Solution of the differential equation

$x\left(\frac{d y}{d x}\right)^2+2 \sqrt{x y} \frac{d y}{d x}+y=0$, is 

$\sqrt{x}+\sqrt{y}=\sqrt{a}$

We have,

$x\left(\frac{d y}{d x}\right)^2+2 \sqrt{x y} \frac{d y}{d x}+y=0$

$\Rightarrow \left(\sqrt{x} \frac{d y}{d x}+\sqrt{y}\right)^2=0$

$\Rightarrow \sqrt{x} \frac{d y}{d x}+\sqrt{y}=0$

$\Rightarrow \frac{1}{\sqrt{x}} d x+\frac{1}{\sqrt{y}} d y=0$

$\Rightarrow 2 \sqrt{x}+2 \sqrt{y}=C \Rightarrow \sqrt{x}+\sqrt{y}=\sqrt{a}$, where $\sqrt{a}=2 C$