Question

Solution of the differential equation $y' = \frac{x^2+y^2}{xy}$, where $y(1) = -2$ is given by

• A. $y^2 = 4x^2 \log x^2 + x^2$
• B. $y^2 = x^2 \log x - x^2$
• C. $y^2 = x \log x^2 + 4x^2$
• D. $y^2 = x^2 \log x^2 + 4x^2$

Hint:

You can easily check your answer by verifying the initial condition. Plug $x = 1$ into the options to see which one yields $y^2 = 4$: For (D): $y^2 = 1^2 \log(1) + 4(1)^2 = 0 + 4 = 4 \Rightarrow y = \pm 2$. This step immediately helps eliminate options that do not satisfy the initial state boundary condition!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Solve the homogeneous differential equation dy/dx = (x² + y²)/(xy) with initial condition y(1) = -2.

Step 2: Key Formula or Approach:
Substitute y = vx and dy/dx = v + x(dv/dx) to convert the homogeneous equation into a separable one.

Step 3: Detailed Explanation:
Substituting: v + x(dv/dx) = (x² + v²x²)/(x·vx) = (1 + v²)/v. Then x(dv/dx) = (1 + v²)/v - v = 1/v. Separating: v dv = (1/x) dx. Integrating: v²/2 = log|x| + c. Substitute back v = y/x: y²/(2x²) = log|x| + c → y² = 2x²log|x| + 2cx² = x²log x² + 2cx². Apply y(1) = -2: 4 = 0 + 2c → 2c = 4. Final particular solution: y² = x²log x² + 4x².

Step 4: Final Answer:
The particular solution matches option (D).