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.^[1] The equation reads as^[2][3]

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

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

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

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

Introducing the transformation ${\displaystyle 4By=(A^{2}-4C-z^{2})x^{2}}$ gives

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

Now, the equation is separable, thus

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

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

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

If ${\displaystyle a\neq b}$, the solution is

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

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

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

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, ${\displaystyle A^{2}+AB-4C=0}$ and the solution is

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

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