[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:
\[
y = \frac{C}{x}
\]