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:

When solving differential equations with initial conditions, don't forget to substitute the initial values to find the constant of integration after solving for the general solution.

Answer 1

Step 1: Rewrite the given equation.
The given differential equation is: \[ 2xy + y^2 - 2x^2 \frac{dy}{dx} = 0. \] Rearrange this equation to express \( \frac{dy}{dx} \) explicitly: \[ \frac{dy}{dx} = \frac{2xy + y^2}{2x^2}. \] Step 2: Separate the variables.
To separate variables, we first divide through by \( x^2 \) and \( y \) (assuming \( y \neq 0 \)): \[ \frac{1}{y} \, dy = \frac{2x + y}{2x^3} \, dx. \] Step 3: Simplify the right-hand side.
We now rewrite the right-hand side of the equation: \[ \frac{2x + y}{2x^3} = \frac{2x}{2x^3} + \frac{y}{2x^3} = \frac{1}{x^2} + \frac{y}{2x^3}. \] Thus, the equation becomes: \[ \frac{1}{y} \, dy = \left( \frac{1}{x^2} + \frac{y}{2x^3} \right) \, dx. \] Step 4: Integrate both sides.
We integrate the left-hand side: \[ \int \frac{1}{y} \, dy = \ln |y|. \] For the right-hand side, we integrate term by term: - The first term is: \[ \int \frac{1}{x^2} \, dx = -\frac{1}{x}. \] - The second term is: \[ \int \frac{y}{2x^3} \, dx = \frac{y}{2} \int x^{-3} \, dx = \frac{y}{2} \cdot \frac{1}{2x^2} = \frac{y}{4x^2}. \] Thus, the equation becomes: \[ \ln |y| = -\frac{1}{x} + \frac{y}{4x^2} + C, \] where \( C \) is the constant of integration. Step 5: Apply the initial condition.
Substitute the given initial condition \( x = 1 \) and \( y = 2 \) into the equation: \[ \ln |2| = -\frac{1}{1} + \frac{2}{4 \times 1^2} + C. \] Simplify: \[ \ln 2 = -1 + \frac{1}{2} + C \quad \Rightarrow \quad \ln 2 = -\frac{1}{2} + C \quad \Rightarrow \quad C = \ln 2 + \frac{1}{2}. \] Step 6: Substitute the value of \( C \).
Substitute the value of \( C \) back into the general solution: \[ \ln |y| = -\frac{1}{x} + \frac{y}{4x^2} + \ln 2 + \frac{1}{2}. \] Conclusion:
The particular solution to the differential equation is: \[ \boxed{\ln |y| = -\frac{1}{x} + \frac{y}{4x^2} + \ln 2 + \frac{1}{2}}. \]