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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b86653ffeb8dc9710092b78366cce7e1ab196fc)
where
are constants, which upon solving for
, gives
![{\displaystyle {\frac {dy}{dx}}=-{\frac {A}{2}}x\pm {\frac {1}{2}}(A^{2}x^{2}-4By-4Cx^{2})^{1/2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/704ecd2cdaa221b06d9fff97f410a104b7e50e7b)
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
gives
![{\displaystyle xz{\frac {dz}{dx}}=A^{2}+AB-4C\pm Bz-z^{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4e3ce52deba54e4aa4b11a3960b229dcc772de8)
Now, the equation is separable, thus
![{\displaystyle {\frac {z\,dz}{A^{2}+AB-4C\pm Bz-z^{2}}}={\frac {dx}{x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e536d7ccf3a14f6bcfa19b36246a83dd20e0b2b)
The denominator on the left hand side can be factorized if we solve the roots of the equation
and the roots are
, therefore
![{\displaystyle {\frac {z\,dz}{(z-a)(z-b)}}={\frac {dx}{x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71c94382aca5bdd278d984315c7871b8f7af6b60)
If
, the solution is
![{\displaystyle x{\frac {(z-a)^{a/(a-b)}}{(z-b)^{b/(a-b)}}}=k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfb325556927535848d65ea719286ef4a4fba7fa)
where
is an arbitrary constant. If
, (
) then the solution is
![{\displaystyle x(z-a)\exp \left[{\frac {a}{a-z}}\right]=k.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b7d10cb12c33ded6cf8ae24b1f17b11c78e56a5)
When one of the roots is zero, the equation reduces to Clairaut's equation and a parabolic solution is obtained in this case,
and the solution is
![{\displaystyle x(z\pm B)=k,\quad \Rightarrow \quad 4By=-ABx^{2}-(k\pm Bx)^{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ebfbb79d64d50c095332635c7ca8f9fa121fef1)
The above family of parabolas are enveloped by the parabola
, therefore this enveloping parabola is a singular solution.
References
- ^ Chrystal G., "On the p-discriminant of a Differential Equation of the First order and on Certain Points in the General Theory of Envelopes Connected Therewith.", Trans. Roy. Soc. Edin, Vol. 38, 1896, pp. 803–824.
- ^ Davis, Harold Thayer. Introduction to nonlinear differential and integral equations. Courier Corporation, 1962.
- ^ Ince, E. L. (1939). Ordinary Differential Equations, London (1927). Google Scholar.