It just occured to me that if we need to evaluate an implicit function given by a nonlinear equation for a range of values, common root-finding methods, like Newton-Raphson, might not be the most economic option. Let's say we have an equation:
$$F(x,y)=C$$
And we want to numerically evaluate $y(x)$ for $x \in [a,b]$.
Then one way to do that is turning the problem into an ODE:
$$F_x dx+F_y dy=0$$
$$\frac{dy}{dx}=-\frac{F_x}{F_y}$$
This equation then could be solved numerically by a suitable finite difference scheme. (On the condition that $F_y \neq 0$ for $x \in [a,b]$).
The initial condition can be obtained by a single application of a root-finding algorithm for the equation:
$$F(a,y)=C$$
If we need more initial conditions (for higher order schemes), they can be obtained by root-finding as well.
Can this method offer a numerical advantage (run-time, complexity, etc.) compared to Newton-Raphson or other algorithms? Does it depend on the nature of $F(x,y)$? Or is this method certainly worse than usual root-finding methods?
The applications I see is computing $y(x)$ for a range of values of $x$ without the need for very high precision (for example, in science applications, when we just need to plot the function).
Here's an example:
$$y^3-x y-1=0$$
The ODE is:
$$\frac{dy}{dx}=\frac{y}{3y^2-x}$$
The initial condition is:
$$y(0)=y_0=\sqrt[3]{1}$$
One useful thing is: by choosing $y_0$ to be one of the three cube roots of unity, we get the three solutions for the original equation. For example:
$$y_0=1$$
gives us precisely the real solution, while $y=e^{2 \pi i/3}$ and $y=e^{4 \pi i/3}$ give the two complex conjugate solutions, as confirmed by computations in Mathematica.
Here's the plot of $y(x)$ with the initial condition $y_0=1$ (blue) obtained from numerically solving the ODE compared to the plot of the exact solution of the cubic equation in radicals (orange):
