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Eyler usuli Eylerning ketma-ket yaqinlashish usuli-masala. Birinchi va ikkinchi takomillashtirish usul bilan quyidagicha
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səhifə | 6/6 | tarix | 01.06.2023 | ölçüsü | 133,32 Kb. | | #114907 |
| 12-ma\'ruza9.2-masala. Birinchi va ikkinchi takomillashtirish usul bilan quyidagicha
differentsial tenglamani h=0,2 qadam bilan u(0)=1 shartni qanoatlantiruvchi yechimni toping.
Yechish.
1. Birinchi takomillashtirish usuli. (9.8) va (9.9) formulalar asosida:
1) i=0, xo=0, yo=1 bo‘lganda quyidagicha hisoblaymiz:
f0=f (x0, y0)=y0-2x0/y0=1
x1/2=xo+h/2=0.1
y1/2=y0+hf (x0,y0)=1+0.1*1=1.1
f1/2=f (x 1/2,y1/2)= y 1/2- 2x1/2/y1/2=0.9182
= hf (x 1/2,y1/2)=0.2*0.9182=0.1836
bu holda birinchi yaqinlashishning birinchi qiymati:
y1=y0+ =1+0.1836=1.1836
2) i=1, x1=0.2 y1=1.183 bo‘lganda quyidagicha hisoblaymiz:
f (x 1,y1)=0.1(y1- )=0.0846
x3/2=0.3
y3/2=y1+ f (x 1,y1)=1.1836+0.0846=1.2682
f (x 3/2,y3/2)= y 3/2- 2x3/2/y3/2=0.7942
y0= h f (x 3/2,y3/2)=0.2*0.7942=0.1590
bu holda birinchi yaqinlashishning ikkinchi qiymati:
y2= y1+y1=1.1836+0.1590=1,3426
shuningdek i=2, 3, 4, 5 lar uchun ham hisoblab, natijalarni quyidagi jadvalga yozamiz.
9.1-jadval
I
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xi
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Yi
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hfi/2
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x i+1/2
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y i+1/2
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yi
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0
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0
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1
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0.1
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0.1
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1.1
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0.1836
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1
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0.2
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1.1836
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0.0846
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0.3
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1.2682
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0.1590
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2
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0.4
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1.3426
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0.0747
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0.5
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1.4173
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0.1424
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3
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0.6
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1.4850
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0.0677
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0.7
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1.5527
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0.1302
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4
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0.8
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1.6152
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0.0625
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0.9
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1.6777
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0.1220
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5
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1.0
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1.7362
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|
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2. Ikkinchi takomillashtirish usuli.
1) i=0, xo=0, yo=1 bo‘lganda quyidagicha hisoblaymiz:
f0=f (x0, y0)=1 , =y0+hf (x0,y0)=1+0.2*1=1.2
f0=0.1 x1=0.2, =1.2
= f (x1, )=0.1(9.2- )=0.0867
y0= (f 0+ )=0.1(9+0.867)=0.1867
bu holda ikkinchi yaqinlashishning birinchi qiymati:
y1=y0+ =1*0,1867=1,1867
2) i=1, x1=0.2 y1=1.1867 bo‘lganda quyidagicha hisoblaymiz:
f1=f (x1, y1)=1.1867- =0.8497
2=1.1867+0.1619 =1.3566
f1=0.0850 x1=0.4 y2=1.3566
= f (x2, 2)=0.0767
y1= (f 1+ )=0.0850+0.0767=0.1617
bu holda ikkinchi yaqinlashishning ikkinchi qiymati:
y2=y1+y1=1.1867+0.1617=1.3484
shuningdek i=3,4,5 lar uchun xam hisoblab, natijalarni quyidagi jadvalga yozamiz.
9.2-jadval
I
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xi
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yi
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hfi/2
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X I+1
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Y i+1
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|
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0
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0
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1
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0.1
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0.2
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1.2
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0.0867
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0.186
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1
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0.2
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1.1867
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0.0850
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0.4
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1.3566
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0.0767
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0.1617
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2
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0.4
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1.3484
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0.0755
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0.6
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1.4993
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0.0699
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0.1415
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3
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0.6
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1.4938
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0.0699
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0.8
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1.6180
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0.0651
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0.1341
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4
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0.8
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1.6272
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0.0645
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1.0
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1.7569
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0.0618
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0.1263
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1.0
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1.7542
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Ushbu misol asosida Eylerning takomillashgan usulida birinchi tartibli differentsial tenglama uchun Koshi masalasini taqribiy yechimini kompyuter yordamida hisoblash quyidagicha dasturida berilgan.
