[latex] \[
\textbf{Q: Solve the differential equation:} \quad xy + 4xy \, dx + x + 5xy \, dy = 0
\]
\[
\textbf{Step 1: Rearranging the equation}
\]
We start by rewriting the given equation:
\[
xy + 4xy \, dx + x + 5xy \, dy = 0
\]
Rearranging it, we get:
\[
4xy \, dx + 5xy \, dy = -x – xy
\]
\[
\textbf{Step 2: Grouping like terms}
\]
Now, factor the terms to make the equation easier to handle:
\[
4xy \, dx + 5xy \, dy = -(x + xy)
\]
Factor out the common terms on the right side:
\[
4xy \, dx + 5xy \, dy = -x(1 + y)
\]
\[
\textbf{Step 3: Separation of variables}
\]
We now attempt to separate the variables \(x\) and \(y\). First, divide through by \(xy\) on both sides:
\[
4 \, dx + 5 \, dy = -\frac{x(1 + y)}{xy}
\]
Simplify the right-hand side:
\[
4 \, dx + 5 \, dy = -\frac{1 + y}{y}
\]
\[
\textbf{Step 4: Integration}
\]
At this point, we can integrate both sides:
\[
\int 4 \, dx + \int 5 \, dy = \int -\frac{1 + y}{y} \, dy
\]
First, integrating the left-hand side:
\[
4x + 5y
\]
For the right-hand side, we split the integrand:
\[
\int -\left(\frac{1}{y} + 1\right) \, dy
\]
Now, integrate:
\[
– \ln |y| – y
\]
Thus, we have:
\[
4x + 5y = -\ln |y| – y + C
\]
\[
\textbf{Step 5: Final Solution}
\]
The final solution is:
\[
4x + 5y + \ln |y| + y = C
\]
where \( C \) is the constant of integration.
This is the implicit solution to the given differential equation.