How to draw 3D implicit function graph with MATLAB

Drawing with isosurface
Example: draw an image of x ^ 2 + y ^ 2-z ^ 2 = 1
[x,y,z]=meshgrid(linspace(-10,10));
val=x.^2+y.^2-z.^2;
isosurface(x,y,z,val,1)
axis equal

Drawing implicit function of three variables with MATLAB
It's not simple, it can be represented by Z directly
For example, (x ^ 2 + y ^ 2 + Z ^ 2) ^ 3 = 27 * Z
How to express x with Z,
Wrong. How to use x, y to express Z?
How to solve Z?

For implicit functions, Z itself can not be expressed by X, y. Syms & nbsp; X & nbsp; Y & nbsp; ZF = (x ^ 2 + y ^ 2 + Z ^ 2) ^ 3-27 * Z; XX = 0:0.1:10; YY = XX; ZZ = sym (zeros (length (XX))); for & nbsp; I = 1: length (XX) & nbsp; & nbsp; & nbsp; & nbsp; F1 = subs (F, x, x

Drawing differential equation curve with MATLAB
X * d2y / DT2 = - 1 / 2 * (1 + (dy / DX) square), the initial condition is that when t = 0, x = - 100, y = 0, the initial value of the first reciprocal of Y is also 0,

Is there a differential equation about x
For example, DX / dt = f (x, y, t)?

How to use matlab to solve complex differential equations
d2x/dt2=(x^2+3x)(dy/dt)
d2y/dt2=-(x^2+3x)(dx/dt)
When t = 0, X (0) = 0, y (0) = 0
(dx/dt)(0)=a,(dy/dt)(0)=a
(d2x/dt2)(0)=a,(d2y/dt2)(0)=a
First, we need to get the expressions of X and y with respect to t,
Then we get the expression of Y with respect to X
The initial conditions are as follows:
When t = 0, X (0) = 0, y (0) = 0
(dx/dt)(0)=a,(dy/dt)(0)=0
(d2x/dt2)(0)=0,(d2y/dt2)(0)=b
Where a and B are constants

[x, y] = resolve ('d2x = (x * x + 3 * x) * dy ','d2y = - (x * x + 3 * x) * DX','x (0) = 0 ','y (0) = 0','dx (0) = a ','dy (0) = a','d2x (0) = a ','d2y (0) = a')%%%%%%%%%% set X (1) = DX / dt; X (2) = x; X (3) = dy / dt; X (4) = Y;%%%%%%% function DX =

How to draw 3D implicit function graph with MATLAB