수치해석 연습문제풀이 정익사

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수치해석 연습문제풀이 정익사에 대한 보고서 자료입니다.

본문내용

i=1,n
k(3,i)=h*f(i)
40 xn(i)=tx(i)+k(3,i)
call dots(tn+h,xn,f)
do 50 i=1,n
50 k(4,i)=h*f(i)
do 60 i=1,n
sk=k(1,i)+2*k(2,i)+2*k(3,i)+k(4,i)
60 xn(i)=tx(i)+sk/6
do 70 i=1,n
70 tx(i)=xn(i)
80 tn=tn+h
return
end
subroutine dots(t,x,f)
real x(10),f(10)
f(1)=x(2)
f(2)=x(3)
f(3)=t+2*x(1)+t*f(1)-t*f(2)
return
end
data : 03xypq
0.0 0.6 0.2 .0001
0.0 1.0 0.0
결과
x-value y-value y-deriv p-value p-deriv q-value q-deriv
0.000 0.0000 1.0000 1.0000 0.0000 0.0000 0.0000
0.200 0.2003 1.0053 1.0053 0.0793 0.0793 0.7857
0.400 0.4042 1.0418 1.0418 0.3095 0.3095 1.5013
0.600 0.6210 1.1380 1.1380 0.6733 0.6733 2.1208
8.15
subroutine dots(t,x,f)
real x(10),f(10)
real k,m
f(1)=x(2)
f(2)=16*cos(8*t)-64*x(1)
return
end
결과
t-value x-value x-deriv p-value p-deriv
0.000 0.0000 0.0000 0.0000 16.0000
0.200 0.1999 0.9529 0.9529 -13.2617
0.400 -0.0233 -3.2529 -3.2529 -14.4786
0.600 -0.5977 -0.5762 -0.5762 39.6522
0.800 0.0932 6.4729 6.4729 9.9240
1.000 0.9894 -0.1746 -0.1746 -65.6465
8.17
c simulataneous differential equations
c solving with modified euler method
external fxp,fyp
open(6,file=\'bnd17.out\',status=\'old\')
c (tl,tu): independent variable bounary value
c (xl,xu): dependent variable initial value
c (ch1,ch2):asumed derivertive value
c h interval increment value
c fxp:function subroutine supplid by user dx/dt=y
c fyp:function subroutine supplied by user dy/dt
tl=0
tu=1.
xl=0
xu=1.9
ch1=1.
ch2=1.50
h=0.1
y0=ch1
r=xu
t0=tl
x0=xl
y0=ch1
tol=0.000001
call euler(t0,tu,h,x0,y0,fxp,fyp,r1,tol)
dch=100
drr=100
while(dch.gt.tol.and.drr.gt.tol) do
t0=tl
x0=xl
y0=ch2
call euler(t0,tu,h,x0,y0,fxp,fyp,r2,tol)
ch= ch1+(ch2-ch1)/(r2-r1)*(r-r1)
ch1=ch2
ch2=ch
dch=abs(ch1-ch2)
drr=abs(r1-r2)
r1=r2
endwhile
stop
end
subroutine euler(t0,tl, h,x0,y0,fxp,fyp,xc,tol)
write(*,15)
15 format(9x,\'x\',9x,\'y\',9x,\'p\')
write(*,25)t0,x0,y0
while(t0.le.tl)do
25 format(5x,6f10.4)
t1=t0+h
xdot=fxp(t0,x0,y0)
ydot=fyp(t0,x0,y0)
xp=x0+h*xdot
yp=y0+h*ydot
dx=10.
dy=10.
while(dx.gt.tol.and.dy.gt.tol)do
xc=x0+(xdot+fxp(t1,xp,yp))*h/2.
yc=y0+(ydot+fyp(t1,xp,yp))*h/2.
dx=abs(xc-xp)
dy=abs(yc-yp)
xp=xc
yp=yc
endwhile
t0=t1
x0=xc
y0=yc
write(*,25)t0,x0,y0
endwhile
return
end
function fxp(x,y,p)
fxp=p
return
end
function fyp(x,y,p)
fyp=4.2*x+ 3*y-x*p
return
end
x y y\'
0.0000 0.0000 1.0000
0.1000 0.1000 1.0310
0.2000 0.2077 1.1238
0.3000 0.3279 1.2787
0.4000 0.4666 1.4958
0.5000 0.6301 1.7750
0.6000 0.8247 2.1163
0.7000 1.0565 2.5198
0.8000 1.3318 2.9855
0.9000 1.6567 3.5132
1.0000 2.0375 4.1031
x y y\'
0.0000 0.0000 1.5000
0.1000 0.1500 1.5360
0.2000 0.3090 1.6437
0.3000 0.4824 1.8236
0.4000 0.6773 2.0757
0.5000 0.9011 2.4000
0.6000 1.1609 2.7964
0.7000 1.4640 3.2650
0.8000 1.8175 3.8057
0.9000 2.2287 4.4186
1.0000 2.7048 5.1036
x y y\'
0.0000 0.0000 0.8970
0.1000 0.0897 0.9269
0.2000 0.1869 1.0166
0.3000 0.2960 1.1664
0.4000 0.4232 1.3763
0.5000 0.5743 1.6462
0.6000 0.7554 1.9762
0.7000 0.9725 2.3663
0.8000 1.2317 2.8164
0.9000 1.5388 3.3267
1.0000 1.9000 3.8969
8.18 답: y

0 0
л/8 0.3851
л/4 0.7108
3л/8 0.9268
л/2 1

키워드

수치해석,   연습문제 ,   정익사