Question

If general solution of the differential equation $\frac{ydx-xdy}{(x-y{)}^2}=\frac{2dx}{\sqrt{1-x^2}}$ is $\frac{x}{x-y}+K\left({\sin }^{-1}x\right)=C,$ then the value of $K$ is
(where $C$ is the constant of integration and $K\in R$ )

Answer 1

$\frac{ydx-xdy}{(x-y{)}^2}=\frac{2dx}{\sqrt{1-x^2}}$
Dividing the numerator and denominator of $L.H.S.$ by $x^2,$ we get
$\frac{\frac{y}{x}\cdot \frac{1}{x}-\frac{1}{x}\cdot \frac{dy}{dx}}{{(1-\frac{y}{x})}^2}=\frac{2}{\sqrt{1-x^2}}$
$\Rightarrow \frac{\frac{y}{x}-\frac{dy}{dx}}{{(1-\frac{y}{x})}^2}=\frac{2x}{\sqrt{1-x^2}}$

Let $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore \frac{v-v-x\frac{dv}{dx}}{(1-v{)}^2}=\frac{2x}{\sqrt{1-x^2}}$
$\Rightarrow \frac{dv}{(1-v{)}^2}=\frac{-2dx}{\sqrt{1-x^2}}$
Integrating both sides, we get
$\frac{1}{1-v}=-2{\sin }^{-1}x+C$
$\Rightarrow \frac{x}{x-y}+2{\sin }^{-1}x=C$