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 $y=xy^{\prime }-{\frac {1}{2}}x^{2}y^{\prime \prime }$

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

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

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.

References

[1] Lass, Harry. "Elements of pure and applied mathematics" New York: McGraw-Hill Companies, 1957.

This entry is a derivative of the Public domain work [1].