The solution of the differential equation $\frac{dy}{dx}=\sqrt{\frac{y}{x}}$ is

$\sqrt{x}-\sqrt{y} = C$: C is an arbitrary constant

The correct answer is Option (2) → $\sqrt{x}-\sqrt{y} = C$: C is an arbitrary constant

Given: $\frac{dy}{dx} = \sqrt{\frac{y}{x}}$

Let $y = v^2$, then $\frac{dy}{dx} = 2v \frac{dv}{dx}$

So: $2v \frac{dv}{dx} = \sqrt{\frac{v^2}{x}} = \frac{v}{\sqrt{x}}$

Cancel $v$ (assuming $v \ne 0$):

$2 \frac{dv}{dx} = \frac{1}{\sqrt{x}}$

$\Rightarrow dv = \frac{1}{2\sqrt{x}} dx$

Integrate both sides:

$\int dv = \int \frac{1}{2\sqrt{x}} dx$

$v = \sqrt{x} + C$

Recall: $v = \sqrt{y}$

So, $\sqrt{y} = \sqrt{x} + C$