Question

Differential equation from $a{x}^2+b{y}^2=1$ is

Answer 1

Given,

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

differentiating on both sides, we get,

$2ax+2by\frac{dy}{dx}=0$

$ax+by\frac{dy}{dx}=0$...(1)

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

again differentiating on both sides (1), we get.

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

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

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

is the required equation.