Matlab can't solve this equation

Secondly, there's nothing wrong with your code, your problem is that your equation cannot be solved explicitly, at least not by Matlab (by me neither ;-). I (probably) have a newer version of the Symbolic Toolbox, and I get

F = solve(x^2+y^2+2*(1-mu)/sqrt((x+mu)^2 + y^2)+2*mu/sqrt((x-1+mu)^2+y^2)-C, y);
Warning: The solutions are parametrized by the symbols:
z1 = RootOf(z^8 %... VERY long expression involving z and x)

and

     F =
0.0000000000000000017347234759768070944119244813919*(576460752303423488.0*x + 576460752303423488.0*z1 - 569456276141304393.0)^(1/2)*(576460752303423488.0*z1 - 576460752303423488.0*x + 569456276141304393.0)^(1/2)
 -0.0000000000000000017347234759768070944119244813919*(576460752303423488.0*x + 576460752303423488.0*z1 - 569456276141304393.0)^(1/2)*(576460752303423488.0*z1 - 576460752303423488.0*x + 569456276141304393.0)^(1/2)

In other words, in order to solve your equation explicitly, Matlab would have to solve an 8-th order polynomial, which is not possible symbolically and so its roots are represented as a parameter z1 in the solution.

I'm not sure where that expression with z and x comes from, but that's probably why there is no call to the numeric solver at this point.

You could try to change your equation by hand a little bit, maybe put it in polar coordinates or something like that?

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Answer 2

Walter Roberson 2012-11-12

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/53466-matlab-can-t-solve-this-equation#answer_65062

在 MATLAB Online 中打开

That equation involves solving a polynomial of degree 8. As that is not ameniable to analytic solution, MATLAB attempts to do a numeric solution, but is unable to come up with one due to the presence of the variable "x" in the equation. MATLAB then returns that no explicit solution was found. It would be better if a RootOf() was returned instead, but that does not happen (especially before R2010b.)

y = (-35683896960000/36567140243929-x^2 + (11947200/6047077)*x + RootOf(126341933667992887917257169865924328125*Z^8 + (751910964962903584397247990989198656250*x - 1180461998766305767714643666362521161125)*Z^6 + 6140637045053743106295028206347312500*Z^5 + (-4690165106043173159677104822135769778500*x + 1491615269667830382072291319994700000000*x^2 + 3243059774492681322198866621654800175507)*Z^4 + (24413320515953573444409323967394625000*x - 31682889022049911181469843314314866000)*Z^3 + (-2650902734132599077567116839128671217943 + 6461052414907953754067607308960084929514*x + 986322148827989529893648034960000000000*x^3 - 4658555017613388317895166544987157600000*x^2)*Z^2 + (-63220176859569683563876554297224357000*x + 24264092851692174463638535109400000000*x^2 + 25067134077408890330125934349515421500)*Z + 149227664261398980770788857994625000*x - 72800592265069399531566165702687500)^2, Z)^(1/2)

Together with the negatives of those values.

In the above, RootOf( f(Z), Z) is the set of values, Z, such the f(Z) becomes 0 -- the roots of the expression.

Warning: a small change to the constants you provided, including adding a single decimal more precision, could result in measurably different solution sets.

Caution: for the test values I tried, near x=1, half of the roots were real valued. You cannot just pick a single root and use that: the other roots meet it and take over, with the original root going complex.