The given differential equation is $$ \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} $$ Separating the variables $x$ and $y$, we get: $$ \frac{1}{\sqrt{y}} dy = \frac{1}{\sqrt{x}} dx $$ Which can be written as: $$ y^{-1/2} dy = x^{-1/2} dx $$
Integrating both sides: $$ \int y^{-1/2} dy = \int x^{-1/2} dx $$ Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $$ \frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C $$ $$ 2\sqrt{y} = 2\sqrt{x} + C $$
Dividing the entire equation by 2: $$ \sqrt{y} = \sqrt{x} + \frac{C}{2} $$ Let $C_1 = \frac{C}{2}$, then: $$ \sqrt{y} - \sqrt{x} = C_1 $$ This matches option (C).
Final Answer: $\sqrt{y}-\sqrt{x}=C$
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