Mavzu: Xusussiy xosilali differensial tenglamalarni sonli yechish Birinchi tartibli oddiy differentsial tenglama uchun Koshi masalasini taqribiy yechish

Birinchi tartibli differentsial tenglama

{Y¹=F(X,Y) uchun}
{Koshi masalasini Eyler usulida}
{taqribiy yechimini topish}
#include
#include
float fne(float x, float y);
{ return x+cos(y/sqrt(5)) ; }
int main()
{ float x,y,y1,h,b,EPS;
int i,n
cin>>x>>y>>h>>b;
n=c((b-x)/h);
for (i=1;i<=n; i++)
{
y=y+h*fne(x,y);
x=x+h;
cout<return 0;}

2. Tenglama yechimini Eylerning ketma-ket yaqinlashish usulida hisoblaymiz. (6) formulada i=0 bo‘lganda

y₁⁽⁰⁾=y₀+hf (x₀;y₀)=y₀+k (x₀+cos(y₀/ ))=
=2.6+ 0.1(8+cos(6/ ))=2.6+0.1(8+0.3968)=2.81968
bo‘ladi. Bu Eyler usulidagi tenglama yechimining birinchi qiymati bo‘ladi.
Endi y₁⁽⁰⁾=2.81968 dan foydalanib (7) formulaga asosan i=1 bo‘lganda
y₁^(k)= y₀+ [f (x₀, y₀)+ f (x₁,y₁^(k-1))]
formulani k = 1,2,3,….. lar uchun ketma-ket
y₁⁽¹⁾, y₁⁽²⁾, y₁⁽³⁾, …, y₁^(k)
larni
y₁^(k-1) – y₁^(k)  < 0.001
shartni qanoatlantirguncha hisoblaymiz.
Demak,
k=1, y₁⁽¹⁾=y₀+ [f (x₀,y₀)+f (x₁,y₁⁽⁰⁾)]=2.6+0.05[2.1968+x₁+cos (y₁⁽⁰⁾/ )] =
=2.6+0.05[2.1968+1.9+0.36486] =2.7102
k=2, y₁⁽²⁾= 2.6+0.05[2.1968+x₁+cos(y₁⁽¹⁾/ )] =
=2.6+0.05[2.1968+1.9+ cos(7102/ )] =2.82239
k=3, y₁⁽³⁾= 2.6+0.05[2.1968+1.9+cos(83339/ )] =
=2.6+0.05[2.1968+1.9+ 0.303709] =2.82002
Endi xatolikni tekshiramiz.
y₁⁽²⁾ – y₁⁽³⁾  =2.8223 – 2.82202=0.0002< 0.001
Bundan 0.001 aniqlikdagi tenglama yechimining birinchi qiymati
y₁ = 2.82000  2.82
bo‘ladi.
Tenglama yechimi y₂ qiymatini topish uchun yuqoridagi qoidani takrorlaymiz.
i=1 uchun (6) formulaga asosan
y₂⁽⁰⁾= y₁ + hf (x₁, y₁) = 2.82 + 0.1(x₁+cos (y₁/ )) =
=2.82+0.1(9+cos(82/ ) = 3.04047
i=1 uchun (7) formulaga asosan
k=1, y₂⁽¹⁾= y₁+ [f (x₁, y₁)+ f (x₂, y₂⁽⁰⁾)] =
=2.82+0.05[2.20471+2+cos(3.0404/ )] =3.0407
k=2, y₂⁽²⁾=y₁+ [f (x₁,y₁)+f (x₂,y₂⁽²⁾)]= y₁+ [2.20471+2+ cos(u₂⁽¹⁾/ )]=
=2.82+0.05[2.20471+2+cos(3.0407/ )] =3.04071
k=3, y₂⁽³⁾= y₁+ [f (x₁, y₁)+ f (x₂, y₂⁽²⁾)] =
=2.82+0.05[2.20471+2+cos (3.0407/ )] =
=2.82+0.05[2.20471+2+cos (3.0407/ )] =3.0407
Endi xatolikni baholaymiz.
y₁⁽²⁾ – y₁⁽³⁾  =3.04071 – 3.04070=0.0001< 0.001
Bundan tenglama yechimining ikkinchi qiymati
y₂ = 3.0407
bo‘ladi.
Bu qoidani i=2,3,…,10 lar uchun ketma-ket davom ettirib tenglama yechimining qolgan qiymatlarini ham topamiz.
y₃=3.261, y₄=3.483, y₅=3.704, y₆=3.926
y₇=4.147, y₈=4.370, y₉=4.593, y₁₀=4.817
Bu usul yordamida hisoblash quyidagicha dastur asosida berilgan.
3. Eylerning takomillashgan usuli.
Quyidagi
y´=f (x,y)
birinchi tartibli differentsial tenglamani
y(x₀)=y₀
boshlang‘ich shartni qanoatlantiruvchi yechimni quyidagi takomillashtirish usullari bilan topamiz.
Birinchi takomillashtirish usuli.
Yuqoridagi masalani yechishda oraliq qiymatlarini aniqlaymiz:
x_i+1/2=x_i+h/2 , y_i+1/2=y_i+hf (x_i,y_i)/2
f _i+1/2=f (x _i+1/2,y _i+1/2), i=0, 1, 2, …..n. (8)
bu takomillashtirish bo‘yicha yechimning qiymatlarini quyidagicha topamiz.
y _i+1=y_i+hf (x _i+1/2,y _i+1/2), i=0, 1, 2, …..n. (9)
Ikkinchi takomillashtirish yoki Eyler – Koshi usuli. Birinchi navbatda aniqligi yaxshi bo‘lmagan yaqinlashishni
=y_i+hf (x _i,y _i), i=0, 1, 2, …..n.
ko‘rinishda topamiz va = ni hisoblaymiz.
Ikkinchi takomillashtirilgan yaqinlashishni quyidagicha topamiz.
y _i+1=y_i+ [f (x _i,y _i)+ ] (10)
Eylerning birinchi va ikkinchi takomillashgan usullarida qoldiq hadnig tartibi, har qadamda O(h³) bo‘ladi.
Har bir nuqtadagi xatolikni baholash takroriy hisob yordamida bo‘ladi, yahni hisob qadam bilan takrorlanadi va ning qiymatini aniqligi ( qadamda) quyidagicha baholanadi:
bu yerda u(x) – differentsial tenglamaning aniq yechimi.
2-MASALA. Birinchi va ikkinchi takomillashtirish usul bilan quyidagicha
y=y-
differentsial tenglamani h=0,2 qadam bilan u(0)=1 shartni qanoatlantiruvchi yechimni toping.
Yechish.
1. Birinchi takomillashtirish usuli. (8) va (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: u₂=u’+y₁=1.1836+0.1590=1,3426
shuningdek i=2, 3, 4, 5 lar uchun ham hisoblab, natijalarni quyidagi jadvalga yozamiz.
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(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.
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.

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