Citations reads

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Agar aniqlikni yanada oshirish lozim boʻlsa, u holda:

Euler^	d	y(t)  1  2  y(t)  t ^2 , y(0)  1, t  1, output  plot, numsteps  50		;
			

 dt			

35

11-rasm. Maple dasturida Eyler usuli bilan olingan natijalar grafigi.


2-misol. Quyidada keltirilgan Koshi masalasini takomillashtirilgan Eyler usuli bilan yechish.


Yechish. Koshi masalasi
y' - 2y + x² = 1, x  [0;1], y(0) = 1.
Faraz qilaylik, n = 10 , h = (1 - 0)/10 = 0,1.

Boshlangʻich nuqta x0 = 0, y0 = 1.

Dastlabki nuqtani hisoblash.

y	 y	0	 h 
1
	1 0,1

f(x			h	;
	0		2

• 
  f(0,05; 1
  

y					h	 f(x		;	y		))		 1  0,1 f(0 					0,1		; 1				0,1	 f(0; 1)) 
		0								0
			2		0													2					2
 0,05 (1 2 1- 0												2		))		 1	 0,1 f(0,05; 1,15) 

1  0,1

(1 2

2

1,15  0,05 )  1,32975

x

1

 x

0

 h  0,1

Keyingi 2, 3, ... ,10 nuqtalar uchun hisoblashlar xiddi shunday.

3-misol. Yuqoridagi 2-misolda keltirilgan Koshi masalasini 4-tartibli

Runge-Kutta usuli bilan yechish.

Yechish. Koshi masalasi

y' - 2y + x^2 = 1, x [0;1],

y(0) = 1.

Faraz qilaylik, n = 10 , h = (1 - 0)/10 = 0,1.

Boshlangʻich nuqta

x0 = 0, y0 = 1.

Dastlab C0, C1, C2, C3larning qiymatlarini hisoblab olamiz:

C

0

 f(x

0

; y

0

)  f(0; 1) 1 2 1 0^2  3

C

 f(x



h

; y

 h 

^K 0

)  f(0,05;1,15) 1 2 1,15  0,05^2  3,2975

1

0

0

2

2

C2

 f(x0

 ^h ;

y

0

 h  ^K^1 )  f(0,05;1,164875)1 2 1,164875 0,05^2  3,32725

2

2

C

3

 f(x

0

 h;

y

0

 h  K

2

)  f(0,1; 1,332725)1 2 1,332725 0,1^2  3,65545

36

C

0

 f(x

0

; y

0

)  f(0; 1) 1 2 1 0^2  3

C

1

 f(x

0

 ^h ; y

0

 h  ^K ^0 )  f(0,05;1,15) 1 2 1,15  0,05^2

 3,2975

2

2

C2

 f(x0



h

;

y0

 h 

K1

)  f(0,05;1,164875)1 2 1,164875 0,05^2

 3,32725

2

2

C3  f(x0  h;

y0  h  K 2 )  f(0,1; 1,332725)1 2 1,332725 0,1^2  3,65545

12-rasm. ODTni 1-tartibli Eyler usuli bilan yechish algoritmi.

37

Dastlabki nuqtani hisoblash:

y	1	 y	0
		1 

• 
  h 6
  

0,1

6

• 
  (C0  2C1  2C2  C3) 
  

• 
  (3  2 3,2975  2 3,32725  3,65545) 1,3317491667
  

Keyingi 2, 3, ... ,10 nuqtalar uchun hisoblashlar xiddi shunday.

1

М_Eiler

2
c0=f(x,y)

с1=f(x+h/2,y+h/2c0)

y=y+c1h

x=x+h

3. 
   
   

Chiqish
Kiritiladigan ma'lumotlar:

h - qadam
x – joriy nuqtaning koordinatasi
Chiqariladigan ma'lumotlar: x – tugunning koordinatasi; y – funksiyaning shu tugundagi qiymati

13-rasm. Takomillashtirilgan Eyler usulining algoritmi.

14-rasm.4-tartibli Runge-Kutta usulining algoritmi.

			2х
4-misol.Ushbu у		 у  у		oddiy differensial tenglamaning	[0,1]
kesmada olingan va y(0)=1 boshlang`ich shartni qanotlantiruvchi					y(x)
				38

yechimining taqribiy qiymatlarini Eyler usuli yordamida h=0,2 qadam bi-lan toping.


Yechish:

f (x, y)  y 	2x	; a  0,b  1, x	 0, y		 1, h  0,2
				0
	y	0

quyidagi hisoblash

1-qator . i=0, x
0
jadvalini tuzamiz.

 0, y	1,0000
0

6. 
   (x0 , y0 )  y0  2x  1  2 * 0  1,000
   

y1

0

y

0

 hf (x

0

, y

0

)  0,2*1

0,2000

y

i1

 y

i

 y

, i  0; y

 y

0

 y

0

1 0,2 1,2000

i

1

2-qator.

i=1

, x1

 0  0,2  0,2; y1

 1,2000;

f (x , y )  y							2x	1,2	2*0,2	 0,8667
			1	1	1		y		1,2
							1
y			 hf (x , y )  0,2 * 0,8667  0,1733
	1				1	1
y	2	 y			 y	 1,2		 0,1733  1,3733
				1	1

i=2,3,4,5 lar uchun hisoblanadi.

i	xi	yi	f(xi ,yi)	yi
1	0,1	1,0000	1,0000	0,200
2	0,2	1,2000	0,8667	0,1733
3	0,4	1,3733	0,7805	0,1561
4	0,6	1,5294	0,7458	0,1492
5	0,8	1,6786	0,7254	0,1451
6	1,0	1,8237

6.