Question

xdy-ydx=x√x²-y²dx
by inspection method ​

Answer 1

Step-by-step explanation:

$xdy - ydx = x \sqrt{ {x}^{2} + {y}^{2} } dx$

$\implies \frac{xdy - ydx}{ {y}^{2} } = \frac{x}{y} \sqrt{ (\frac{x}{y} )^{2} - 1} dx$

$\implies \frac{d(\frac{x}{y})}{ \frac{x}{y} \sqrt{( \frac{x}{y}) ^{2} - 1} } = dx$

Integratigrating both sides,

$\implies \int \frac{d(\frac{x}{y})}{ \frac{x}{y} \sqrt{( \frac{x}{y}) ^{2} - 1} } = \int dx$

$\implies \sec^{ - 1} ( \frac{x}{y} ) = x + c$

$\implies \: y = x \cos(x + c)$