Question

Form the differential equation representing the family of ellipses having foci on $x$ -axis and centre at the ofigin.

Answer 1

Equation of ellipse (having foci at x-axis & centre at origin)

(i) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ $b^2=a^2(1-e^2)$

$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$

no. of constraints are 2,

So order = 2

Differentiating with respect to $x,$ we get

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

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

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

Differentiating with respect to $x,$ we get

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

$\frac{d}{dx}(y.\frac{1}{x}\frac{dy}{dx})=0$

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

$\frac{y}{x}\frac{d^{2}y}{d{x}^2}+\frac{1}{x}(\frac{dy}{dx}{)}^2=\frac{y}{x^2}(\frac{dy}{dx})$

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

(ii) $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ (equation of ellipse with foci at y-axis)

Differentiating with respect to $x,$ we get

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

differentiating with respect to x again,

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

Substituting this in eq - (2)

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