PlanetPhysics/Differential Equation of the Family of Parabolas

To find the differential equation of the family of parabolas

$y=ax+b{x}^2$

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 $y=ax+b{x}^2$. The result is the [[../DifferentialEquations/|differential equation]] $y=x{y}^{\prime }-\frac{1}{2}{x}^2{y}^{\prime \prime }$

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 [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] of homogeneous linear equations in $a$,$b$,$1$. The solution $(a,b,1)$ is nontrivial, and hence the [[../Determinant/|determinant]] of the coefficients vanishes.

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

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 [[../Bijective/|domain]] [[../Work/|work]] [1].