数值分析课程设计实验六.doc

61一、题目： 用欧拉公式和四阶龙格-库塔法分别求解下列初值问题；二、算法分析：Euler公式为：四阶龙格-库塔法为：这里我们设h=0.1三、程序设计及运行结果：（1）f11.mglobal y;%定义y为数组global x;a=0;b=1;h=0.1;alpha=1;f=inline(0.9*y/(1+2*x);x=a:h:b;n=(b-a)/h;y(1)=alpha;disp(euler公式求得的解为：)for i=2:n+1 y(i)=y(i-1)+h*feval(f,x(i-1),y(i-1); sprintf(x=%3.1f,y=%9.7f,x(i),y(i)end disp(四阶龙格-库塔法求得的解为：)for i=2:n+1 k1=feval(f,x(i-1),y(i-1); k2=feval(f,x(i-1)+h/2,y(i-1)+k1*h/2); k3=feval(f,x(i-1)+h/2,y(i-1)+k2*h/2); k4=feval(f,x(i-1)+h,y(i-1)+k3*h); y(i)=y(i-1)+h/6*(k1+2*k2+2*k3+k4); sprintf(x=%3.1f,y=%9.7f,x(i),y(i)end运行结果为：euler公式求得的解为：x=0.1,y=1.0900000x=0.2,y=1.1717500x=0.3,y=1.2470768x=0.4,y=1.3172249x=0.5,y=1.3830861x=0.6,y=1.4453250x=0.7,y=1.5044519x=0.8,y=1.5608688x=0.9,y=1.6148989x=1.0,y=1.6668064四阶龙格-库塔法求得的解为：x=0.1,y=1.0855051x=0.2,y=1.1634777x=0.3,y=1.2355334x=0.4,y=1.3027862x=0.5,y=1.3660420x=0.6,y=1.4259056x=0.7,y=1.4828446x=0.8,y=1.5372290x=0.9,y=1.5893579x=1.0,y=1.6394763xeuler公式求得的解四阶龙格-库塔法求得的解x=0.1y=1.0900000y=1.0855051x=0.2y=1.1717500y=1.1634777x=0.3y=1.2470768y=1.2355334x=0.4y=1.3172249y=1.3027862x=0.5y=1.3830861y=1.3660420x=0.6y=1.4453250y=1.4259056x=0.7y=1.5044519y=1.4828446x=0.8y=1.5608688y=1.5372290x=0.9y=1.6148989y=1.5893579x=1.0y=1.6668064y=1.6394763（2）f12.mglobal y;%定义y为数组global x;a=0;b=1;h=0.1;alpha=2;f=inline(-x*y/(1+x2);x=a:h:b;n=(b-a)/h;y(1)=alpha;disp(euler公式求得的解为：)for i=2:n+1 y(i)=y(i-1)+h*feval(f,x(i-1),y(i-1); sprintf(x=%3.1f,y=%9.7f,x(i),y(i)end disp(四阶龙格-库塔法求得的解为：)for i=2:n+1 k1=feval(f,x(i-1),y(i-1); k2=feval(f,x(i-1)+h/2,y(i-1)+k1*h/2); k3=feval(f,x(i-1)+h/2,y(i-1)+k2*h/2); k4=feval(f,x(i-1)+h,y(i-1)+k3*h); y(i)=y(i-1)+h/6*(k1+2*k2+2*k3+k4); sprintf(x=%3.1f,y=%9.7f,x(i),y(i)end运行结果为：euler公式求得的解为：x=0.1,y=2.0000000x=0.2,y=1.9801980x=0.3,y=1.9421173x=0.4,y=1.8886645x=0.5,y=1.8235382x=0.6,y=1.7505966x=0.7,y=1.6733644x=0.8,y=1.5947500x=0.9,y=1.5169573x=1.0,y=1.4415285四阶龙格-库塔法求得的解为：x=0.1,y=1.9900743x=0.2,y=1.9611612x=0.3,y=1.9156523x=0.4,y=1.8569530x=0.5,y=1.7888539x=0.6,y=1.7149854x=0.7,y=1.6384634x=0.8,y=1.5617372x=0.9,y=1.4865879x=1.0,y=1.4142132xeuler公式求得的解四阶龙格-库塔法求得的解x=0.1y=2.0000000y=1.9900743x=0.2y=1.9801980y=1.9611612x=0.3y=1.9421173y=1.9156523x=0.4y=1.8886645y=1.8569530x=0.5y=1.8235382y=1.7888539x=0.6y=1.7505966y=1.7149854x=0.7y=1.6733644y=1.6384634x=0.8y=1.5947500y=1.5617372x=0.9y=1.5169573y=1.4865879x=1.0y=1.4415285y=1.4142132四、精度分析： Euler公式的精度阶为一，而四阶龙格-库塔法的精度阶为四。所以四阶龙格-库塔法的精度更高。62一、题目： 用求解常微分方程初值问题的方法计算积分上限函数的值，实际上是将上面表达式两端求导化为常微分方程形式，并用初值条件。试用MATLAB中的指令ode23解决定积分计算问题二、算法分析：因为，其中则该问题就转化成了求解常微分方程初值问题，下面调用指令ode23求解三、程序及运行结果f2.mdisp(调用edo23函数求得的微分值为：)x,y=ode23(eq1,4 5,0)计算结果为：X的值对应的积分值4.00000000000000004.0000009328078290.0000008000002804.0000055968469730.0000048000100744.0000289170426940.0000248002689314.0001455180212950.0001248068105194.0007285229143010.0006249707274674.0036435473793290.0031290738332074.0182186697044700.0157320909047894.0910942813301740.0808623831691534.1910942813301740.1762809728777944.2910942813301740.2792485988765694.3910942813301730.3904243050480034.4910942813301730.5105261233539644.5910942813301730.6403365242301624.6910942813301720.7807083753594514.7910942813301720.9325714572793864.8910942813301711.0969395888905755.0000000000000001.291468550321271四、精度分析：本题运用的ode23方法计算的，精度阶最多只能达到3阶63一、题目： 将下列高阶常微分方程初值问题化为一阶常微分方程组的初值问题，然后用MATEAB指令ode23求数值解。二、算法分析：将下列高阶常微分方程初值问题化为一阶常微分方程组的初值问题为：三、程序及运行结果f3.mx,Y=ode23( cdq1,0,10,0 0 0 0)运行结果为：x = 0 0.000001000000000 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