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Question Number 29442 by prof Abdo imad last updated on 08/Feb/18

$$splvethed.e{xy}^{{}^{\prime }}=\sqrt{x^2+y^2}+ywithx>0$$

Answered by mrW2 last updated on 09/Feb/18

$${xy}^{{}^{\prime }}=\sqrt{x^2+y^2}+y$$ $$y^{{}^{\prime }}=\sqrt{1+{\left(\frac{y}{x}\right)}^2}+\left(\frac{y}{x}\right)$$ $$t=\frac{y}{x}$$ $$y=tx$$ $$y^{\prime }=t+x\frac{dt}{dx}$$ $$\Rightarrow t+x\frac{dt}{dx}=\sqrt{1+t^2}+t$$ $$\Rightarrow x\frac{dt}{dx}=\sqrt{1+t^2}$$ $$\Rightarrow \int \frac{dt}{\sqrt{1+t^2}}=\int \frac{dx}{x}$$ $$\Rightarrow \ln (t+\sqrt{1+t^2})=\ln x+C$$ $$\Rightarrow \ln \frac{t+\sqrt{1+t^2}}{x}=C$$ $$\Rightarrow \frac{t+\sqrt{1+t^2}}{x}=c$$ $$\Rightarrow \frac{y}{x}+\sqrt{1+{\left(\frac{y}{x}\right)}^2}=cx$$ $$\Rightarrow y+\sqrt{x^2+y^2}={cx}^2$$

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