专题3-4 插值与拟合+数值微分积分(8)

Nonlinear rows= 4 Nonlinear vars= 15 Nonlinear constraints= 4

Nonzeros= 73 Constraint nonz= 58 Density=0.304

No. < : 4 No. =: 0 No. > : 10, Obj=MIN Single cols= 0

Optimal solution found at step: 28970 Objective value: 28.00000 Branch count: 2946

Variable Value Reduced Cost X1 10.00000 0.0000000 X2 10.00000 0.0000000 X3 8.000000 0.0000000 R11 3.000000 0.0000000 R12 2.000000 0.0000000 R13 0.0000000 0.0000000 R21 0.0000000 0.0000000 R22 1.000000 0.0000000 R23 0.0000000 0.0000000 R31 1.000000 0.0000000 R32 1.000000 0.0000000 R33 0.0000000 0.0000000 R41 0.0000000 0.0000000 R42 0.0000000 0.0000000 R43 2.000000 0.0000000

Row Slack or Surplus Dual Price 1 28.00000 0.0000000 2 0.0000000 0.0000000 3 0.0000000 0.0000000 4 0.0000000 0.0000000 5 1.000000 0.0000000 6 1.000000 0.0000000 7 0.0000000 0.0000000 8 3.000000 0.0000000 9 2.000000 0.0000000 10 3.000000 0.0000000 11 0.0000000 0.0000000 12 2.000000 0.0000000 13 3.000000 0.0000000 14 0.0000000 0.0000000 15 2.000000 0.0000000 程序12：

Max 0.1y1-0.2226x1+0.1833x2+0.2618x3+0.1695x4+0.1571y2+0.0196y3 st

1.5x1+2x2+x3+3x4<=14.4 x1+x2+x3<=5 x4<=2

y2-x1-2x2-4x4+y1=0

y3-10x1-4x2-16x3-5x4+2y1=0 y1-x1-2x2-4x4<=0

y1-5x1+2x2+8x3+2.5x4<=0 end

ROW SLACK OR SURPLUS DUAL PRICES

2) 0.000000 0.241577 3) 0.284615 0.000000 4) 0.000000 0.055237 5) 0.000000 0.157100 6) 0.000000 0.019600 7) 15.369231 0.000000 8) 0.000000 0.046373

NO. ITERATIONS= 3

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE Y1 -0.100000 0.342673 INFINITY X1 -0.222600 0.034507 1.570250 X2 0.183300 0.038297 0.028418 X3 0.261800 0.037162 INFINITY X4 0.169500 INFINITY 0.055237 Y2 0.157100 INFINITY 0.024155 Y3 0.019600 0.001725 0.078512

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 14.400000 0.528571 6.900000 3 5.000000 INFINITY 0.284615 4 2.000000 1.840000 0.187342 5 0.000000 INFINITY 15.369231 6 0.000000 INFINITY 41.230770 7 0.000 13.400000 7.400000

第八次 微分方程数值解

P120实验内容四：

1.a :

欧拉和R-K和精确解的比较 function ydot=el1(x,y) ydot=y+2*x;

x=0:0.1:1; n=length(x); y(1)=1;

for i=2:n

y(i)=y(i-1)+0.1*(y(i-1)+2*x(i-1)) end

y1=3*exp(x)-2*x-2;

[x,y2]=ode45('el1',[0:0.1:1],1)

plot(x,y,'g',x,y1,'r',x,y2,'b+')

dsolve('Dy=y+2*x','y(0)=1','x')

dsolve('x^2*D2y+x*Dy+(x^2-1/4)*y=0','y(pi/2)=2,Dy(pi/2)=-2/pi','x')

2: 1）C小题

function ydot=el3(x,y)

ydot=[y(2);-y(2)/x-(1-1/(4*x*x))*y(1)];

[x,y]=ode45('el3',[pi/2:pi/100:3*pi],[2,-2/pi]); x1=pi/2:pi/10:3*pi

y1=sin(x1).*(2*pi./x1).^(1/2); plot(x1,y1,'r*',x,y(:,1))

4. 单摆：

一根长为l的 (无弹性)细线上端固定，下端系着一个质量为m的小球。在重力的

作用下小球处于平衡位置。使小球偏离平衡位置一个小角度，然后让它自由，小球

就会沿着圆弧摆动。在不考虑空气阻力的情况下，小球将作周期一定的简谐运动.

(1)试建立模型来描述单摆运动规律(初始偏角分别为

5??0.0873,30??0.5236).

(2)用matlab编写求解程序.(取l?25cm,g?9.8m?s?2) (1)分析与建立模型：

如图，以??0为平衡位置，以右边为正方向建立摆角?的坐标系.在小球摆动过程中的任一位置?,小球所受重力沿运动轨迹方向的分力为?mgsin?(负号表示力的方向与?的正向相反)，

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