y 2 2xy dx x 2dy 0 (Solution of differential equation)

[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.