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 provided differential equation is:\[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}.\]1. Equation Rewriting: Divide numerator and denominator by \( x^2 \) to yield: \[ \frac{dy}{dx} = \frac{1 + \left(\frac{y}{x}\right)^2}{2 \cdot \frac{y}{x}}. \] Substitute \( v = \frac{y}{x} \), which implies \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). The equation becomes: \[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}. \]2. Equation Simplification: Isolate \( \frac{dv}{dx} \): \[ x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v. \] Simplify the right side: \[ x\frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}. \] Divide by \( x \): \[ \frac{dv}{dx} = \frac{1 - v^2}{2vx}. \]3. Variable Separation: Separate \( v \) and \( x \) terms: \[ \frac{2v}{1 - v^2} \, dv = \frac{1}{x} \, dx. \]4. Integration: Integrate both sides. - Left side: \[ \int \frac{2v}{1 - v^2} \, dv = \int \frac{-d(1 - v^2)}{1 - v^2} = -\ln|1 - v^2|. \] - Right side: \[ \int \frac{1}{x} \, dx = \ln|x|. \] Combining these yields: \[ -\ln|1 - v^2| = \ln|x| + C, \] where \( C \) is the constant of integration.5. Simplification: 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. Back Substitution: Substitute \( v = \frac{y}{x} \): \[ 1 - \left(\frac{y}{x}\right)^2 = \frac{K}{x}. \] Multiply by \( x^2 \): \[ x^2 - y^2 = Kx. \] General Solution: \[x^2 - y^2 = Kx,\]where \( K \) is an arbitrary constant.