Differensial tenglamalarni taqribiy yechimlarini aniqlash. Eyler usuli


-masala. Birinchi va ikkinchi takomillashtirish usul bilan quyidagicha



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9.2-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

xi

Yi

hfi/2

x i+1/2

y i+1/2

yi

0

0

1

0.1

0.1

1.1

0.1836

1

0.2

1.1836

0.0846

0.3

1.2682

0.1590

2

0.4

1.3426

0.0747

0.5

1.4173

0.1424

3

0.6

1.4850

0.0677

0.7

1.5527

0.1302

4

0.8

1.6152

0.0625

0.9

1.6777

0.1220

5

1.0

1.7362















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

xi

yi

hfi/2

X I+1

Y i+1





0

0

1

0.1

0.2

1.2

0.0867

0.186

1

0.2

1.1867

0.0850

0.4

1.3566

0.0767

0.1617

2

0.4

1.3484

0.0755

0.6

1.4993

0.0699

0.1415

3

0.6

1.4938

0.0699

0.8

1.6180

0.0651

0.1341

4

0.8

1.6272

0.0645

1.0

1.7569

0.0618

0.1263




1.0

1.7542
















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:



  1. Birinchi tartibli differentsial tenglamalari uchun Koshi masalasi

  2. Birinchi tartibli differentsial tenglama taqribiy uchimini Eyler usuli yordamida qanday topiladi?

  3. Birinchi tartibli differentsial tenglama taqribiy uchimi Eylerning ketma-ket yaqinlashish usuli yordamida qanday topiladi?

  4. Birinchi tartibli differentsial tenglama taqribiy uchimini topishda Eyler va Eylerning ketma-ket yaqinlashish usulidan farqini tushintiring?

  5. Echim aniqligini baholashni aniqlovchi shartni yozing.



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