如何用matlab循环解方程,比如解方程3b-c=0,b是变量,取值从0到10,变化量为0.5,求方程的解c的每个值。

2025-03-28 23:02:49
推荐回答(3个)
回答(1):

sol=[];
syms c
for i=1:21
b=(i-1)*0.5;
eq1=3*b-c;
sol(i)=solve(eq1,'c');
end
sol

sol =
Columns 1 through 3
0 1.5000 3.0000
Columns 4 through 6
4.5000 6.0000 7.5000
Columns 7 through 9
9.0000 10.5000 12.0000
Columns 10 through 12
13.5000 15.0000 16.5000
Columns 13 through 15
18.0000 19.5000 21.0000
Columns 16 through 18
22.5000 24.0000 25.5000
Columns 19 through 21
27.0000 28.5000 30.0000

回答(2):

%你的错误原因,只定义了符号c,而不是符号数组c(i)
syms b c
for i=1:21
b(i)=(i-1)*0.5;
eq1=3*b(i)-c;
c1=solve(eq1==0);%我的版本是2012,solve函数的格式==
cnum(i)=eval(c1);%c1是求解出来的符号c,采用eval函数把符号常数变为数值常数
end
如果按照下面编程,非常简单
clear all
syms b c
csolve=solve(3*b-c==0,c);
bvalue=0:0.5:10;
cvalue=subs(csolve,b,bvalue)%help subs,查看用法

回答(3):

你非要用这么不靠谱的程序那只能这么改了
clear;clc;
syms b c
for i=1:21
b(i)=(i-1)*0.5;
eq1=3*b(i)-c;
[d(i)]=solve(eq1);
end

完全可以有更好的做法,需要的话追问

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