Vazifalar ketma-ketligi Последовательность задач
Algoritmlash VAZIFALAR
- Bu sahifa navigatsiya:
- 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
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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:
xi=xi-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
yi+1=yi+ h f(xi;yi), i=1,2,...,10
formulaga asosan berilgan differentsial tenglama yechimining qiymatlarini quyidagicha topamiz.
y1=y0+hf (x0, 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.
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y'=x/(x+y) y(0)=1, Xϵ [0,1] |
y'-2y=3ex y(0,3)=1,415 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)=0.27 Xϵ [0;1] |
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6 |
7 |
8 |
9 |
10 |
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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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11 |
12 |
13 |
14 |
15 |
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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, [0;1] |
y'=ex-y2 y(0)=0 Xϵ [0;0.4] |
y'=2x+cosy y(0)=0 Xϵ [0;0.1] |
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16 |
17 |
18 |
19 |
20 |
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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, Xϵ [1;2] |
y'= y(1)=1, Xϵ [1;2] |
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21 |
22 |
23 |
24 |
25 |
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y'=excosy/x y(1)=1 , Xϵ [1;2] |
y'=exsiny/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] |
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26 |
27 |
28 |
29 |
30 |
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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(0.348)=2 Xϵ [0;1] |
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31 |
32 |
33 |
34 |
35 |
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y'=x/(x+y) y(0)=3, Xϵ [0,1] |
y'-2y=3ex 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] |
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36 |
37 |
38 |
39 |
40 |
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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] |
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41 |
42 |
43 |
44 |
45 |
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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'=ex-y2 y(0)=0 [0;0.4] |
y'=2x+cosy y(0)=0 [0;0.1] |
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46 |
47 |
48 |
49 |
50 |
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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] |
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51 |
52 |
53 |
54 |
55 |
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y'=excosy/x y(1)=1 , Xϵ [1;2] |
y'=exsiny/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] |
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56 |
57 |
58 |
59 |
60 |
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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] |
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61 |
62 |
63 |
64 |
65 |
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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] |
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66 |
67 |
68 |
69 |
70 |
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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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71 |
72 |
73 |
74 |
75 |
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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] |
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76 |
77 |
78 |
79 |
80 |
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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] |
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81 |
82 |
83 |
84 |
85 |
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y'=excosy/x y(1)=1 , Xϵ [1;2] |
y'=exsiny/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] |
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86 |
87 |
88 |
89 |
90 |
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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] |
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91 |
92 |
93 |
94 |
95 |
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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] |
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96 |
97 |
98 |
99 |
100 |
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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] |
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101 |
102 |
103 |
104 |
105 |
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y'=x3+y2 y(1)=0.5 Xϵ [0;0.5] |
y'=x*y3-y y(1)=1 Xϵ [0;1] |
y'=y2ex-2y y(0)=1 Xϵ [0;1] |
y'= y(1)=0, Xϵ [0.5;2] |
y'= y(1)=0, Xϵ [1;2] |
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106 |
107 |
108 |
109 |
110 |
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y'=excosy/x y(1)=1 , Xϵ [1;2] |
y'=exsiny/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] |
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111 |
112 |
113 |
114 |
115 |
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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] |
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116 |
117 |
118 |
119 |
120 |
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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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