[latex]
\[
\textbf{Q: Solve the following differential equation:}
\]
\[
x \cdot 4y^3 \, dx + x \cdot 6y^5 \, dy = 0
\]
\[
\textbf{Solution:}
\]
We begin by simplifying the given equation:
\[
x \cdot 4y^3 \, dx + x \cdot 6y^5 \, dy = 0
\]
Factor out \(x\) from both terms:
\[
x \left( 4y^3 \, dx + 6y^5 \, dy \right) = 0
\]
Since \(x \neq 0\), the equation reduces to:
\[
4y^3 \, dx + 6y^5 \, dy = 0
\]
Now, we separate the variables:
\[
\frac{dx}{x} = – \frac{6y^5}{4y^3} \, dy
\]
Simplify the right side:
\[
\frac{dx}{x} = – \frac{3y^2}{2} \, dy
\]
Integrate both sides:
\[
\int \frac{dx}{x} = \int – \frac{3y^2}{2} \, dy
\]
The integrals are:
\[
\ln |x| = – \frac{y^3}{2} + C
\]
Thus, the general solution is:
\[
\ln |x| = – \frac{y^3}{2} + C
\]
Or equivalently:
\[
x = e^{- \frac{y^3}{2} + C}
\]