9.1.2. Eylerning ketma-ket yaqinlashish usuli
Berilgan (9.1) tenglama uchun Koshi masalasini yechishdagi Eyler usulini mukammal-lashtirish
mumkin. Ya’ni
x
i+1
= x
i
+ (i+1)*h
y
i+1
(0)
= y
i
+ h f (x
i
,y
i
) i=0, 2,….n-1
(9.6)
taqribiy yechimga asosan yechim qiymatlarini aniqligini oshirish uchun quyidagi formulani
tuzamiz.
)]
,
(
)
,
(
[
2
)
1
(
1
1
)
(
1
k
i
i
i
i
i
k
i
y
x
f
y
x
f
h
y
y
(9.7)
У
y
i +1
y
i
x
i +1
x
i
b
Х
a=х
0
y
0
k=1, 2, 3,….n, i=1, 2, 3,….n-1
bu formula bo‘yicha hisoblashni i+1 yechim qiymatiga k -yaqinlashish
)
(
1
к
i
у
va k+1 -
yaqinlashish
)
1
(
1
k
i
y
lar ustma-ust tushguncha davom ettiramiz. Natijada yechimning taqribiy
qiymati
y
i+1
=
)
1
(
1
k
i
y
bo‘ladi. Yuqoridagi usul bilan yechimning
y
i+2
qiymatini hisoblaymiz va x.k, h qadamni
yetarlicha kichik tanlanishi yechimga tez yaqinlashtiradi..
9.1-masala. Quyidagi.
birinchi tartibli differentsial tenglamaning
x
0
=1.8
y
0
=2.6
boshlang’ich shartni qanoatlantiruvchi [1.8, 2.8] oraliqda yechimini h=0.1 qadami bilan,
e=0.001 aniqlikda:
1.
Eyler usuli;
2.
Eylerning ketma-ket yaqinlashish usuli;
3.
Runge – Kutta usuli bilan hisoblang.
Yechish.
1
. Berilgan differentsial tenglamani Eyler usulida yechamiz.
Buning uchun [1.8, 2.8] oraliqni
10
1
.
0
8
.
1
8
.
2
h
a
b
n
ya’ni, n=10 ta bo‘lakka ajratamiz. Bo‘linish nuqtalarini:
x
i
=x
i-1
+h,
i=1,2,...,10
formulaga asosan topamiz.
x
1
=x
0
+h=1.8+0.1=1.9
x
2
=x
1
+h=1.9+0.1=2.0
shuningdek
x
3
=2.1, x
4
=2.2, x
5
=2.3, x
6
=2.4, x
7
=2.5, x
8
=2.6, x
9
=2.7, x
10
=2.8
Berilgan tenglamaning o‘ng tomonidagi
F(x;y)=x+cos(y/
5
)
funktsiyaga asosan, Eyler qoidasi bilan quyidagi
y
i+1
=y
i
+ h f(x
i
;y
i
),
i=1,2,...,10
formulaga asosan berilgan differentsial tenglama yechimining qiymatlarini quyidagicha topamiz.
y
1
=y
0
+hf (x
0,
y
0
)=y
0
+h (x
0
+cos(y
0
/
5
))=
=2.6+ 0.1(1.8+cos(26/
5
))=2.6+0.1(18+0.3968)=2.81968
y
2
=y
1
+h f (x
1
,y
1
)=y
1
+h(x
1
+cos(y
1
/
5
))=
=2.819+ 0.1(1.9+cos(9.819/
5
))=2.819+0.1(1.9+0.3968)=3.03948
SHuningdek, quyidagilarni topamiz:
y
3
=3.261, y
4
=3.4831, y
5
=3.7045, y
6
=3.926
y
7
=4.1478, y
8
=4.3701, y
9
=4.5931, y
10
=4.8173
Bu usul yordamida hisoblash quyidagicha dastur asosida berilgan.
)
5
cos(
y
x
y
2
. Tenglama yechimini Eylerning ketma-ket yaqinlashish usulida hisoblaymiz
. (9.6)
formulada i=0 bo‘lganda
y
1
(0)
=y
0
+hf (x
0
;y
0
)=y
0
+k (x
0
+cos(y
0
/
5
))=
=2.6+ 0.1(9.8+cos(9.6/
5
))=2.6+0.1(9.8+0.3968)=2.81968
bo‘ladi. Bu Eyler usulidagi tenglama yechimining birinchi qiymati bo‘ladi.
