Question

Form the differential equation by eliminating arbitrary constants from the relation $A{x}^2+B{y}^2=1$ or $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Answer 1

Given: Equation of ellipse$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$Concept:$

Here we see that there are two arbitrary constants($a$ and $b$) in equation of ellipse thus we have to get a differential equation of 2nd order for eliminating all constants.

$Solution:$

Differentiating given equation with respect to $x$, we get:

$\frac{2x}{a^2}+\frac{2y}{b^2}\left(\frac{dy}{dx}\right)=0$

$\Rightarrow \frac{y}{x}\left(\frac{dy}{dx}\right)=\frac{-b^2}{a^2}$

Again differentating with respect to $x$, we get:

$\Rightarrow \frac{y}{x}\left(\frac{d^{2}y}{d{x}^2}\right)+\frac{x\frac{dy}{dx}-y}{x^2}\frac{dy}{dx}=0$

$\Rightarrow xy\frac{d^{2}y}{d{x}^2}+x(\frac{dy}{dx}{)}^2-y\frac{dy}{dx}=0$

which is the required differential equation.