Oliy matematika va axborot texnologiyalari

№	X k	yk	f (X k;yk)
f (X k ;yk1)h
0	X 0	y0	f (X 0;y0)	y0= f (X 0 ;y0)h
1	X 1	y1=y0+ y0	f (X 1;y1)	y1= f (X1 ;y1)h
2	X 2	y2=y1+ y1	f (X2;y2)	y2= f (X 2 ;y21)h
…	….	………….….	…………	……………………
n-1	X n-1	yn-1=yn-2+ yn-2	f (Xn-1;yn-1)	yn-1= f (X n-1 ;yn-1)h
n	X n	yn=yn-1+ yn-1	f (Xn;yn)

Misol: tеnglama bеrilgan. [0;1] kеsmada х_о=0, у_о=1 boshlang’ich shartni qanoatlantiruvchi yеchim Teylor usuli yordamida topilsin. Qadamlar h₁=0,1 vа h₂=0,2 dеb olinsin. Hisob vеrguldan so’ng uch raqam bilan olib bеrilsin.

Yechish: Hisoblash qulay bo’lishi uchun jadval tuzib olamiz (h₁=0,1)

1. Boshlang’ich shartlar bo’yicha jadvalning birinchi satrini to’ldirib olamiz.

х₀=0 y₀=1 dа bo’ladi

2. Jadvalning ikkinchi satrini to’ldiramiz.

dа

va hokazo.

Hisoblashlarni toki [0,I] oraliqni o’tib bo’lmaguncha davom ettiramiz.

№	x i=x 0+ih	yi+1=yi+yi	y^2i	x ^3i
0	0	1,000	1,000	0	1,000	0,100
1	0,1	1,100	1,210	0,001	1,000	0,100
2	0,2	1,210	1,440	0,008	1,006	0,101
3	0,3	1,301	1,692	0,027	1,016	0,102
4	0,4	1,402	1,966	0,064	1,033	0,103
5	0,5	1,506	2,256	0.125	1,055	0,105
6	0,6	1,611	2,595	0.216	1,083	0,108
7	0,7	1,719	2,956	0.343	1,116	0,112
8	0,8	1,830	3,350	0.512	1,153	0,115
9	0,9	1,946	3,768	0.729	1,26	0,126
10	1,0	2,060	-	-	-	-

Endi h=0,2 qadam uchun yuqoridagi kabi hisoblashlarni bajarib, jadvalni to’ldiramiz.

№	x i	yi	y^2i	x ^3i
0	0	1,000	1,000	0	1,000	0,200
1	0,2	1,200	1,440	0,098	1,006	0,200
2	0,4	1,400	1,960	0,054	1,030	0,210
3	0,6	1,610	2,592	0,216	1,080	0,220
4	0,8	1,830	3,350	0,512	1,100	0,210
5	1,0	2,050	-	-	-	-

Dеmak, ,

Shunday qilib absolyut xato nisbiy xato

Koordinatalar tеkisligida (x₀,y₀), (x₁,y₁) ,…, (x _n,y_n) nuqtalarni to’g’ri chiziq kеsmalari bilan tutashtirib, Teylor siniq chizig’ini hosil qilamiz:
у

2,0

1,8

1,6

1,4

1,2

1

0 0,2 0,4 0,5 0,8 1,0 1,2 х

15-chizma
An’anamizga ko’ra yuqorida namuna sifatida keltirilgan misolni dasturlab ko’ramiz.

Eyler – Koshi usuli uchun Pascalda maxsus tuzilgan dastur.

uses Crt;

const c:array[1..4] of real = (0,0.5,0.5,1);

type

coef=array[0..4] of real;

var

i,j,m:integer;

a,b,h,x,y,y1,y2:real;

k0,k:coef;

ch:char;

{ ------------------------------------------------

Подпрограммы }

{ Праvaя часть дифференциального уравнения функции- f(x,y)}

function f(x,y:real):real;

begin

f:= x+y;

end;

{ -------------------------------------------------------- }

procedure Pausa;

begin

writeln;

writeln (' Davom ettirish uhun PLAY ni bosing...');

repeat ch:=readkey

until ch <> '';

end;

{ ------------------------------------------------

Общая программа }

begin

clrscr;

writeln ('Oraliq qiymatlarini kiriting * [a,b]');

read (a,b);

writeln (' y0 ning x=x0 bo"lgandagi boshlang"ich funksiyaning qiymatini kiriting');

read (y);

writeln (' [a,b] oralig"idagi qiymatlar sonini kiriting');

read (m);

x:=a;

h:=(b-a)/m;

y1:=y;

y2:=y;

writeln (' Eyler - Koshi usuli');

writeln ('x=',x:5:2,'y1=',y1:9:6,'y2=',y2:9:6);

for i:=1 to m do

begin

y1:=y1+h*f(x,y); { <-----------| Eyler usuli}

for j:=1 to 2 do

k0[j]:=h*f(x+2*c[j]*h, y2+2*c[j]*k0[j-1]);

y2:=y2+(k0[1]+k0[2])/2; { <-----------| Eyler - Koshi usuli}

x:=x+h;

writeln ('x=',x:5:2,' y1=',y1:9:6,' y2=',y2:9:6);

end;

pausa;

end.
Dasturni ishga tushiramiz.

Xatolik nisbatan kattaroq, ammo natija qoniqarli.

Birinchi tartibli diffеrеnsial tеnglamalarni Eylеr usuliga turbo paskalda tuzilgan yana bir dastur
program aslon;

uses crt;

const h1=0.1;

h2=0.2;

var

x,y,z:array [0..10] of real;

i:integer;

S:real;

begin

clrscr;

x[0]:=0.3;

y[0]:=1;

for i:=0 to 5 do begin

x[i+1]:=x[i]+((i+1)*h2);

end;

for i:=0 to 5 do begin

y[i+1]:=y[i]+((sqrt(1+(sqr(x[i])+(exp(3*ln(y[i])))))*h2));

end;

for i:=0 to 5 do begin

writeln('x[',i,']=',x[i],' ','y[',i,']=',y[i]);

end;

writeln;

begin

for i:=0 to 10 do begin

x[i+1]:=x[i]+((i+1)*h1);

end;

for i:=0 to 10 do begin

y[i+1]:=y[i]+((sqrt(1+(sqr(x[i])+(exp(3*ln(y[i]))))))*h1);

end;

for i:=0 to 10 do begin

writeln('x[',i,']=',x[i],' ','y[',i,']=',y[i]);

end;

S:=((y[10]-y[5])/y[5])*100;

if S<0.001 then writeln('To`g`ri','S=',S)

else writeln('Noto`g`ri', 'S=',S);

end;

end.
Hisob grafik ishi variantlari

№	Tеnglama	Boshlang’ich qiymat	Kеsma	qadamlar
1
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5
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15
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