Algoritma program Pascal untuk menyelesaikan masalah nilai awal pada persamaan diferensial biasa orde satu menggunakan Metode Adams-Bashforth-Moulton.
program AdamsBashforthMoulton;
uses wincrt;
var
n : integer;
vinterval1, vinterval2, h, k1, k2, k3, k4 : real;
x, y, yaksen, yABM, yaksenABM : array [0..100] of real;
function fungsi(x,y:real):real;
begin
fungsi:= y-x;
end;
label 1, 2, 3, 4, 5;
begin
{Input diketahui}
writeln('Program Adam Bashforth Moulton.');
writeln('Diketahui : y` = y - x');
writeln(' Pada interval [a,b]');
write(' a = ');read(vinterval1);x[0]:= vinterval1;
write(' b = ');readln(vinterval2);
write(' y(',vinterval1:0:0,') = ');readln(y[0]);
write(' n = ');readln(n);h:= (vinterval2-vinterval1)/n;
{Pencarian 3 nilai awal Runge-Kutta}
n:= 0;
1:
x[n+1]:= x[n] + h;
if x[n+1] < vinterval2 then
begin
n:= n+1;
goto 1;
end;
for n:= 0 to 2 do
begin
k1:= h*fungsi(x[n],y[n]);
k2:= h*fungsi((x[n]+(1/2)*h),(y[n]+(1/2)*k1));
k3:= h*fungsi((x[n]+(1/2)*h),(y[n]+(1/2)*k2));
k4:= h*fungsi((x[n]+h),(y[n]+k3));
y[n+1]:= y[n] + (1/6)*(k1+2*k2+2*k3+k4);
end;
{Penghitungan nilai y aksen awal}
begin
for n:= 0 to 3 do
begin
yaksen[n]:= fungsi(x[n],y[n]);
end;
end;
{Penghitungan nilai ABM, catatan n=3}
n:= 3;
begin
2:
yABM[n+1]:= y[n] + (h/24) * (55*yaksen[n] - 59*yaksen[n-1] + 37*yaksen[n-2] - 9*yaksen[n-3]);
yaksenABM[n+1]:= fungsi(x[n+1],yABM[n+1]);
y[n+1]:= y[n] + (h/24) * (9*yaksenABM[n+1] + 19*yaksen[n] - 5*yaksen[n-1] + yaksen[n-2]);
yaksen[n+1]:= fungsi(x[n+1],y[n+1]);
if x[n+1] > vinterval2 then goto 3;
n:= n+1;goto 2;
end;
3:
{Pencetakan hasil ABM}
n:= 0;
begin
writeln;
writeln('| x[n] | py[n] | y[n] |');
4:
if n <= 3 then
begin
writeln('| ',x[n]:0:2,' | - | ',y[n]:0:7,' |');
end;
if n > 3 then
begin
writeln('| ',x[n]:0:2,' | ',yABM[n]:0:7,' | ',y[n]:0:7,' |');
end;
if x[n] < vinterval2 then
begin
n:= n+1;goto 4;
end;
if x[n] > vinterval2 then goto 5;
end;
5:
end.
Preview program:
Preview program:
Tidak ada komentar:
Posting Komentar