Differensial tenglamalarni taqribiy yechimlarini aniqlash. Eyler usuli

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, x_o=0, y_o=1 bo‘lganda quyidagicha hisoblaymiz:
f₀=f (x₀, y₀)=y₀-2x₀/y₀=1
x_1/2=x_o+h/2=0.1
y_1/2=y₀+hf (x₀,y₀)=1+0.1*1=1.1
f_1/2=f (x _1/2,y_1/2)= y _1/2- 2x_1/2/y_1/2=0.9182
= hf (x _1/2,y_1/2)=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
x_3/2=0.3
y_3/2=y₁+ f (x ₁,y₁)=1.1836+0.0846=1.2682
f (x _3/2,y_3/2)= y _3/2- 2x_3/2/y_3/2=0.7942
y₀= h f (x _3/2,y_3/2)=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, x_o=0, y_o=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.