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Question Number 108053 by john santu last updated on 14/Aug/20

$$\frac{⋏\mathcal{J}\mathbb{S}⋏}{{}^{\rightrightarrows }}$$$$(\frac{y}{x}+\sqrt{\frac{x^2+y^2}{x^2}})dx=dy$$

Answered by bemath last updated on 14/Aug/20

Answered by Dwaipayan Shikari last updated on 14/Aug/20

$$(v+\sqrt{1+v^2})=\frac{dy}{dx}v=\frac{y}{x},\frac{dy}{dx}=v+\frac{dv}{dx}x$$$$v+\frac{dv}{dx}x=v+\sqrt{1+v^2}$$$$\int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}$$$$log(v+\sqrt{1+v^2})=logx+logC$$$$\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}=Cx$$$$y+\sqrt{x^2+y^2}={Cx}^2$$

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