How can I solve this implicit function in MatLab?

I have the implicit function f1 = @(x,y) a*log(y) - alpha*y + b*log(x) - beta*x - k1 and with the fimplicit command it is possible to plot a graph. I need to create an output with the x and y values ​​used to plot this graph. Is there a way to obtain these values? a, alpha, b, beta and k1 are known.

Answer 1

[John D'Errico]

Trivial, in some eyes. ;-) This is one of the simpler implicit problems you might write down, since a direct solution exists in the form of the Wright omega function.

syms a y x alpha k1 b beta

ysol(x,a,b,alpha,beta,k1) = solve(a*log(y) - alpha*y + b*log(x) - beta*x - k1,y)

ysol(x, a, b, alpha, beta, k1) = 

And of course, you have not provided any of the constants. But if you did, I could even evaluate it.

ysol(3,4,5,6,7,8)

ans = 

double(ysol(1,2,3,4,5,6))

ans = -2.2686 + 1.3091i

I'd bet for some sets of those parameters, the result will even be real. I don't need to change a lot.

double(ysol(1,-2,3,4,5,6))

ans = 0.0041

[João Pedro]

Thank you very much, this answer helped a lot. I learned one more thing. What I needed most was a y(x) function. Just one more question, how can the omega variable that appears be determined? Below I leave the values ​​that I am using in the problem.

a = 12;

b = 6;

alpha = 4;

beta = 2.5;

x0 = 2.5;

y0 = 1.5;

k1 = a*log(y0) - alpha*y0 + b*log(x0) - beta*x0;

f1 = @(x,y) a*log(y) - alpha*y + b*log(x) - beta*x - k1;

fp = fimplicit(f1,[0 7 0 6]);

Answer 2

[Steven Lord]

Let's take a simpler example and call fimplicit with an output argument, the handle of the graphics object that fimplicit plots.

f1 = @(x,y) x.^2-y.^2+sin(x).*cos(y);
h = fimplicit(f1);

Among the properties the object h returned from fimplicit has are XData and YData properties.

X = h.XData;
Y = h.YData;

Let's plot the X and Y data to see if it gives us the same plot.

figure
plot(X, Y)

[João Pedro]

Thanks for the answer, getting the entire set of points used to build the graph with the fp.XData and fp.YData commands helped a lot too.