Endi y
1
(0)
=2.81968 dan foydalanib (9.7) formulaga asosan i=1 bo‘lganda
y
1
(k)
= y
0
+
2
h
[f (x
0
, y
0
)+ f (x
1
,y
1
(k-1)
)]
formulani k = 1,2,3,….. lar uchun ketma-ket
y
1
(1)
, y
1
(2)
, y
1
(3)
, …, y
1
(k)
larni
y
1
(k-1)
– y
1
(k)
< 0.001
shartni qanoatlantirguncha hisoblaymiz.
Demak,
k=1, y
1
(1)
=y
0
+
2
h
[f (x
0
,y
0
)+f (x
1
,y
1
(0)
)]=2.6+0.05[2.1968+x
1
+cos (y
1
(0)
/
5
)] =
=2.6+0.05[2.1968+1.9+0.36486] =2.7102
k=2, y
1
(2)
= 2.6+0.05[2.1968+x
1
+cos(y
1
(1)
/
5
)] =
=2.6+0.05[2.1968+1.9+ cos(9.7102/
5
)] =2.82239
k=3, y
1
(3)
= 2.6+0.05[2.1968+1.9+cos(9.83339/
5
)] =
=2.6+0.05[2.1968+1.9+ 0.303709] =2.82002
Endi xatolikni tekshiramiz.
y
1
(2)
– y
1
(3)
=
2.8223 – 2.82202
=
0.0002
< 0.001
Bundan 0.001 aniqlikdagi tenglama yechimining birinchi qiymati
y
1
= 2.82000
2.82
bo‘ladi.
Tenglama yechimi y
2
qiymatini topish uchun yuqoridagi qoidani takrorlaymiz.
i=1 uchun (9.6) formulaga asosan
y
2
(0)
= y
1
+ hf (x
1
, y
1
) = 2.82 + 0.1(x
1
+cos (y
1
/
5
)) =
=2.82+0.1(9.9+cos(9.82/
5
) = 3.04047
i=1 uchun (9.7) formulaga asosan
k=1, y
2
(1)
= y
1
+
2
h
[f (x
1
, y
1
)+ f (x
2
, y
2
(0)
)] =
=2.82+0.05[2.20471+2+cos(3.0404/
5
)] =3.0407
k=2, y
2
(2)
=y
1
+
2
h
[f (x
1
,y
1
)+f (x
2
,y
2
(2)
)]= y
1
+
2
h
[2.20471+2+ cos(u
2
(1)
/
5
)]=
=2.82+0.05[2.20471+2+cos(3.0407/
5
)] =3.04071
k=3, y
2
(3)
= y
1
+
2
h
[f (x
1
, y
1
)+ f (x
2
, y
2
(2)
)] =
=2.82+0.05[2.20471+2+cos (3.0407/
5
)] =
=2.82+0.05[2.20471+2+cos (3.0407/
5
)] =3.0407
Endi xatolikni baholaymiz.
y
1
(2)
– y
1
(3)
=
3.04071 – 3.04070
=
0.0001
< 0.001
Bundan tenglama yechimining ikkinchi qiymati
y
2
= 3.0407
bo‘ladi.
Bu qoidani i=2,3,…,10 lar uchun ketma-ket davom ettirib tenglama yechimining qolgan
qiymatlarini ham topamiz.
y
3
=3.261, y
4
=3.483, y
5
=3.704, y
6
=3.926
y
7
=4.147, y
8
=4.370, y
9
=4.593, y
10
=4.817
3.Runge – Kutta usuli
Birinchi tartibli (9.1) tenglamani (9.2) shartni qanoatlantiruvchi yechimning taqribiy
qiymatini Runge – Kutta usuli bilan quyidagicha topamiz. Berilgan [x
0
,b] kesmani n ta teng
bo‘lakka bo‘lib bo‘linish nuqtalari orasidagi qadam
h
=(
b-x
0
)/
n
bo‘lganda,
x
i
=x
i -1
+ h
(
x=x
0
; y=y
0
)
)
(
1
i
Q
=
h f
(x
i
,y
i
)
)
2
/
,
2
(
)
(
1
)
(
2
i
i
i
i
Q
y
h
x
hf
Q
)
2
/
,
2
(
)
(
2
)
(
3
i
i
i
i
Q
y
h
x
hf
Q
(9.11)
)
;
(
)
(
3
)
(
4
i
i
i
i
Q
y
h
x
hf
Q
6
/
)
2
2
(
)
(
4
)
(
3
)
(
2
)
(
1
i
i
i
i
i
Q
Q
Q
Q
y
y
i+1
= y
i
+
i
y
i=0, 1, 2, 3……n.