9.3- dastur
10 DEF FNE (X,Y)=Y-2*X/Y
12 PRINT: PRINT
14 PRINT “Birinchi tartibli differentsial tenglamasi uchun ”
15 PRINT “Koshi masalasining taqribiy yechimini”
16 PRINT “Eylerning takomillashgan usulida”
18 PRINT “ hisoblash ”
20 PRINT
18 PRINT TAB(98) “1-takomillashtirish 2-takomillashtirish”
20 PRINT
22 REM boshlang’ich qiymatlar ,bo‘linish soni, berilgan kesma yuqori chegarasi:
40 READ X, Y, N, B
42 REM boshlang’ich qiymatlar , bo‘linish soni, berilgan kesma yuqori chegarasi qiymatlari:
44 DATA 0,1,5,1
50 H=(B-X)/N: Y3=Y
60 FOR I=1 TO N
80 Y1=Y+H*FINE(X,Y)/2
120 Y2=Y+H*(FINE(X,Y)/2, Y1)
130 Y4=Y3+H*FINE(X,Y3)
140 Y5=Y3+(H/2)*(FINE(X,Y3)+ FINE(X+H,Y4))
102 X=X+H; Y=Y2; Y3=Y5
120 PRINT X “(“;USING “###.###”;I:
122 PRINT ”)=“;USING “###.###”;X;
130 PRINT Y2 “(“I”)=“;USING “###.###”;Y2;
140 PRINT Y5 “(“I”)=“;USING “###.###”;Y5
180 NEXT I
200 END
RUN
Birinchi tartibli differentsial tenglamasi uchun
Koshi masalasining taqribiy yechimini
Eylerning takomillashgan usulida
hisoblash.
1-takomillashtirish 2-takomillashtirish
X(1)=0.20 Y2(1)=1.1836 Y5 (1)=1.1867
X(2)= 0.40 Y2(2)= 1.3427 Y5 (2)=1.3483
X(3)= 0.60 Y2 (3)= 1.4850 Y5 (3)=1.4937
X(4)= 0.80 Y2 (4)= 1.6152 Y5 (4)=1.6279
X(5)= 1.0 Y2 (5)= 1.7362 Y5 (5)=1.7542
9.3- Paskal tili dasturi
{Birinchi tartibli differentsial tenglama }
{Y1=F(X,Y) uchun}
{Koshi masalasini Eylerning takomillashgan usulida}
{taqribiy yechimini topish}
program AILER3(output,INPUT);
function fne(x,y:real):real;
begin fne:=Y-2*X/Y; end;
var
x,y,y1,y2,y3,y4,y5,b,h:real;
i,n:integer;
begin
writeln(' x=','y=',' n=',' b='); readln(x,y,n,b);
h:=(b-x)/n;
for i:=1 to n
BEGIN
y1:=y+h*fne(x,y)/2;
y2:=y+h*fne(x+h/2,y1);
y4:=y3+h*fne(x,y3);
y5:=y3+h*(fne(x,y3)+fne(x+h,y4))/2;
x:=x+h;
y:=y2;
y3:=y5;
writeln;
write(' x(',I:2,')=',x:8:4);
WRITE(' y2(',I:2,')=',Y2:8:4);
write(' y5(',I:2,')=',Y5:8:4);
END;
end.
O‘z-o‘zini tekshirishuchun savollar:
Birinchi tartibli differentsial tenglamalari uchun Koshi masalasi
Birinchi tartibli differentsial tenglama taqribiy uchimini Eyler usuli yordamida qanday topiladi?
Birinchi tartibli differentsial tenglama taqribiy uchimi Eylerning ketma-ket yaqinlashish usuli yordamida qanday topiladi?
Birinchi tartibli differentsial tenglama taqribiy uchimini topishda Eyler va Eylerning ketma-ket yaqinlashish usulidan farqini tushintiring?
Birinchi tartibli differentsial tenglama taqribiy uchimi Runge-Kutta usuli yordamida qanday topiladi?
Taqribiy uchim hato;igini baholashni tushintirib bering?
Birinchi tartibli differentsial tenglama taqribiy uchimini topishda Eyler va Runge-Kutta usulidagi yechim aniqniqlashdagi baholashni tushintirib bering?
Echim aniqligini baholashni aniqlovchi shartni yozing.
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