6- tajriba mashg’ulot

100 FOR I=1 TO N
110 Y=Y+H*FNE(X,Y):Y(I)=Y
120 X=X+H:X(I)=X
130 PRINT " y(";USING "##.##";X;
140 PRINT ")=";USING "###.#####";Y
150 NEXT I
160 FOR I=1 TO N
170 PSET(X(I)*60+250,-Y(I)*60*.64+136),3
180 NEXT I
190 FOR X1=X(1) TO X(N) STEP .01
200 S=0
210 FOR I=1 TO N :P=1 : FOR J=1 TO N
220 IF I=J THEN 240
230 P=P*(X1-X(J))/(X(I)-X(J))
240 NEXT J
250 S=S+P*Y(I)
260 NEXT I
270 Y1(X1)=S:FOR R=1 TO 35000!:NEXT R
280 PSET(X1*60+250,-Y1(X1)*60*.64+136),5
290 NEXT X1
300 END

9.1- Paskal tili dasturi
{Birinchi tartibli differentsial tenglama }
{Y¹=F(X,Y) uchun}
{Koshi masalasini Eyler usulida}
{taqribiy yechimini topish}
Uses Crt;
function fne(x,y:real):real;
begin fne:=x+cos(y/sqrt(5)) end;
var x,y,y1,h,b,EPS:real;
i,n:integer;
begin
write(' x=',' y=',' h=',' b=');readln(x,y,h,b);
n:=trunc((b-x)/h);
for i:=1 to n do
begin
y:=y+h*fne(x,y);
x:=x+h;
writeln;
write(' x=(',i:2,')=',x:8:4);
write(' y=(',i:2,')=',x:8:4);
end;
end.

2. Tenglama yechimini Eylerning ketma-ket yaqinlashish usulida hisoblaymiz. (9.6) formulada i=0 bo‘lganda
y₁⁽⁰⁾=y₀+hf (x₀;y₀)=y₀+k (x₀+cos(y₀/ ))=
=2.6+ 0.1(9.8+cos(9.6/ ))=2.6+0.1(9.8+0.3968)=2.81968
bo‘ladi. Bu Eyler usulidagi tenglama yechimining birinchi qiymati bo‘ladi.
Endi y₁⁽⁰⁾=2.81968 dan foydalanib (9.7) 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(9.7102/ )] =2.82239
k=3, y₁⁽³⁾= 2.6+0.05[2.1968+1.9+cos(9.83339/ )] =
=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 (9.6) formulaga asosan
y₂⁽⁰⁾= y₁ + hf (x₁, y₁) = 2.82 + 0.1(x₁+cos (y₁/ )) =
=2.82+0.1(9.9+cos(9.82/ ) = 3.04047
i=1 uchun (9.7) 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.

8 ‘------- 9.2- dastur------------
10 DEF FNE (X,Y)=X+COS(Y/SQR (5))
14 RRINT “Birinchi tartibli differentsial tenglama ”
15 PRINT “ Y¹=F(X,Y) uchun”
16 PRINT “Koshi masalani Eylerning ketma-ket yaqinlashish”
18 PRINT “ usulida taqribiy yechimini topish”
20 REM “boshlang’ich qiymat ,qadam, berilgan kesma yuqori chegarasi:”
22 READ X, Y, H, B: EPS=0.0001
24 REM “boshlang’ich qiymat ,qadam, berilgan kesma yuqori chegarasi qiymatlari:”
26 DATA 1.8, 2.6, 0.1, 2.8
50 N=(B-X)/H
60 FOR I=1 TO N
70 Y1=Y
80 Y=Y+H*FNE(X,Y)
90 Z=Y
100 Y=Y1+(FNE(X,Y1)+FNE(X+H,Z))*H/2
102 X=X+H
110 IF ABS (Z-Y)>EPS THEN 80
120 PRINT “X (“;USING “###.###”;I
130 PRINT “)=”;USING “###.###”;X;
140 PRINT “ Y(“;USING “###.###”;I;
150 PRINT “)=”;USING “###.###”;Y
160 PRINT “ Z(“;USING “###.###”;I;
170 PRINT “)=”;USING “###.###”;Z
180 NEXT I
200 END


Birinchi tartibli differentsial tenglama
Y¹=F(x,y) uchun
Koshi masalasini Eylerning ketma-ket yaqinlashish
usulida taqribiy yechimini topish.
X(1)=1.900 Y(1)=2.8197 Z(1)=2.8197