How to solve non linear equation for one variable?
显示 更早的评论
Hi, I am trying to solve a non-linear equation for one variable but it's not working. Can somebody help me in this regards? your help would be highly appreciated. Thanks
if
temp=[24 78 139 194 297 397];
yieldstress=[45 36 37 33 30 28];
syms DD positive
solve((6.046.*sqrt(DD))./(1e6)+(9.79.*(1-((-4.7872e-5.*temp).*(log(6.667e-4./(286.*sqrt(DD))))).^(2/3)).^(3/2)).*(3.06) == yieldstress);
end
采纳的回答
If you want to solve for multiple values of temp and yieldstress, you will have to use a for loop.
for i=1:numel(temp)
for j=1:numel(yieldstress)
out{i,j} = solve((6.046.*sqrt(DD))./(1e6)+(9.79.*(1-((-4.7872e-5.*temp(i)).*(log(6.667e-4./(286.*sqrt(DD))))).^(2/3)).^(3/2)).*(3.06) == yieldstress(j));
end
end
The output variable "out" contains the values of DD that solve the equations as symbolic numbers (so long as the equation can be solved), and these can be evaluated using the command eval().
3 个评论
Farhan Ashraf
2018-4-24
njj1
2018-4-24
If you type in the code exactly as I have then it should work. It will take a while but it should work.
for i=1:numel(temp)
for j=1:numel(yieldstress)
out{i,j} = solve((6.046.*sqrt(DD))./(1e6)+(9.79.*(1-((-4.7872e-5.*temp(i)).*(log(6.667e-4./(286.*sqrt(DD))))).^(2/3)).^(3/2)).*(3.06) == yieldstress(j));
dd(i,j) = eval(out{i,j}); %DO NOT put DD here
end
end
FYI, I've run this code for the first few values of temp and yieldstress that you posted above, and Matlab is not able to find an explicit solution.
Farhan Ashraf
2018-4-24
更多回答（1 个）
Torsten
2018-4-24
0 个投票
The left-hand side of your equation (6.046.*sqrt(...)...) is a scalar, the right-hand side (yieldstress) is a vector. This is not compatible.
Best wishes
Torsten.
类别
在 帮助中心 和 File Exchange 中查找有关 Programming 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
