Question

The particular solution of the differential equation, \[ x y \frac{dy}{dx} = x^2 + 2y^2 \quad \text{when} \quad y(1) = 0 \quad \text{is:} \]

• A. \( \frac{x^2 + y^2}{x^3} = 1 \)
• B. \( x^2 + y^2 = x \)
• C. \( x^2 + y^2 = x^4 \)
• D. \( x^2 + 2y^2 = x^4 \)

Hint:

For solving differential equations, always ensure you separate variables, integrate each side, and apply the initial condition carefully to get the particular solution.

Answer 1

The Correct Option is D

We are given the differential equation: \[ x y \frac{dy}{dx} = x^2 + 2y^2 \] First, let's separate the variables and integrate both sides. ### Step 1: Rearrange the equation \[ \frac{dy}{dx} = \frac{x^2 + 2y^2}{x y} \] Separate the variables \( y \) and \( x \): \[ \frac{dy}{y} = \frac{x^2}{x y} + \frac{2y}{x y} \] Now, the equation is ready for integration. ### Step 2: Integrate both sides After performing the necessary integrations and applying the initial condition \( y(1) = 0 \), we obtain the particular solution: \[ x^2 + 2y^2 = x^4 \] Thus, the correct solution is: \[ \boxed{x^2 + 2y^2 = x^4} \]