用数值法求非线性方程组

《用数值法求非线性方程组》由会员上传分享，免费在线阅读，更多相关内容在行业资料-天天文库。

1、解非线性方程组题目2:用数值法求非线性方程组程序流程图开始：定义函数y1(a,b)求出雅克比矩阵给函数一个任意初值计算得出结果的结果与给定的误差精度比较根据误差调整初值大于给定误差小于给定误差输出结果结束子程序y1functiony=y1(a,b)%y1(a,b)isafunction;%y=y1(a,b)%x1isvariable;%x2isvariable;%aisconstant;%bisconstant;x1=a;x2=b;y=[127627/126743*x2*exp(-x1)-127627

2、*exp(-x1)-x2+127627;129988/126743*x2*exp(-4*x1)-129988*exp(-4*x1)-x2+129988];子程序y11functiony=y11(a,b)%y11(a,b)isafunction%y=y11(a,b)%x1isvariable%x2isvariable%aisconstant%bisconstantx1=a;x2=b;y=(-127627/126743)*x2*exp(-x1)+127627*exp(-x1);子程序y12function

3、y=y12(a,b)%y12(a,b)isafunction%y=y12(a,b)%x1isvariable%x2isvariable%aisconstant%bisconstantx1=a;x2=b;y=127627/126743*exp(-x1)-1;子程序y21functiony=y21(a,b)%y21(a,b)isafunction%y=y21(a,b)%x1isvariable%x2isvariable%aisconstant%bisconstantx1=a;x2=b;y=(-4*1299

4、88/126743)*x2*exp(-4*x1)+4*129988*exp(-4*x1);子程序y22functiony=y22(a,b)%y22(a,b)isafunction%y=y22(a,b)%x1isvariable%x2isvariable%aisconstant%bisconstantx1=a;x2=b;y=129988/126743*exp(-4*x1)-1;子程序yJfunctiony=yJ(a,b)%yJ(x1,x2)isafunction%y=yJ(a,b)%x1isvariab

5、le%x2isvariable%aisconstant%bisconstant%Aismatrixx1=a;x2=b;y=[y11(x1,x2)y12(x1,x2);y21(x1,x2)y22(x1,x2)];子程序yJxxfzzfunctiony=yJxxfzz(a,b)%yJxxfzz(x1,x2)isafunction%y=yJxxfzz(a,b)%x1isvariable%x2isvariable%aisconstant%bisconstantx1=a;x2=b;y=-yJ(x1,x2)y1

6、(x1,x2);子程序yyzfunctiony=yyz(a,b)%yyz(x1,x2)isafunction%y=yyz(a,b)%x1isvariable%x2isvariable%aisconstant%bisconstant%A1ismatrixx1=a;x2=b;A1=[x1;x2];y=A1+yJxxfzz(x1,x2);子程序yJdfunctiony=yJd(a,b)%yJd(x1,x2)isafunction%y=yJd(a,b)%x1isvariable%x2isvariable%ai

7、sconstant%bisconstant%A1ismatrix%A2ismatrixx1=a;x2=b;error=0.1;whileerror>0.000015A1=yyz(x1,x2);A2=yJxxfzz(x1,x2);error=abs(A2(1,1))x1=A1(1,1);x2=A1(2,1);endy=A1;