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wanyshenger

金虫 (正式写手)

引用回帖:
9楼: Originally posted by 断剑残雪 at 2015-07-30 13:08:44
给定矩阵,每一次让它自己提取即可。...

我刚刚试了下,用你的code跑出来的为什么是四个同样的解呀?
x =

(K1*a + K2*b + 2*K2*c)/(K1 - K2) - (2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[1]^3)/(K1 - K2) + (2*c*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[1])/(K1 - K2) - (2*K2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[1]^2)/(K1 - K2)
(K1*a + K2*b + 2*K2*c)/(K1 - K2) - (2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[2]^3)/(K1 - K2) + (2*c*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[2])/(K1 - K2) - (2*K2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[2]^2)/(K1 - K2)
(K1*a + K2*b + 2*K2*c)/(K1 - K2) - (2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[3]^3)/(K1 - K2) + (2*c*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[3])/(K1 - K2) - (2*K2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[3]^2)/(K1 - K2)
(K1*a + K2*b + 2*K2*c)/(K1 - K2) - (2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[4]^3)/(K1 - K2) + (2*c*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[4])/(K1 - K2) - (2*K2*RootOf(z^4 + z^3*(K1 + K2) - z^2*(c - K1*K2) - z*((K1*a)/2 + (K2*b)/2 + K1*c + K2*c) - K1*K2*c - (K1*K2*b)/2 - (K1*K2*a)/2, z)[4]^2)/(K1 - K2)
而且我刚刚试了把我上一个回复的复杂一点的方程组放进去解,解出来是八个,好像也是都一样,请问那怎么选择呢?
USAwaitforme
11楼2015-07-30 13:35:01
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WWJ1991

木虫 (正式写手)

引用回帖:
5楼: Originally posted by 断剑残雪 at 2015-07-30 08:34:25
最终版:
syms z a b c  K1 K2 x y;
a=1 ;b=1 ;c=1  ;K1=1 ;K2=1; %比如已知它们的值,要求x,y。那么把前面的注释去掉再运行即可,它们的值也可以自己设定。
=solve(x*sqrt(c+0.5*(x+y))/(a-x)==K1 , y*sqrt(c ...

同学你的程序中出现z1的求解时为什么?
syms z a b c  K1 K2 x y;
a=1 ;b=1 ;c=1  ;K1=1 ;K2=1;
[x,y]=solve(x*sqrt(c+0.5*(x+y))/(a-x)==K1 , y*sqrt(c+0.5*(x+y))/(b-y)==K2,x,y);
x=double(x)
y=double(y)
这样不是能直接求出数值解吗?
12楼2015-07-31 18:25:10
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