Let, $ \sqrt {x + y} = t$

So, $ \frac{1}{{2\sqrt {x + y} }}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$

Or, $\frac{1}{{2t}}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$

$\therefore \frac{{dy}}{{dx}} = 2t\frac{{dt}}{{dx}} – 1$

From the original differential equation,

$2t\frac{{dt}}{{dx}} – 1 = t$

$\therefore 2t\frac{{dt}}{{dx}} = 1 + t$

Or, $\frac{{2t}}{{1 + t}}dt = dx$

$ \Rightarrow 2\int {\frac{{1 + t – 1}}{{1 + t}}dt} = \int {dx} $

$ \Rightarrow 2\int {\left( {1 – \frac{1}{{1 + t}}} \right)dt} = \int {dx} $

$ \Rightarrow 2t – 2\ln (1 + t) = x + C$

$ \Rightarrow 2\sqrt {x + y} – 2\ln (1 + \sqrt {x + y} ) = x + C$

Or, $\sqrt {x + y} – \ln (1 + \sqrt {x + y} ) = \frac{x}{2} + C’$