Let's solve the problem step by step: ### Step 1: Show that the given D.E. is homogeneous The given differential equation is: \[ dx - \left( \frac{x}{y} + \sqrt{\frac{x}{y}} \right) dy = 0 \] or, rearranged, \[ \frac{dx}{dy} = \frac{x}{y} + \sqrt{\frac{x}{y}} \] A differential equation of the form \(\frac{dx}{dy} = f\left(\frac{x}{y}\right)\) is called homogeneous if the right side is a function of \(\frac{x}{y}\) only. Here, \[ f\left(\frac{x}{y}\right) = \frac{x}{y} + \sqrt{\frac{x}{y}} \] This is indeed a function of \(\frac{x}{y}\) only. **Explanation:** The equation is homogeneous because the right-hand side depends only on the ratio \(\frac{x}{y}\). --- ### Step 2: Solve the homogeneous D.E. Let \(v = \frac{x}{y}\), so \(x = vy\). Differentiate both sides with respect to \(y\): \[ \frac{dx}{dy} = v + y\frac{dv}{dy} \] Substitute into the original equation: \[ v + y\frac{dv}{dy} = v + \sqrt{v} \] \[ y\frac{dv}{dy} = \sqrt{v} \] \[ \frac{dv}{\sqrt{v}} = \frac{dy}{y} \] Integrate both sides: \[ \int v^{-1/2} dv = \int \frac{dy}{y} \] \[ 2\sqrt{v} = \ln|y| + C \] Substitute back \(v = \frac{x}{y}\): \[ 2\sqrt{\frac{x}{y}} = \ln|y| + C \] **Final Answer:** \[ 2\sqrt{\frac{x}{y}} = \ln|y| + C \]