what does $dy/dx = (y^2-x^2)/xy$ model?

Question

Good morning,

Can someone help me in knowing what does the ODE $dy/dx = (y^2-x^2)/xy$ or $dy/dx = (y^2+x^2)/xy$model? I know how to solve it, I just want to know what physical phenomenon it models.

Thank you

Answer 1

There are probably many different things that are modeled by those equations.

Rewrite the first equation as $(xy)dy-(y^2-x^2)dx=0$

$$(xy)dy+(x^2-y^2)dx=0$$

Can be represented as the product of a vector function valued function with a displacement.

$$<x^2-y^2,xy>\cdot<dx,dy>=0$$

So the differential implies a vector field perpendicular to the displacement vector $d\vec{s}=<dx,dy>$.

If the vector field can be represented as a gradient, then it is the gradient of a scalar and the vector field is conservative. That has implications in fluid mechanics and electromagnetism.

One method for solving Maxwell's Equations in 2 dimensions,(e.g. practical if a cross section is infinitely long in the z direction), is to set up differential equations along these lines.Typically the function multiplying dx represents a curve intersecting the curve represented by the function multiplying dy at right angles. In this case, one function represents equipotential curves, the other represents points of constant electric field.

These are necessarily orthogonal. Sometimes one is easier to figure out than the other.

The surface of a conductor is an equipotential surface. The electric field is orthogonal to the surface and zero on the inside of a conductor.

The gradient of a scalar points in the direction of maximum increase. Traveling perpendicular to the gradient will yield a new point in space having the same value as the scalar at the previous point. This can be used to design certain objects with optimal geometries.