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0 ПродажOddiy differensial tenglamalarni sonli yechish eyler usulining matlab dasturidagi tadbig’i
![1.3. Oddiy differensial tenglamalarni sonli yechish
?
.bn Quyidagi y'= f(x,y)
(1)
birinchi tartibli differentsial tenglamaning [a,b] kesmada boshlang’ich shart
x=x
0 bo`lgan hol uchun y=y
0 ni qanoatlantiruvchi echimi topilishi lozim bo`lsin.
[a,b] kesmani x
0 , x
1 ,
x
2 ,…, x
n nuqtalar bilan n ta teng bo`lakchalarga ajratamiz;
bunda
хi= x0+ih ( i = 0,1,2,… n ), h= b− a
n - qadam.
(1) tenglamani [a,b] kesmaga tegishli bo`lgan biror [x
k , x
k+1 ] kesmada
integrallasak,
∫
xk
xk+1
f(x,y)dx = ∫xk
xk+1
y'dx = y(x)|xk
xk+1
= y(xk+1)− y(xk)= yk+1− yk
ya`ni,
yk+1= yk+∫
xk
xk+1
f(x,y)dx
(2)
Bu erda integral ostidagi funktsiyani x=x
k nuqtada boshlang’ich o`zgarmas
qiymatiga teng deb qabul qilinsa, quyidagini hosil qilamiz:
∫
xk
xk+1
f(x,y)dx = f(xk,yk)¿x |xk
xk+1
= f(xk,yk)⋅(xk+1− xk)= yk'¿h
U holda (2) dan
yk+1= yk+yk
'h
(3)
yk+1− yk= Δy k
ya`ni yk
'h= Δy k deb belgilasak,
yk+1= yk+Δy k
(4)
Ushbu jarayonni [a,b] ga tegishli bo`lgan har bir kesmacha uchun takrorlab,
(1) ning echimini ifodalovchi jadvalini to`zamiz. eyler usulining geometrik
ma`nosi shundayki, bunda (1) ning echimini ifodalovchi integral egri chiziq siniq
(II) chiziqlar bilan almashtiriladi (10 - rasm).
15](https://docx.uz/documents/cba5fbfc-1c79-48cc-8967-4066878bcf2f/page_14.png?v=1)
![10 – rasm
Quyidagi tizim{y'=f1(x,y,z)¿¿¿¿
(5)
uchun
x=x
0 da y=y
0 , z=z
0 (6)
boshlang’ich shart berilgan. (5) ning taqribiy echimlari quyidagi formulalar
orqali topiladi:
yi+1= yi+ Δy i, zi+1= zi+ Δz i
bu erda
Δy i= hf 1(xi,yi,zi); Δz i= hf 2(xi,yi,zi) (i= 0,1,2 ,...)
Misol. eyler usuli yordamida
у'= у−
2 х
у differentsial tenglamaning
[0,1] kesmada olingan va u (0) = 1 boshlang’ich shartni qanotlantiruvchi u(x)
echimining taqribiy qiymatlarini h =0,2 qadam bilan toping.
Echish :
f(x,y)= y− 2x
y
; a= 0, b= 1, x0= 0, y0= 1, h= 0,2
Quyidagi hisoblash jadvalini to`zamiz.
16 x0y
x
0 x
1 x
2 x
3 x
n-1 x
4y
0 - II](https://docx.uz/documents/cba5fbfc-1c79-48cc-8967-4066878bcf2f/page_15.png?v=1)
![2-rasm. Koshi masalasi yechimining grafigi.
Endi Koshi masalasining yechimini darajali qator ko’rinishida topamiz
hamda sonli yechim va olingan darajali qatorning grafigini ular mosroq tushishi
mumkin bo’lgan interval uchun yasaymiz (2-rasm).
> dsolve({eq, cond}, y(x), series);
> convert(%, polynom):p:=rhs(%):
> p1:=odeplot(de,[x,y(x)],-3..3, thickness=2,
color=black):
> p2:=plot(p,x=-3..3,thickness=2,linestyle=3,
color=blue):
> display(p1,p2);
Yechimning darajali qator bilan juda yaqin qiymatlari 1 < x < 1 ekanligi
grafikdan ko’rinib turibdi.
Agar bu kabi masalalrni oddiy matematik usulda echish, hamda uning
grafigini hosil qilish zarur bo’lsa, bu talabalardan, ilmiy xodim va o’qituvchilardan
ko’p vaqt va malaka talab etadi. Yuqoridagi masaladan ko’rinib turibdiki , uni
Maple muhitida oson yechish va bir paytda uning grafigini ham hosil qilish
mumkin ekan.
