用数值法求非线性方程组.doc

解非线性方程组题目2:用数值法求非线性方程组127627- y1+y126743-1e-x=0129988- y1+y126743-1e-4x=0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y1function y=y1(a,b)% y1(a,b) is a function;% y=y1(a,b)% x1 is variable;% x2 is variable;% a is constant;% b is constant;x1=a;x2=b;y=127627/126743*x2*exp(-x1)-127627*exp(-x1)-x2+127627;129988/126743*x2*exp(-4*x1)-129988*exp(-4*x1)-x2+129988;子程序y11function y=y11(a,b)% y11(a,b) is a function% y=y11(a,b)% x1 is variable% x2 is variable% a is constant% b is constantx1=a;x2=b;y=(-127627/126743)*x2*exp(-x1)+127627*exp(-x1);子程序y12function y=y12(a,b)% y12(a,b) is a function% y=y12(a,b)% x1 is variable% x2 is variable% a is constant% b is constantx1=a;x2=b;y=127627/126743*exp(-x1)-1;子程序y21function y=y21(a,b)% y21(a,b) is a function% y=y21(a,b)% x1 is variable% x2 is variable% a is constant% b is constantx1=a;x2=b;y=(-4*129988/126743)*x2*exp(-4*x1)+4*129988*exp(-4*x1);子程序y22function y=y22(a,b)% y22(a,b) is a function% y=y22(a,b)% x1 is variable% x2 is variable% a is constant% b is constantx1=a;x2=b;y=129988/126743*exp(-4*x1)-1;子程序yJfunction y=yJ(a,b)% yJ(x1,x2) is a function% y=yJ(a,b)% x1 is variable% x2 is variable% a is constant% b is constant% A is matrixx1=a;x2=b;y=y11(x1,x2) y12(x1,x2);y21(x1,x2) y22(x1,x2);子程序yJxxfzzfunction y=yJxxfzz(a,b)% yJxxfzz(x1,x2) is a function% y=yJxxfzz(a,b)% x1 is variable% x2 is variable% a is constant% b is constantx1=a;x2=b;y=-yJ(x1,x2)y1(x1,x2);子程序yyzfunction y=yyz(a,b)% yyz(x1,x2) is a function% y=yyz(a,b)% x1 is variable% x2 is variable% a is constant% b is constant% A1 is matrixx1=a;x2=b;A1=x1;x2;y=A1+yJxxfzz(x1,x2);子程序yJdfunction y=yJd(a,b)% yJd(x1,x2) is a function% y=yJd(a,b)% x1 is variable% x2 is variable% a is constant% b is constant% A1 is matrix% A2 is matrixx1=a;x2=b;error=0.1;while err