solve the given differential equation -y dx+(x + √(xy))dy = 0

Explanation:

Step1: Rewrite the differential equation

We have (-y\mathrm{d}x+(x + \sqrt{xy})\mathrm{d}y = 0), which can be rewritten as (\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{x+\sqrt{xy}}{y}). Let (x = vy), then (\frac{\mathrm{d}x}{\mathrm{d}y}=v + y\frac{\mathrm{d}v}{\mathrm{d}y}).

Step2: Substitute (x = vy) into the differential equation

Substituting gives (v + y\frac{\mathrm{d}v}{\mathrm{d}y}=\frac{vy+\sqrt{v y\cdot y}}{y}=v+\sqrt{v}).

Step3: Separate variables

We get (y\frac{\mathrm{d}v}{\mathrm{d}y}=\sqrt{v}), and then (\frac{\mathrm{d}v}{\sqrt{v}}=\frac{\mathrm{d}y}{y}).

Step4: Integrate both sides

Integrating (\int\frac{\mathrm{d}v}{\sqrt{v}}=\int\frac{\mathrm{d}y}{y}). We know that (\int\frac{\mathrm{d}v}{\sqrt{v}} = 2\sqrt{v}+C_1) and (\int\frac{\mathrm{d}y}{y}=\ln|y|+C_2). So (2\sqrt{v}=\ln|y| + C).

Step5: Substitute back (v=\frac{x}{y})

We have (2\sqrt{\frac{x}{y}}=\ln|y|+C).

Answer:

(2\sqrt{\frac{x}{y}}=\ln|y| + C)