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Solve equation for one unknown with one known parameter

3 Ansichten (letzte 30 Tage)
clear all
clc
a=0.5
x=linspace(-2*a,2*a,100)
eqn=@(x,y,a)(1/2*a^2)*[x^2./(2-(1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))-y^2./((1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))]-1==0;
solve(eqn,y)
Answer is
Unrecognized function or variable 'y'.
Error in naca (line 6)
solve(eqn,y)
how can I solve y for x matrix value
  2 Kommentare
onur karakurt
onur karakurt am 18 Mai 2021
Also
ı dont want to use syms subprogram
value must be numerical
Star Strider
Star Strider am 18 Mai 2021
The actual equations are:
I = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/621828/image.png');
figure
imshow(I)
.

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Star Strider
Star Strider am 18 Mai 2021
Try something like this —
a=0.5;
% x=linspace(-2*a,2*a,100)
x=linspace(-2*a,2*a,20);
eqn=@(x,y,a)(1/2*a^2)*[x^2./(2-(1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))-y^2./((1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))]-1;
for k = 1:numel(x)
yv(k,:) = fsolve(@(y)eqn(x(k),y,a), 10);
end
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
Results = table(x', yv)
Results = 20×2 table
Var1 yv _________ ______ -1 1.9833 -0.89474 2.1056 -0.78947 2.2028 -0.68421 2.2812 -0.57895 2.3444 -0.47368 2.3946 -0.36842 2.4335 -0.26316 2.4619 -0.15789 2.4805 -0.052632 2.4897 0.052632 2.4897 0.15789 2.4805 0.26316 2.4619 0.36842 2.4335 0.47368 2.3946 0.57895 2.3444
It might be necessary to use the uniquetol function to eliminate duplicate (or near-duplicate) values of ‘yv’ and thier associated ‘x’ values.
.
  9 Kommentare
Star Strider
Star Strider am 19 Mai 2021
As always, my pleasure!

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