Question

Show that the differential equation of the family of circles having their centre at the origin and radius $a$ is
$x+y\frac{dy}{dx}=0$

Answer 1

$y=e^{aco{s}^{-1}x}$

$-1\le x\le 1$

$\Rightarrow y_1=\frac{dy}{dx}=\frac{d}{dx}\left(e^{aco{s}^{-1}x}\right)=e^{aco{s}^{-1}x}.\frac{-1.a}{\sqrt{1-x^2}}=\frac{-a}{\sqrt{1-x^2}}{e}^{aco{s}^{-1}x}$

$y_2=\frac{d^{2}y}{d{x}^2}=\frac{d{y}_1}{dx}=\frac{d}{dx}\left\{\frac{-a}{\sqrt{1-x^2}}{e}^{aco{s}^{-1}x}\right\}$

$=\frac{-a}{\sqrt{1-x^2}}.e^{aco{s}^{-1}x}.\frac{-a}{\sqrt{1-x^2}}+e^{a}co{s}^{-1}x.\frac{(-a)\{a-\frac{2x}{2\sqrt{1-x^2}}\}}{1-x^2}$

$=\frac{a^2}{(1-x^2)}{e}^{aco{s}^{-1}x}-\frac{ax}{(1-x^2)3/2}{e}^{aco{s}^{-1}x}$

$(1-x^2)y_2=a^2{e}^{aco{s}^{-1}x}-\frac{ax}{\sqrt{1-x^2}}{e}^{aco{s}^{-1}x}$

$x{y}_1=\frac{-ax}{\sqrt{1-x^2}}{e}^{aco{s}^{-1}x}$

$a^{2}y=a^2{e}^{aco{s}^{-1}x}$

$\therefore (1-x^2)y_2-x{y}_1-a^{2}y=a^2{e}^{aco{s}^{-1}x}-\frac{ax}{\sqrt{1-x^2}}{e}^{aco{s}^{-1}x}+\frac{ax}{\sqrt{1-x^2}}{e}^{aco{s}^{-1}x}$

$-a^2{e}^{aco{s}^{-1}x}$

$=0$

$\therefore (1-x^2)y_2-x{y}_1-a^{2}y=0$ (proved)