作业帮matlab中solve的用法我的函数如下function s=sss(p)syms x y;equation1=sym('c*x^2+c*x*y+(c*g-c*a-c*b+1)*x-a-e=0');equation2=sym('d*y^2+d*x*y+(d*g-d*a-d*b+1)*y-b-f=0');[x0,y0]=solve(equation1,equation2,'x','y');a=p(1);b=p(2);c=p(3);d=p(4);e=p(5)
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![matlab中solve的用法我的函数如下function s=sss(p)syms x y;equation1=sym('c*x^2+c*x*y+(c*g-c*a-c*b+1)*x-a-e=0');equation2=sym('d*y^2+d*x*y+(d*g-d*a-d*b+1)*y-b-f=0');[x0,y0]=solve(equation1,equation2,'x','y');a=p(1);b=p(2);c=p(3);d=p(4);e=p(5)](/uploads/image/z/580585-49-5.jpg?t=matlab%E4%B8%ADsolve%E7%9A%84%E7%94%A8%E6%B3%95%E6%88%91%E7%9A%84%E5%87%BD%E6%95%B0%E5%A6%82%E4%B8%8Bfunction+s%3Dsss%28p%29syms+x+y%3Bequation1%3Dsym%28%27c%2Ax%5E2%2Bc%2Ax%2Ay%2B%28c%2Ag-c%2Aa-c%2Ab%2B1%29%2Ax-a-e%3D0%27%29%3Bequation2%3Dsym%28%27d%2Ay%5E2%2Bd%2Ax%2Ay%2B%28d%2Ag-d%2Aa-d%2Ab%2B1%29%2Ay-b-f%3D0%27%29%3B%5Bx0%2Cy0%5D%3Dsolve%28equation1%2Cequation2%2C%27x%27%2C%27y%27%29%3Ba%3Dp%281%29%3Bb%3Dp%282%29%3Bc%3Dp%283%29%3Bd%3Dp%284%29%3Be%3Dp%285%29)
matlab中solve的用法我的函数如下function s=sss(p)syms x y;equation1=sym('c*x^2+c*x*y+(c*g-c*a-c*b+1)*x-a-e=0');equation2=sym('d*y^2+d*x*y+(d*g-d*a-d*b+1)*y-b-f=0');[x0,y0]=solve(equation1,equation2,'x','y');a=p(1);b=p(2);c=p(3);d=p(4);e=p(5)
matlab中solve的用法
我的函数如下
function s=sss(p)
syms x y;
equation1=sym('c*x^2+c*x*y+(c*g-c*a-c*b+1)*x-a-e=0');
equation2=sym('d*y^2+d*x*y+(d*g-d*a-d*b+1)*y-b-f=0');
[x0,y0]=solve(equation1,equation2,'x','y');
a=p(1);
b=p(2);
c=p(3);
d=p(4);
e=p(5);
f=p(6);
g=p(7);
x0=eval(x0);
y0=eval(y0);
s=[x0 y0];
取一个特殊的p=[0 0 0.1 0.2 0 0 3]
这时应该有一个[0 0]的解,可是matlab算出来的里面没有
只有
q =
-0.0000 -8.0000
-8.0000 0 - 0.0000i
-8.0000 0 + 0.0000i
matlab中solve的用法我的函数如下function s=sss(p)syms x y;equation1=sym('c*x^2+c*x*y+(c*g-c*a-c*b+1)*x-a-e=0');equation2=sym('d*y^2+d*x*y+(d*g-d*a-d*b+1)*y-b-f=0');[x0,y0]=solve(equation1,equation2,'x','y');a=p(1);b=p(2);c=p(3);d=p(4);e=p(5)
matlab的运行结果是：
0 -8.0000
-8.5000 - 0.8660i -0.0000 + 0.0000i
-8.5000 + 0.8660i -0.0000 - 0.0000i
上面的结果自身就是不正确的,我分析的原因是：matlab先求出了解析解,接着代入系数值,由于matlab运算精度低,系数又太多,公式极其复杂,从而导致结果失真过于严重
mathematica运行结果显示：mathematica先求出解析解,接着精确代入系数值计算,也出现了数量级为-16的误差,有部分失真,但是这个失真很容易发现,而得到修正
mathematica执行消去y的运算结果显示原方程组只能由三组解
借助mathematica的结果分析,可以得出：matlab的运行结果中,并没有丢失一组解,而是解的失真太为严重达到了0.1,
可以用下面的程序执行数值
function s=sss(p)
syms x y
a=p(1);
b=p(2);
c=p(3);
d=p(4);
e=p(5);
f=p(6);
g=p(7);
a1=simplify(c*x^2+c*x*y+(c*g-c*a-c*b+1)*x-a-e);
a2=simplify(d*y^2+d*x*y+(d*g-d*a-d*b+1)*y-b-f);
[x0,y0]=solve(a1,a2);
s=[x0 y0]
这个程序运行速度快且结果准确,
结果为
[ 0,0]
[ 0,-8]
[ -13,0]
注：可以看到新的程序输出结果是矩阵形式,而老程序的输出不是,问题很可能和eval的使用有关.