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

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

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

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

This equation is a generalization particular cases of Clairaut's equation since it reduces to a form of Clairaut's equation under 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

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

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 $a,b=\pm [B+\sqrt{(2A+B{)}^2-16C}]/2$, therefore

$\frac{zdz}{(z-a)(z-b)}=\frac{dx}{x}.$

If $a\ne b$, 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 a special-case of Clairaut's equation and a parabolic solution is obtained in this case, $A^2+AB-4C=0$ and the solution is

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

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