CODEWITHSUNDEEP
matlabequationsymbolic-math
user3375111
user3375111

Reputation: 141

How to get only the positive solution to a quadratic equation using the MATLAB Symbolic Toolbox?

I want to solve this equation in Matlab.

e=2;
while (e<30)
    syms v;
    solve('(v-e) = -(0.5*0.2*1.276*v^2*0.3)');
    e=e+1;
end

When I write for example "solve('(v-10) = -(0.5*0.2*1.276*v^2*0.3)');" it works. But I need variable "e" in this equation. In some cases, this equation has 2 solutions (negative and positive), but I need only positive solutions. What is the correct syntax? Thank you.

Upvotes: 1

Views: 2986

Answers (2)

horchler
horchler

Reputation: 18484

There's no need to use strings unless you're using a really old version of Matlab. This is the modern and preferred way of using solve:

syms v positive;
e = 2:29;
s = zeros(length(e),1);
for i=1:length(e)
    s(i) = double(solve(v-e(i)==-0.5*0.2*1.276*v^2*0.3,v));
end

However, since this is just a polynomial you can use the roots function:

e = 2:29;
s = zeros(length(e),1);
for i=1:length(e)
    r = roots([0.5*0.2*1.276*0.3 1 -e(i)]);
    s(i) = r(r>=0);
end

Upvotes: 1

Etienne
Etienne

Reputation: 1012

To add e to your equation you can concact it as a number to your equation: ['equation part one', num2str(e), 'end of your equation'].

To only have the positive solution, you can add a condition to your equation ( v>=0 ).

Here is an example of a complete solution to your problem:

ans = zeros(1,size(2:29,2));
i = 0;
syms v;
for e = 2:29
    i = i+1;
    ans(i) = solve(['(v-',num2str(e),') = -(0.5*0.2*1.276*v^2*0.3) and v>=0']);
end

Upvotes: 1

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