Question

Form the differential equation representing the family of ellipses having foci on $y$-axis and center at the origin.

Answer 1

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

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

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

let

$\frac{dy}{dx}=y^{\prime }\text{and}\frac{d^{2}y}{d{x}^2}=y"$

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

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

$\frac{d}{dx}[\frac{x^2}{a^2}+\frac{y^2}{b^2}]=\frac{d\left(1\right)}{dx}$

$\frac{1}{a^2}\times \frac{d\left(x^2\right)}{dx}+\frac{1}{b^2}\times \frac{d\left(y^2\right)}{dx}=0$

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

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

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

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

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

$\frac{y{y}^{\prime }}{x}=\frac{-b^2}{a^2}$

Again, differentiating both sides w.r.t. x
we get,
Quotient rule

$\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

$\frac{d\left(\frac{y}{x}\right)}{dx}.y^{\prime }+\frac{y}{x}\frac{d\left(y^{\prime }\right)}{dx}=\frac{d}{dx}\left(\frac{-b^2}{a^2}\right)$

$\frac{[\frac{dy}{dx}.x-y.\frac{dx}{dx}]}{x^2}{y}^{\prime }+\frac{y}{x}\times y"=0$

$\frac{[y^{\prime }x-y]}{x^2}{y}^{\prime }+\frac{y}{x}\times y"=0$

$[y^{\prime }x-y]y^{\prime }+xyy"=0$

$xyy"+x(y^{\prime }{)}^2-y{y}^{\prime }=0$

Final Answer:
Hence, the required differential equation is

$xyy"+x(y^{\prime }{)}^2-y{y}^{\prime }=0$