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- .MCD 20000 0
- .CMD PLOTFORMAT logs=0,0 subdivs=1,1 size=5,15 type=l
- .CMD FORMAT rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge
- .CMD SET ORIGIN 0
- .CMD SET TOL 0.001000
- .CMD MARGIN 0
- .CMD LINELENGTH 78
- .CMD SET PRNCOLWIDTH 8
- .CMD SET PRNPRECISION 4
- .TXT 0 42 1 39
- a1,38,39,37
- Copyright (c) 1988 by MathSoft, Inc.
- .TXT 1 -41 1 45
- a1,44,78,43
- SOLVING A SYSTEM OF DIFFERENTIAL EQUATIONS
- .TXT 0 80 1 13
- a1,12,78,11
- /equations
- .TXT 2 -80 3 37
- a3,36,35,97
- x and y are functions of the time
- variable t, and F and G are the
- derivatives dx/dt and dy/dt.
- .EQN 0 45 1 24
- F(t,x,y)~1.5*x-x*y
- .EQN 2 0 1 25
- G(t,x,y)~-3*y+2*x*y
- .EQN 4 -45 1 12
- startt~0
- .EQN 0 15 1 11
- endt~15
- .EQN 0 13 1 9
- n~100
- .TXT 0 10 1 12
- a1,11,40,10
- intervals
- .EQN 0 13 1 11
- initx~1
- .EQN 0 12 1 11
- inity~3
- .EQN 1 18 1 24
- k1(t,x,y,h)~F(t,x,y)
- .EQN 1 -69 13 45
- U&L&x[k,y[k,0{1,1,11,35,ll}@endt&startt&t[k
- .EQN 0 44 1 9
- U=?
- .EQN 1 25 1 24
- K1(t,x,y,h)~G(t,x,y)
- .EQN 2 0 1 69
- k2(t,x,y,h)~F(t+.5*h,x+.5*h*k1(t,x,y,h),y+.5*h*K1(t,x,y,h))
- .TXT 1 -81 2 12
- a2,11,77,19
- graph of
- solution
- .EQN 1 81 1 69
- K2(t,x,y,h)~G(t+.5*h,x+.5*h*k1(t,x,y,h),y+.5*h*K1(t,x,y,h))
- .EQN 2 0 1 69
- k3(t,x,y,h)~F(t+.5*h,x+.5*h*k2(t,x,y,h),y+.5*h*K2(t,x,y,h))
- .EQN 2 0 1 69
- K3(t,x,y,h)~G(t+.5*h,x+.5*h*k2(t,x,y,h),y+.5*h*K2(t,x,y,h))
- .EQN 1 -25 1 10
- L=?
- .EQN 1 25 1 60
- k4(t,x,y,h)~F(t+h,x+h*k3(t,x,y,h),y+h*K3(t,x,y,h))
- .EQN 2 0 1 60
- K4(t,x,y,h)~G(t+h,x+h*k3(t,x,y,h),y+h*K3(t,x,y,h))
- .EQN 2 0 3 77
- rk(t,x,y,h)~h/6*(k1(t,x,y,h)+2*k2(t,x,y,h)+2*k3(t,x,y,h)+k4(t,x,y,h))
- .EQN 4 0 3 77
- rK(t,x,y,h)~h/6*(K1(t,x,y,h)+2*K2(t,x,y,h)+2*K3(t,x,y,h)+K4(t,x,y,h))
- .EQN 3 33 3 19
- h~(endt-startt)/n
- .EQN 1 -33 1 15
- j~0;n-1
- .EQN 0 17 2 13
- t[0~startt
- .EQN 0 40 2 15
- t[(j+1)~t[j+h
- .EQN 3 -57 4 16
- ({2,1}÷y[0÷x[0)~({2,1}÷inity÷initx)
- .EQN 0 27 4 32
- ({2,1}÷y[(j+1)÷x[(j+1))~({2,1}÷y[j+rK(t[j,x[j,y[j,h)÷x[j+rk(t[j,x[j,y[j,h))
- .EQN 5 -27 1 51
- L~floor(if(min(x)<min(y),min(x),min(y))-.5)
- .EQN 0 63 1 11
- k~0;n
- .EQN 1 -63 1 50
- U~ceil(if(max(x)>max(y),max(x),max(y))+.5)
-