Bu Runge – Kutta usuli Koshi masalasi yechimning qiymatini to‘rtinchi tartibli aniqlikda
hisoblaydi.
9.2-masala
da
differentsial tenglama uchun Koshi masalasi yechimning qiymatlarini
Runge-Kutta usulida hisoblaymiz. Buning uchun tenglama yechimini topish uchun quyidagi
hisoblash ketma-ketligini bajaramiz.
i =0 bo‘lganda x
0
=1.3
u
0
=2.6 lar uchun yechimning birinchi qiymatini hisoblaymiz.
Q
1
0
=h
f
(x,y)=0.1(x
0
+cos(y
0
/
5
)=0.1(1.8+cos(2.6/
5
)=0.2196
Q
2
0
=
h
f
(x
0
+h/ 2, y
0
+ Q
1
0
/2)=0.201245
Q
3
0
=
h
f
(x
0
+h/ 2, y
0
+ Q
2
0
/2)=0.2205
Q
4
0
=
h
f
(x
0
+h, y
0
+ Q
3
0
)=0.2927
y
1
=y
0
+( Q
1
0
+ 2Q
2
0
+ 2Q
3
0
+ Q
4
0
)/ 6=2.02596
Demak, berilgan tenglamaning birinchi qiymati
y
1
=2.02596
bo‘ladi. Yuqoridagi qoidani i=1, x
1
=1.9, y
1
=2.02596 lar uchun qo‘llab
y
2
=3.0408
ni topamiz. SHuningdek, i=2,3,…,10 lar uchun tenglama yechimini qolgan qiymatlarini topamiz.
y
3
=3.2619
y
14
=3.4831
y
5
=3.7045
y
6
=3.9260
y
7
=4.1478
y
8
=4.370
y
9
=4.5931
y
10
=4.9172
MUSTAQIL ISHLASH
UCHUN TOPSHIRIQLAR
Quyidagi birinchi tartibli differentsial tenglamalar uchun Koshi masalasini ko'rsatilgan kesmada
h=0,1 bo‘lganda:
1.
Eyler usulida.
2.
Eylerning ketma-ket yaqinlashish usulida.
3.
Runge-Kutta usulida.
Taqribiy yechimini toping va dasturini tuzing.
1
2
3
4
5
y'=x/(x+y)
y(0)=1,
[0,1]
y'-2y=3e
x
y(0,3)=1,415
[0,1;0,5]
y'=x+y
2
y(0)=0,
[0;0,3]
y'=y
2
-x
2
y(1)=1,
[1;2]
y'=x
2
+y
2
y(0)=0.27
[0;1]
6
7
8
9
10
y'+xy(9-y
2
)=0
y(0)=0.5
[0;1]
y'=x
2
-xy+y
2
y(0)=0.1
[0;1]
y'=(2y-x)/y
y(1)=2
[1;2]
y'=x
2
+xy+y
2
+1
y(0)=0
[0;1]
y'+y=x
3
y(1)=-1
[1;2]
11
12
13
14
15
y'=xy+e
y
y(0)=0
[0;0.1]
y'=2xy+x
2
y(0)=0
[0;0.5]
y'=x+
3
sin
y
y(0)=1, [0;1]
y'=e
x
-y
2
y(0)=0
[0;0.4]
y'=2x+cosy
y(0)=0
[0;0.1]
16
17
18
19
20
y'=x
3
+y
2
y(0)=0.5
[0;0.5]
y'=xy
3
-y
y(0)=1
[0;1]
y'=y
2
e
x
-2y
y(0)=1
[0;1]
y'=
х
у
2
1
y(1)=0, [1;2]
y'=
x
e
x
1
2
y(1)=1, [1;2]
21
22
23
24
25
y'=e
x
cosy/x
y(1)=1 ,
[1;2]
y'=e
x
siny/x
y(1)=1
[1;2]
y'cosx-ysinx=2x
y(0)=0
[0;1]
y’=ytgx-
x
3
cos
1
y(0)=0 , [0;1]
y'+ycosx=cosx
y(0)=0
[0;1]
26
27
28
29
30
y’=
y
x
tg
x
y
y(0)=0, [0;1]
y'=(9+
x
y
2
1
)
2
y(1)=1, [1;2]
xy'-
1
x
y
-x=0
y(1)=1/2, [1;2]
y'=
x
y
(9+lny-lnx)
y(1)=e , [1;2]
y
3
xdx=(x
2
y+2)dy
y(0.348)=2
[0;1]
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