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
( d y d x ) 2 + A x d y d x + B y + C x 2 = 0 {\displaystyle \left({\frac {dy}{dx}}\right)^{2}+Ax{\frac {dy}{dx}}+By+Cx^{2}=0}
where A , B , C {\displaystyle A,\ B,\ C} are constants, which upon solving for d y / d x {\displaystyle dy/dx} , gives
d y d x = − A 2 x ± 1 2 ( A 2 x 2 − 4 B y − 4 C x 2 ) 1 / 2 . {\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 of some particular cases of Clairaut's equation since it reduces to a form of Clairaut's equation under certain conditions as given below.
Solution Introducing the transformation 4 B y = ( A 2 − 4 C − z 2 ) x 2 {\displaystyle 4By=(A^{2}-4C-z^{2})x^{2}} gives
x z d z d x = A 2 + A B − 4 C ± B z − z 2 . {\displaystyle xz{\frac {dz}{dx}}=A^{2}+AB-4C\pm Bz-z^{2}.}
Now, the equation is separable, thus
z d z A 2 + A B − 4 C ± B z − z 2 = d x x . {\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 A 2 + A B − 4 C ± B z − z 2 = 0 {\displaystyle A^{2}+AB-4C\pm Bz-z^{2}=0} and the roots are a , b = ± [ B + ( 2 A + B ) 2 − 16 C ] / 2 {\displaystyle a,\ b=\pm \left[B+{\sqrt {(2A+B)^{2}-16C}}\right]/2} , therefore
z d z − ( z − a ) ( z − b ) = d x x . {\displaystyle {\frac {z\,dz}{-(z-a)(z-b)}}={\frac {dx}{x}}.}
If a ≠ b {\displaystyle a\neq b} , the solution is
x ( z − a ) a / ( a − b ) ( z − b ) b / ( a − b ) = k {\displaystyle x{\frac {(z-a)^{a/(a-b)}}{(z-b)^{b/(a-b)}}}=k}
where k {\displaystyle k} is an arbitrary constant. If a = b {\displaystyle a=b} , ( ( 2 A + B ) 2 − 16 C = 0 {\displaystyle (2A+B)^{2}-16C=0} ) then the solution is
x ( z − a ) exp [ a a − z ] = k . {\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, A 2 + A B − 4 C = 0 {\displaystyle A^{2}+AB-4C=0} and the solution is
x ( z ± B ) = k , ⇒ 4 B y = − A B x 2 − ( k ± B x ) 2 . {\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 4 B y = − A B x 2 {\displaystyle 4By=-ABx^{2}} , therefore this enveloping parabola is a singular solution.
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