How to solve forth order Polynomial (equation) with one independent variable and two constants.
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Dear MATLAB experts,
I have an forth order polynomial (equation) as follow:
(A^2*y^4) + (4*x*A^2*y^3) + ((4*B-4-2*A^2)*y^2) + ((8-8*x-4*A^2*x)*y) + (8*x-4+A^2-4*x^2*B) = 0;
Where, A and B are constants and y is function of x. The independent variable x varies between 0 and 1 (0<=x<=1).
My questions are:
- Is there any way to define A and B as a constants instead of set them as a numbers since the equation is to be used in iterative code and thus A and B change from time to time.
- Need to find the function y=f(x).
I have tried
Equation=((A^2*y^4)+(4*x*A^2*y^3)+(((4*B)-4-(2*A^2))*y^2)+((8-(8*x)-(4*A^2*x))*y)+((8*x)-4+(A^2)-(4*x^2*B)));
solve(Equation,y);
The MATLAB version i have is 7.10.0 (R2010a), but not an issue if answers in newer version.
I really appreciate any suggestions
Thank you
Aziz
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回答 (1 件)
Walter Roberson
2016 年 2 月 1 日
If you have the Symbolic Toolbox, you can use
syms A B x y
Equation=((A^2*y^4)+(4*x*A^2*y^3)+(((4*B)-4-(2*A^2))*y^2)+((8-(8*x)-(4*A^2*x))*y)+((8*x)-4+(A^2)-(4*x^2*B)));
sol := solve(Equation, y);
The solution you get will be a RootOf() a quartic (polynomial of order 4), so there are 4 solutions, some of which might be imaginary at portions of the range x = 0 to 1. You can convert this to a function by using
Y = matlabFunction(sol, 'vars', [x A B]);
after which Y will be a function handle of a function that takes 3 parameters in the order x, A, B, and which should accept a vector of x values.
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