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

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

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