Question

Find the general solution of the differential equation: \[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}. \]

Hint:

For solving homogeneous differential equations, use the substitution \( v = \frac{y}{x} \), which simplifies the equation into separable variables.

Answer 1

The given differential equation is: \[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}. \] 1. Rewrite the Equation: Divide numerator and denominator by \( x^2 \) to simplify: \[ \frac{dy}{dx} = \frac{1 + \left(\frac{y}{x}\right)^2}{2 \cdot \frac{y}{x}}. \] Let \( v = \frac{y}{x} \), so \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute into the equation: \[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}. \] 2. Simplify the Equation: Rearrange to isolate \( \frac{dv}{dx} \): \[ x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v. \] Simplify the right-hand side: \[ x\frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}. \] Divide through by \( x \): \[ \frac{dv}{dx} = \frac{1 - v^2}{2vx}. \] 3. Separate Variables: Separate \( v \) and \( x \): \[ \frac{2v}{1 - v^2} \, dv = \frac{1}{x} \, dx. \] 4. Integrate Both Sides: - The left-hand side: \[ \int \frac{2v}{1 - v^2} \, dv = \int \frac{-d(1 - v^2)}{1 - v^2}. \] Using substitution, this becomes: \[ \int \frac{-d(1 - v^2)}{1 - v^2} = -\ln|1 - v^2|. \] - The right-hand side: \[ \int \frac{1}{x} \, dx = \ln|x|. \] Combine the results: \[ -\ln|1 - v^2| = \ln|x| + C, \] where \( C \) is the constant of integration. 5. Simplify: Rewrite the equation: \[ \ln|1 - v^2| = -\ln|x| - C. \] Exponentiate both sides: \[ |1 - v^2| = \frac{K}{x}, \] where \( K = e^{-C} \) is a constant. 6. Substitute Back \( v = \frac{y}{x} \): Replace \( v \) with \( \frac{y}{x} \): \[ 1 - \left(\frac{y}{x}\right)^2 = \frac{K}{x}. \] Multiply through by \( x^2 \): \[ x^2 - y^2 = Kx. \] General Solution: \[ x^2 - y^2 = Kx, \] where \( K \) is an arbitrary constant.