Question

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:

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