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O'zbekiston Respublikasi Axborot texnologiyalari va kommunikatsiyalarini rivojlantirish vazirligi
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| bet | 3/3 | | Sana | 16.04.2023 | | Hajmi | 91.28 Kb. | | #51772 |
Bog'liq Baxtsiz hodisalarni tahlil etish va shikastlanishlar, kompleks-sonlar-mavzusini-o-qitishda-bumerang-texnologiyasi (1), Pisa test informatika 5-6-7 sinflar uchun, модул 1, Файзуллаев Ёдгоржон 2000, hjhjhgy, 222, konstitutsiya uz, Taqdimoti boshlang’ich sinif o’quvchilari diqqatining ayrim xususiyatlari, dvigatel111, 1-dars, bayonnoma 31.03.2023, Baxromov M, ApplicationFile | X
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U
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u’=f(x,y)
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K=hf(x,y)
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u
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1
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2
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3
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4
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5
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6
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x0
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y0
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f(x0 ,y0)
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K1(0)
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K1(0)
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x0+h/2
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y0+K1(0)/2
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f(x0+h/2; y0+K1(0)/2)
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K2(0)
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2K2(0)
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0
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x0+h/2
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y0+K2(0)/2
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f(x0+h/2; y0+K2(0)/2)
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K3(0)
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2K3(0)
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x0+h
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y0+K3(0)
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f(x0+h; y0+K3(0))
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K4(0)
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K4(0)
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x1
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y1=y0+ y0
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f(x1 ,y1)
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K1(0)
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K1(0)
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x1+h/2
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y1+K1(1)/2
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f(x1+h/2; y1+K1(1)/2)
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K2(0)
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2K2(0)
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1
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x1+h/2
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y1+K2(1)/2
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f(x1+h/2; y1+K2(1)/2)
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K3(0)
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2K3(0)
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x1+h
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y1+K3(1)
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f(x1+h; y1+K3(1))
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K4(0)
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K4(0)
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2
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x2
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y2=y1+ y1
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Misol. Runge-Kutta usuli yordamida quyidagi differensial tenglamaga qo’yilgan boshlang’ich masalaning
y’= , u(1)=0 yechimi [1;1,5] kesmada h=0,1 qadam bilan topilsin.
Yechish. Yechimlar va xisobiy qiymatlar 2-jadvalda keltirilgan.
2-Jadval
i
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xi
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yi
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f(xi, yi)
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K=hf(xi, yi)
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y1
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0
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1
1,05
1,05
1,1
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0
0,05
0,057262
0,115907
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1
1,145238
1,159071
1,310740
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0,1
0,114524
0,115907
0,131074
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0,1
0,229048
0,231814
0,131074
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|
|
|
|
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0,115323
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1
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1,1
1,15
1,15
1,20
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0,115323
0,180807
0,188546
0,263114
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1,309678
1,464447
1,477905
1,638523
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0,130968
0,146445
0,147791
0,163852
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0,130968
0,292889
0,295581
0,163852
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|
|
|
|
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0,147215
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2
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1,2
1,25
1,25
1,3
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0,262538
0,344416
0,352591
0,443953
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1,637563
1,801066
1,814146
1,983005
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0,163756
0,180107
0,181415
0,198301
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0,163756
0,360213
0,362829
0,198301
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|
|
|
|
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0,180805
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3
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1,3
1,35
1,35
1,4
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0,443388
0,524495
0,551073
0,660028
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1,982135
2,153696
2,166404
2,342897
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0,198214
0,215370
0,216640
0,234290
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0,198214
0,430739
0,443281
0,234290
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|
|
|
|
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0,216087
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4
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1,4
1,45
1,45
1,50
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0,659475
0,776580
0,785532
0,912824
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2,342107
2,521146
2,533493
2,717099
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0,234211
0,252115
0,253349
0,271710
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0,234211
0,504229
0,506700
0,271711
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0,252808
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5
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1,5
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0,912283
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Xulosa.
Men bu mavzudan shuni tushindimki Runge-kutta usuli boshqa usullardan aniqroq hisoblashga yordam berar ekan. Differensial tenglamalarni yuqori bo’limlardagidek aniq yechimini topish juda kamdan kam hollardagina mumkin bo’ladi. Amaliyotda uchraydigan ko’plab masalalarga aniq yechish usullarini qo’lashning iloji bo’lmaydi. Shuning uchun bunday differensial tenglamalarni taqribiy yoki sonli usular yordamida yechishga to’g’ri keladi.
Foydalanilgan adabiyotlar:
1.A.A.Abduqodirov “Hisoblash matematikasi”
2.F.B.Badalov “Optimallash nazariyasi”
3.K.Safoyeva, N.Beknazarova “O’quv qo’llanma”
Saytlar Ro’yxati:
1.www.tuit.uz
2.www.edu.uz
3.www.amazon.com
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