Question

Solve $\frac{ydx-xdy}{(x-y{)}^2}=\frac{dx}{\sqrt[2]{21-x^2}}$, given that $y=2$, when $x=1$.

Answer 1

$\frac{ydx-xdy}{(x-y{)}^2}=\frac{dx}{\sqrt[2]{21-x^2}}=\frac{y-x\frac{dy}{dx}}{(x-y{)}^2}=\frac{1}{\sqrt[2]{21-x^2}}$

Divide the numerator and denominator of L. H. S by $x^2$

$\Rightarrow \frac{\frac{y}{x}.\frac{1}{x}-\frac{1}{x}.\frac{dy}{dx}}{(1-\frac{y}{x}{)}^2}=\frac{1}{\sqrt[2]{21-x^2}}$

$\Rightarrow \frac{\frac{dy}{dx}-\frac{y}{x}}{(1-\frac{y}{x}{)}^2}=\frac{-x}{\sqrt[2]{21-x^2}}$

$y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow \frac{v+x\frac{dv}{dx}-v}{(1-v{)}^2}=\frac{-x}{\sqrt[2]{21-x^2}}$

$\Rightarrow \int \frac{dv}{(1-v{)}^2}=\int \frac{dc}{\sqrt[2]{21-x^2}}$

$\Rightarrow \frac{1}{1-v}=\frac{{\sin }^{-1}\left(x\right)}{2}+c$

$\Rightarrow \frac{1}{1-\frac{y}{x}}=\frac{{\sin }^{-1}\left(x\right)}{2}+c$

At $x=1\to y=2$

$\Rightarrow \frac{1}{1-2}=\frac{{\sin }^{-1}\left(1\right)}{2}+c$

$\Rightarrow -1=\frac{\pi /2}{2}+c$

$\Rightarrow c=-(\frac{\pi }{4}+1)$

$\Rightarrow \frac{x}{x-y}=\frac{{\sin }^{-1}\left(x\right)}{2}-(\frac{\pi }{4}+1)$