Question

The general solution of the differential equation $\frac{ydx-xdy}{(x-y{)}^2}=\frac{dx}{\sqrt{1-x^2}}$ is
(where $c$ is the constant of integration)

• A. $\frac{x}{x-y}+\frac{1}{2}{\sin }^{-1}x=c$
• B. $\frac{1}{x+y}-{\sin }^{-1}x=c$
• C. $\frac{y}{x+y}-{\sin }^{-1}x=c$
• D. $\frac{x}{x-y}+{\sin }^{-1}x=c$

Answer 1

The correct option is D $\frac{x}{x-y}+{\sin }^{-1}x=c$

$\frac{ydx-xdy}{(x-y{)}^2}=\frac{dx}{\sqrt{1-x^2}}$
Dividing the numenator and denominator of $L.H.S.$ by $x^2$
$\Rightarrow \frac{\frac{y}{x}\cdot \frac{1}{x}-\frac{1}{x}\cdot \frac{dy}{dx}}{{(1-\frac{y}{x})}^2}=\frac{1}{\sqrt{1-x^2}}$
$\Rightarrow \frac{\frac{y}{x}-\frac{dy}{dx}}{{(1-\frac{y}{x})}^2}=\frac{x}{\sqrt{1-x^2}}$
Let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow \frac{v-v-x\frac{dv}{dx}}{(1-v{)}^2}=\frac{x}{\sqrt{1-x^2}}$
$\Rightarrow \frac{dv}{(1-v{)}^2}=\frac{-dx}{\sqrt{1-x^2}}$
Integrating both sides, we get
$\frac{1}{1-v}=-{\sin }^{-1}x+c$
$\Rightarrow \frac{x}{x-y}+{\sin }^{-1}x=c$