Question

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

• A. \( y^2 = 2x^2\ln|x| + Cx^2 \)
• B. \( y = x\ln|x| + C \)
• C. \( y^2 = x^2 + C \)
• D. \( y = e^x + Cx \)

Hint:

When verifying solutions for homogeneous equations, check the polynomial degrees across your final terms. Every term in the solution should maintain a balanced quadratic degree distribution when factoring variables out.

Answer 1

The Correct Option is A

Concept: A first-order homogeneous differential equation can be solved by substituting \( y = vx \). Differentiating this product with respect to \( x \) using the product rule transforms the derivative component: \[ \frac{dy}{dx} = v + x\frac{dv}{dx} \] This substitution transforms the homogeneous structure into a basic separable format.

Step 1: Apply the substitution parameters to rewrite the equation.
Divide both terms in the numerator by the denominator to simplify the expression: \[ \frac{dy}{dx} = \frac{x^2}{xy} + \frac{y^2}{xy} = \frac{x}{y} + \frac{y}{x} \] Substitute \( y = vx \) (which means \( \frac{y}{x} = v \) and \( \frac{x}{y} = \frac{1}{v} \)), and replace \( \frac{dy}{dx} \): \[ v + x\frac{dv}{dx} = \frac{1}{v} + v \]

Step 2: Separate the variables and integrate both sides.
Subtract the common scalar variable \( v \) from both sides to isolate the differential terms: \[ x\frac{dv}{dx} = \frac{1}{v} \quad \Rightarrow \quad v\,dv = \frac{1}{x}\,dx \] Set up and evaluate the indefinite integrals on both sides: \[ \int v\,dv = \int \frac{1}{x}\,dx \quad \Rightarrow \quad \frac{v^2}{2} = \ln|x| + C' \]

Step 3: Substitute original variables back to find the final equation.
Replace \( v \) with its original fractional definition, \( v = \frac{y}{x} \): \[ \frac{\left(\frac{y}{x}\right)^2}{2} = \ln|x| + C' \quad \Rightarrow \quad \frac{y^2}{2x^2} = \ln|x| + C' \] Multiply through by \( 2x^2 \) to isolate our dependent variable terms: \[ y^2 = 2x^2\ln|x| + 2C'x^2 \] Combine the constant terms by letting \( C = 2C' \) to find the final general solution: \[ y^2 = 2x^2\ln|x| + Cx^2 \]