[latex] \[
\textbf{Q: Solve the differential equation:} \quad y^2 + 2xy \, dx + x^2 \, dy = 0
\]
\[
\textbf{Step 1: Rearranging the equation}
\]
First, rewrite the given equation:
\[
y^2 + 2xy \, dx + x^2 \, dy = 0
\]
Separate the terms involving \( dx \) and \( dy \):
\[
x^2 \, dy = – y^2 – 2xy \, dx
\]
Now, divide through by \( x^2 \):
\[
dy = \left( – \frac{y^2}{x^2} – \frac{2y}{x} \right) \, dx
\]
\[
\textbf{Step 2: Use separation of variables}
\]
We now separate the variables \( x \) and \( y \) to integrate:
\[
\frac{dy}{y^2 + 2y} = – \frac{dx}{x^2}
\]
We can factor the left-hand side:
\[
\frac{dy}{y(y + 2)} = – \frac{dx}{x^2}
\]
\[
\textbf{Step 3: Integration}
\]
Now, integrate both sides:
\[
\int \frac{dy}{y(y + 2)} = – \int \frac{dx}{x^2}
\]
We will use partial fraction decomposition to break down the left-hand side:
\[
\frac{1}{y(y + 2)} = \frac{A}{y} + \frac{B}{y + 2}
\]
Multiplying both sides by \( y(y + 2) \), we get:
\[
1 = A(y + 2) + B y
\]
Expanding and equating the coefficients:
\[
1 = A y + 2A + B y
\]
\[
1 = (A + B)y + 2A
\]
For the coefficients to be equal, we have the system:
\[
A + B = 0 \quad \text{and} \quad 2A = 1
\]
Solving for \( A \) and \( B \), we get:
\[
A = \frac{1}{2}, \quad B = -\frac{1}{2}
\]
Thus, we can rewrite the integral as:
\[
\int \frac{1}{2y} – \frac{1}{2(y + 2)} \, dy = – \int \frac{dx}{x^2}
\]
Now, integrate both sides:
\[
\frac{1}{2} \ln|y| – \frac{1}{2} \ln|y + 2| = \frac{1}{x} + C
\]
\[
\textbf{Step 4: Final Solution}
\]
Simplify the equation:
\[
\frac{1}{2} \ln \left( \frac{|y|}{|y + 2|} \right) = \frac{1}{x} + C
\]
Multiply both sides by 2:
\[
\ln \left( \frac{|y|}{|y + 2|} \right) = \frac{2}{x} + C_1
\]
Exponentiate both sides:
\[
\frac{|y|}{|y + 2|} = e^{\frac{2}{x} + C_1}
\]
\[
\textbf{Final Solution:}
\]
The final solution is:
\[
\frac{|y|}{|y + 2|} = C e^{\frac{2}{x}}
\]
where \( C = e^{C_1} \) is a constant.