y 2 xy dx x 2 dy 0 homogeneous

[latex] \[ \textbf{Q: Solve the homogeneous equation:} \quad y^2 x \, dx + x^2 \, dy = 0 \] \[ \textbf{Step 1: Rearrange the equation} \] Start by isolating the terms involving \( dy \) and \( dx \): \[ y^2 x \, dx = – x^2 \, dy \] \[ \textbf{Step 2: Separate variables} \] Now, separate the variables \( x \) and \( y \): \[ \frac{y^2}{x^2} \, dx = – \frac{dy}{y} \] \[ \textbf{Step 3: Integrate both sides} \] Integrate both sides of the equation: \[ \int \frac{y^2}{x^2} \, dx = – \int \frac{dy}{y} \] For the left side: \[ \int \frac{y^2}{x^2} \, dx = \int y^2 \cdot x^{-2} \, dx \] We can treat \( y^2 \) as a constant since we are integrating with respect to \( x \), so the integral becomes: \[ y^2 \int x^{-2} \, dx = – y^2 \cdot \frac{1}{x} \] For the right side: \[ – \int \frac{dy}{y} = – \ln|y| \] \[ \textbf{Step 4: Combine the results} \] Now, combine the results of both integrals: \[ y^2 \cdot \left(- \frac{1}{x}\right) = – \ln|y| + C \] \[ \textbf{Step 5: Final Solution} \] Finally, solve for \( y \): \[ y^2 \cdot \frac{1}{x} = \ln|y| + C \] or \[ y = \frac{1}{x} e^{C} \] \[ \textbf{Final Solution:} \] The solution to the homogeneous equation is: \[ y = \frac{C}{x} \]