Question

Find the differential equation of all ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a$ and $b$ are constant.

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Differentiating both sides, w.r.t $x$ we get

$\frac{1}{a^2}2x+\frac{1}{b^2}2y\frac{dy}{dx}=0$

$\Rightarrow \frac{1}{b^2}2y\frac{dy}{dx}=-\frac{1}{a^2}2x$

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

Differentiating both sides, w.r.t $x$ we get

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

$\Rightarrow \frac{y}{x}\frac{d^{2}y}{d{x}^2}+\frac{1}{x}{\left(\frac{dy}{dx}\right)}^2-\frac{y}{x^2}\frac{dy}{dx}=0$

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

$\Rightarrow xy\frac{d^{2}y}{d{x}^2}+x{\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}=0$ is the required differential equation.