$\displaystyle xy'=\sqrt{x^2-y^2}+y$

So far I've rearranged it to get:

$\displaystyle \frac{dy}{dx}=\frac{\sqrt{x^2-y^2}+y}{x}$

Thanks

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- Aug 1st 2011, 07:52 AMdwally89Which technique should I use to solve this ODE?
$\displaystyle xy'=\sqrt{x^2-y^2}+y$

So far I've rearranged it to get:

$\displaystyle \frac{dy}{dx}=\frac{\sqrt{x^2-y^2}+y}{x}$

Thanks - Aug 1st 2011, 10:46 AMKrizalidRe: Which technique should I use to solve this ODE?
Now the RHS is rewritten as $\displaystyle \sqrt{1-{{\left( \frac{y}{x} \right)}^{2}}}+\frac{y}{x}.$