Chrystal's equation

In mathematics, Chrystal's equation is a first order nonlinear ordinary differential equation, named after the mathematician George Chrystal, who discussed the singular solution of this equation in 1896. The equation reads as


 * $$\left(\frac{dy}{dx}\right)^2 + Ax \frac{dy}{dx} + By + Cx^2 =0$$

where $$A,\ B, \ C$$ are constants, which upon solving for $$dy/dx$$, gives


 * $$\frac{dy}{dx} = -\frac{A}{2} x \pm \frac{1}{2} (A^2 x^2 - 4By - 4Cx^2)^{1/2}.$$

This equation is a generalization of Clairaut's equation since it reduces to Clairaut's equation under certain condition as given below.

Solution
Introducing the transformation $$4By=(A^2-4C-z^2)x^2$$ gives


 * $$xz\frac{dz}{dx} = A^2 + AB - 4C \pm Bz - z^2.$$

Now, the equation is separable, thus


 * $$\frac{z \, dz}{A^2 + AB - 4C \pm Bz - z^2} = \frac{dx}{x}.$$

The denominator on the left hand side can be factorized if we solve the roots of the equation $$A^2 + AB - 4C \pm Bz - z^2=0$$ and the roots are $$a,\ b = \pm \left[ B +\sqrt{(2A+B)^2 - 16C} \right]/2$$, therefore


 * $$\frac{z \, dz}{(z-a)(z-b)} = \frac{dx}{x}.$$

If $$a\neq b$$, the solution is


 * $$x \frac{(z-a)^{a/(a-b)}}{(z-b)^{b/(a-b)}} = k$$

where $$k$$ is an arbitrary constant. If $$a=b$$, ($$(2A+B)^2 - 16C=0$$) then the solution is


 * $$x(z-a) \exp \left[\frac a {a-z}\right]=k.$$

When one of the roots is zero, the equation reduces to Clairaut's equation and a parabolic solution is obtained in this case, $$A^2+ AB -4C=0$$ and the solution is


 * $$x(z\pm B)=k, \quad \Rightarrow \quad 4By = - AB x^2 - (k\pm Bx)^2.$$

The above family of parabolas are enveloped by the parabola $$4By=-ABx^2$$, therefore this enveloping parabola is a singular solution.