Question

Find the particular solution of the differential equation given by: \[ 2xy + y^2 - 2x^2 \frac{dy}{dx} = 0; \quad y = 2, \, \text{when } x = 1. \]

Hint:

For initial value problems, always substitute the given values after solving for the general solution to find the constant of integration.

Answer 1

Step 1: Rearrange the given equation.
The given equation is: \[ 2xy + y^2 - 2x^2 \frac{dy}{dx} = 0. \] Rearrange to express \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{2xy + y^2}{2x^2}. \] Step 2: Separate variables.
Divide through by \( y^2 \) (assuming \( y \neq 0 \)): \[ \frac{1}{y^2} \, dy = \frac{2x + y}{2x^2} \, dx. \] Step 3: Integrate both sides.
- For the left-hand side: \[ \int \frac{1}{y^2} \, dy = \int y^{-2} \, dy = -\frac{1}{y}. \] - For the right-hand side: Split the fraction: \[ \int \frac{2x + y}{2x^2} \, dx = \int \frac{2x}{2x^2} \, dx + \int \frac{y}{2x^2} \, dx. \] - The first term: \[ \int \frac{2x}{2x^2} \, dx = \int \frac{1}{x} \, dx = \ln|x|. \] - The second term (not dependent on \( y \)): \[ \int \frac{y}{2x^2} \, dx = \frac{y}{2} \int x^{-2} \, dx = \frac{y}{2} \cdot \left(-\frac{1}{x}\right) = -\frac{y}{2x}. \] Step 4: Combine results.
The general solution is: \[ -\frac{1}{y} = \ln|x| - \frac{y}{2x} + C, \] where \( C \) is the constant of integration. Step 5: Apply the initial condition.
When \( x = 1 \) and \( y = 2 \): \[ -\frac{1}{2} = \ln|1| - \frac{2}{2 \cdot 1} + C. \] Simplify: \[ -\frac{1}{2} = 0 - 1 + C \quad \Rightarrow \quad C = \frac{1}{2}. \] Step 6: Write the particular solution.
Substitute \( C = \frac{1}{2} \) into the general solution: \[ -\frac{1}{y} = \ln|x| - \frac{y}{2x} + \frac{1}{2}. \] Conclusion:
The particular solution is: \[ \boxed{-\frac{1}{y} = \ln|x| - \frac{y}{2x} + \frac{1}{2}}. \]