Solve the differential equation, $(x - y)\frac{d}{{dx}}\sqrt {\frac{y}{x}} = x + y + 2\sqrt {xy} ;x > 0,y > 0$
Solution
We have, $\left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)\frac{d}{{dx}}\sqrt {\frac{y}{x}} = {\left( {\sqrt x + \sqrt y } \right)^2}$
$ \Rightarrow \frac{d}{{dx}}\sqrt {\frac{y}{x}} = \frac{{\sqrt x + \sqrt y }}{{\sqrt x - \sqrt y }}$
$ \Rightarrow \frac{d}{{dx}}\sqrt {\frac{y}{x}} = \frac{{1 + \sqrt {\frac{y}{x}} }}{{1 - \sqrt {\frac{y}{x}} }}$
$ \Rightarrow \frac{{dt}}{{dx}} = \frac{{1 + t}}{{1 - t}};t = \sqrt {\frac{y}{x}} $
$ \Rightarrow \frac{{1 - t}}{{1 + t}}dt = dx$
$ \Rightarrow \int {\frac{{2 - (t + 1)}}{{1 + t}}} dt = \int {dx} $
$ \Rightarrow \int {\left( {\frac{2}{{1 + t}} - 1} \right)} dt = \int {dx} $
$ \Rightarrow 2\ln |1 + t| - t = x + C$
$ \Rightarrow 2\ln \left| {1 + \sqrt {\frac{y}{x}} } \right| - \sqrt {\frac{y}{x}} = x + C$