Question

Form the differential equation representing the family of ellipse having foci on x-axis as center at the origin.

Answer 1

According to question the given curve is Ellipse whose foci is on x-axis and center at origin is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let
$\frac{dy}{dx}=y^{\prime }$ and $\frac{d^{2}y}{d{x}^2}=y"$
Differentiating both sides w.r.t. $x$ we get,
$\frac{d}{dx}[\frac{x^2}{a^2}+\frac{y^2}{b^2}]=\frac{d\left(1\right)}{dx}$
$\frac{1}{a^2}\times \frac{d\left(x^2\right)}{dx}+\frac{1}{b^2}\times \frac{d\left(y^2\right)}{dx}=0$
$\frac{1}{a^2}\times 2x+\frac{1}{b^2}\times (2y.\frac{dy}{dx})=0$
$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$
$\frac{2y}{b^2}\frac{dy}{dx}=\frac{-2x}{a^2}$
$\frac{y}{b^2}\frac{dy}{dx}=\frac{-x}{a^2}$
$\frac{y}{x}\frac{dy}{dx}=\frac{-b^2}{a^2}$
$\frac{y}{x}{y}^{\prime }=\frac{-b^2}{a^2}$
Again,differentiatig both sides w.r.t. $x$ we get,
Product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$
$\frac{d\left(\frac{y}{x}\right)}{dx}.y^{\prime }+\frac{y}{x}\frac{d\left(y^{\prime }\right)}{dx}=\frac{d}{dx}\left(\frac{-b^2}{a^2}\right)$
$\frac{[\frac{dy}{dx}.x-y.\frac{dx}{dx}]}{x^2}{y}^{\prime }+\frac{y}{x}\times y^{\prime \prime }=0$
$\frac{[y^{\prime }x-y]}{x^2}{y}^{\prime }+\frac{y}{x}\times y^{\prime \prime }=0$
$[y^{\prime }x-y]y^{\prime }+xyy"=0$
$xy{y}^{\prime \prime }+x(y^{\prime }{)}^2-y{y}^{\prime }=0$
$xy\frac{d^{2}y}{d{x}^2}+x{\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}=0$

Final answer:
Hence, therequired differential equation is $xy\frac{d^{2}y}{d{x}^2}+x{\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}=0$