The differential equation \(y^{\prime}=\frac{y}{x+\sqrt{x y}}\) has general solution given by (Where \(C\) is a constant of integration.)

\(\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x+\sqrt{x y}} \text { let } y=v x \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=v+x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x} \\ & \Rightarrow v+x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{v x}{x+\sqrt{x \cdot v x}}=\frac{v}{1+\sqrt{v}} \\ & \Rightarrow x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{v}{1+\sqrt{v}}-v=\frac{-v \sqrt{v}}{1+\sqrt{v}} \\ & \Rightarrow-\frac{1+\sqrt{v}}{v \sqrt{v}} \mathrm{~d} v=\frac{\mathrm{d} x}{x} \\ & \Rightarrow-\int\left(v^{\frac{-3}{2}}+v^{-1}\right) \mathrm{d} v=\int \frac{\mathrm{d} x}{x} \\ & \Rightarrow 2 v \frac{-1}{2}-\log v=\log x+\log c^{\prime} \\ & \Rightarrow \frac{2 \sqrt{x}}{\sqrt{y}}-\log \frac{y}{x}=\log c^{\prime} x\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{2 \sqrt{x}}{\sqrt{y}}=\log \left(c^{\prime} x \cdot \frac{y}{x}\right) \\ & \Rightarrow c \cdot e^{\frac{2 \sqrt{x}}{\sqrt{y}}}\end{aligned}\)