PlanetPhysics/Differential Equation of the Family of Parabolas

To find the differential equation of the family of parabolas

$$y = ax + bx^2$$

we differentiate twice to obtain

$$ y^{\prime} = a + 2bx$$ $$ y^{\prime \prime} = 2b$$

The last equation is solved for $$b$$, and the result is substituted into the previous equation. This equation is solved for $$a$$, and the expressions for $$a$$ and $$b$$ are substituted into $$y = ax + bx^2$$. The result is the differential equation $$y = xy^{\prime} - \frac{1}{2}x^2y^{\prime \prime}$$

The elimination of the constants $$a$$ and $$b$$ can also be obtained by considering the equations

$$xa + x^2b + (-y)1 = 0$$ $$a + 2xb + (-y^{\prime})1 = 0$$ $$2b +(-y^{\prime \prime})1 = 0$$

as a system of homogeneous linear equations in $$a$$,$$b$$,$$1$$. The solution $$(a,b,1)$$ is nontrivial, and hence the determinant of the coefficients vanishes.

$$ \left| \begin{matrix} x & x^2 & -y \\ 1 & 2x & -y^{\prime} \\ 0 & 2 & -y^{\prime \prime} \end{matrix} \right| = 0 $$

Expansion about the third column yields the result above.