1. To`rda approksimatsiya xatoligi
18](https://docx.uz/documents/cba5fbfc-1c79-48cc-8967-4066878bcf2f/page_17.png?v=1)
![1.4. Eyler usulining matlab dasturidagi tadbig’i
Quyidagi y′ + 2y = 2 − e−4x , y (0) = 1 Koshi masalasini Eyler usuli yordamida h
= 0.1 qadam uchun x = 0.1, 0.2, 0.3, 0.4, 0.5 qiymatlardagi taqribiy qiymatini
toping.
% Parametrlar
h = 0.1; % Qadam uzunligi
x = 0:h:0.5; % x = 0.1, 0.2, 0.3, 0.4, 0.5
y = zeros(1, length(x)); % y massivini boshlang'ich qiymatlar bilan to'ldirish
% Boshlang'ich shart
y(1) = 1; % y(0) = 1
% Differential tenglama: y' + 2y = 2 - e^(-4x)
% y' = 2 - e^(-4x) - 2y
for i = 1:length(x)-1
f = 2 - exp(-4*x(i)) - 2*y(i); % f(x, y)
y(i+1) = y(i) + h*f; % Eylier usuli
end
% Natijalarni chiqarish
disp('x va y qiymatlari:');
disp([x', y']);
h = 0.1 - bu qadam uzunligi.
x = 0:h:0.5 - x qiymatlari 0 dan 0.5 gacha 0.1 qadam bilan hisoblanadi.
y(1) = 1 - boshlang'ich shart
Differential tenglama : y ' = 2 − e − 4 x − 2 y
bu yerda f(x,y) funksiyasini topish va
E y ler usulidan foydalanib, har bir qadamda yangi yy qiymatini hisoblash.
28](https://docx.uz/documents/cba5fbfc-1c79-48cc-8967-4066878bcf2f/page_27.png?v=1)
![2-Misol
Differensial tenglama :
% Parametrlar
h = 0.1; % Qadam uzunligi
x = 0:h:0.5; % x = 0.1, 0.2, 0.3, 0.4, 0.5
y = zeros(1, length(x)); % y massivini boshlang'ich qiymatlar bilan to'ldirish
% Boshlang'ich shart
y(1) = 1; % y(0) = 1
% Differential tenglama: y' = x + y
for i = 1:length(x)-1
f = x(i) + y(i); % f(x, y)
y(i+1) = y(i) + h*f; % Eylier usuli
end
% Natijalarni chiqarish
disp('x va y qiymatlari:');
disp([x', y']);
Misol 2 : y'= x+y tenglamasini yechish uchun Eylier usuli qo'llanilgan. Bu oddiy
chiziqli tenglama bo'lib, u ko'plab amaliy masalalarda uchraydi.
x y ( x )
0.0 1.0000
0.1 1.1100
0.2 1.2321
0.3 1.3665
0.4 1.5134
0.5 1.6730
29](https://docx.uz/documents/cba5fbfc-1c79-48cc-8967-4066878bcf2f/page_28.png?v=1)
![3- Misol:
y'= x2− y2y'= x2− y2y'= x2− y2,y(0)=1y(0)=1y(0)=1
% Parametrlar
h = 0.1; % Qadam uzunligi
x = 0:h:0.5; % x = 0.1, 0.2, 0.3, 0.4, 0.5
y = zeros(1, length(x)); % y massivini boshlang'ich qiymatlar bilan to'ldirish
% Boshlang'ich shart
y(1) = 1; % y(0) = 1
% Differential tenglama: y' = x^2 - y^2
for i = 1:length(x)-1
f = x(i)^2 - y(i)^2; % f(x, y)
y(i+1) = y(i) + h*f; % Eylier usuli
end
% Natijalarni chiqarish
disp('x va y qiymatlari:');
disp([x', y']);
Misol 3 :
y ' = x 2
− y 2
tenglamasi non-linеar va qiyinroq bo'lishi mumkin, lekin
Eylier usuli yordamida uning yechimini ham topish mumkin.
x y ( x )
0.0 1.0000
0.1 1.0099
0.2 1.0398
0.3 1.0898
0.4 1.1600
0.5 1.2505
30](https://docx.uz/documents/cba5fbfc-1c79-48cc-8967-4066878bcf2f/page_29.png?v=1)
Oddiy differensial tenglamalarni sonli yechish eyler usulining matlab dasturidagi tadbig’i
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