Question

pls solve this question​

Answer 1

Answer:

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$ydx - xdy = \sqrt{ {x}^{2} + {y}^{2} } dx \\ ydx + \sqrt{ {x}^{2} + {y}^{2} } dx = xdy \\ (y + \sqrt{ {x}^{2} + {y}^{2} } )dx = xdy \\ \frac{dy}{dx} = \frac{(y + \sqrt{ {x}^{2} + {y}^{2} } )}{x} \\ \small\bold\red{put \: y \: = vx} \\ \frac{d}{dx} vx = \frac{vx + \sqrt{ {x}^{2} + {v}^{2} {x}^{2} } }{x} \\ v \times 1 + x \frac{dv}{dx} = \frac{vx + x \sqrt{1 + {v}^{2} } }{x} \\ v + x \frac{dv}{dx} = v + \sqrt{1 + {v}^{2} } \\ x \frac{dv}{dx} = \sqrt{1 + {v}^{2} } \\ \small\bold\red{(on \: seperating \: the \: variables)} \\ \frac{dv}{ \sqrt{1 + {v}^{2} } } = \frac{dx}{x} \\ \small\bold\red{(on \: integrating \: both \: sides)} \\ log((v + \sqrt{1 + {v}^{2} } ) = log(x) + log(c) \\ = > \: v + \sqrt{1 + {v}^{2} } = xc \\ \small\bold\red{(put \: v = \frac{y}{x} )} \\ \frac{y}{x} + \sqrt{1 + \frac{ {y}^{2} }{ {x}^{2} } } = xc \\ \frac{y + \sqrt{ {x}^{2} + {y}^{2} } }{x} = xc \\ y + \sqrt{ {x}^{2} + {y}^{2} } = c {x}^{2}$