1)=e , Xϵ [1;2]		y³xdx=(x³y+2)dy y(0.48)=2 Xϵ [0;1]
61	62			63		64		65
y'=x/(x+y) y(0)=1, Xϵ [0,1]	y'-2y=3ex y(0,3)=1,15 Xϵ [0,1;0,5]			y'=x+y² y(0)=0, Xϵ [0;0,3]		y'=y²-x² y(1)=1, Xϵ [1;2]		y'=x²+y² y(0)=0.7 Xϵ [0;1]
66	67			68		69		70
y'+xy(9-y²)=0 y(0)=0.5 Xϵ [0;1]	y'=x²-xy+y² y(0)=0.1 Xϵ [0;1]			y'=(2y-x)/y y(1)=2 Xϵ [1;2]		y'=x²+xy+y²+1 y(0)=0 Xϵ [0;1]		y'+y=x³ y(1)=-1 Xϵ [1;2]
71	72			73		74		75
y'=xy+ey y(0)=0 Xϵ [0;0.1]	y'=2xy+x² y(0)=0 Xϵ [0;0.5]			y'=x+ y(0)=1, Xϵ [0;1]		y'=ex-y² y(0)=0 Xϵ [0;0.4]		y'=2x+cosy y(0)=0 Xϵ [0;0.1]
76	77			78		79		80
y'=x³+y² y(0)=0.5 Xϵ [0;0.5]	y'=x*y³-y y(0)=1 Xϵ [0;1]			y'=y²ex-2y y(0)=1 Xϵ [0;1]		y'= y(1)=0, Xϵ [1;2]		y'= y(1)=1, Xϵ [1;2]
81	82			83		84		85
y'=e^xcosy/x y(1)=1 , Xϵ [1;2]	y'=e^xsiny/x y(0)=1 Xϵ [1;2]			y'cosx-ysinx=2x y(0)=0 Xϵ [0;1]		y’=ytgx- y(0)=0 , Xϵ [0;1]		y'+ycosx=cosx y(0)=0 Xϵ [0;1]
86	87			88		89		90
y’= y(0)=0, Xϵ [0;1]	y'=(9+ )2 y(1)=1, Xϵ [1;2]			xy'- -x=0 y(1)=1/2, Xϵ [1;2]		y'= (9+lny-lnx) y(1)=e , Xϵ [1;2]		y³xdx=(x²y+2)dy y(0.5)=2 Xϵ [0;1]
91	92			93		94		95
y'+xy(9-y2)=0 y(0)=5 Xϵ [0;1]	y'=x²-xy+y2 y(0)=1 Xϵ [0;1]			y'=(2y-x)*y y(1)=2 Xϵ [1;2]		y'=x2+xy+y2+1 y(1)=5 Xϵ [0;1]		y'+y=x³ y(1)=-5 Xϵ [1;2]
96	97			98		99		100
y'=xy+ey y(1)=0 Xϵ [0;0.1]	y'=2xy+x² y(0)=1 Xϵ [0;0.5]			y'=x+ y(0)=1, [0;1]		y'=ex-y² y(0)=0 Xϵ [0;0.5]		y'=2x+cosy y(0)=0 Xϵ [0;0.1]
101	102			103		104		105
y'=x³+y² y(1)=0.5 Xϵ [0;0.5]	y'=x*y³-y y(1)=1 Xϵ [0;1]			y'=y²e^x-2y y(0)=1 Xϵ [0;1]		y'= y(1)=0, Xϵ [0.5;2]		y'= y(1)=0, Xϵ [1;2]
106	107			108		109		110
y'=e^xcosy/x y(1)=1 , Xϵ [1;2]	y'=e^xsiny/x y(1)=1 Xϵ [1;2]			y'cosx-ysinx=2x y(0)=0 Xϵ [0;1]		y’=ytgx- y(0)=0 , Xϵ [0;1]		y'+ycosx=cosx y(0)=0 Xϵ [0;1]
111	112			113		114		115
y'=x/(x+y) y(0)=1, Xϵ [0,1]	y'-2y=3ex y(0,3)=1,4 Xϵ [0,1;0,5]			y'=x+y² y(0)=1, Xϵ [0;0,3]		y'=y²-x² y(1)=0, Xϵ[1;2]		y'=x²+y² y(1)=0.5 Xϵ [0;1]
116	117			118		119		120
y'+xy(9-y²)=0 y(0)=0.5 Xϵ [0;1]	y'=x²-xy+y² y(0)=0.1 Xϵ [0;1]			y'=(2y-x)/y y(1)=2 Xϵ [1;2]		y'=x²+xy+y²+1 y(0)=0 Xϵ [0;1]		y'+y=x³ y(1)=-1 Xϵ [1;2]

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