Examples

Family of curves x squared plus y squared equals cx

Family of curves $x^2+y^2=cx$

For example, start with the family given by $x^2+y^2=cx$.

Differentiate both sides: $2x+2y\frac{dy}{dx}=c$. Now plug in $c=\frac{1}{x}(x^2+y^2)$ from the original equation, and we get the differential equation:

$$2x+2y\frac{dy}{dx}=\frac{1}{x}(x^2+y^2)\text{which simplifies to}2xy\frac{dy}{dx}+x^2=y^2.$$

Conversely, let us solve this differential equation. Write it as:

$$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}.$$

This is a homogeneous equation, so we can substitute $z=y/x$:

$$\frac{dz}{dx}x+z=\frac{z^2-1}{2z}\text{which implies}\frac{-2z}{z^2+1}dz=\frac{dx}{x}.$$

Integrating both sides, applying $e^{-}$, and substituting $z=y/x$ gives:

$$|x|=\frac{c}{(y/x{)}^2+1}c>0.$$

Now multiply the fraction by $\frac{x^2}{x^2}$ and cross multiply to obtain $|x|(x^2+y^2)=c{x}^2$, which is equivalent to $(x^2+y^2)=c|x|$ assuming $x\ne 0$. Allowing $c\in \mathbb{R}$, we have the original equation.

Circles and rays

Circles and rays

We verify that the family of rays emanating from the origin is the orthogonal family to the family of circles centered at the origin. The family of circles is given by $x^2+y^2=c$. This corresponds to the differential equation $\frac{dy}{dx}=-\frac{x}{y}$. The orthogonal family is then given by $\frac{dy}{dx}=\frac{y}{x}$. Separating variables: $\frac{dy}{y}=\frac{dx}{x}$ so $\ln |y|=\ln |x|+c$ and thus $y=Cx$ for $C\in \mathbb{R}$. This is the family of rays.