Question

Let \( y = y(x) \) be the solution of the differential equation: \[ (x^2 + 4)^2 \, dy + \left( 2x^3 y + 8xy - 2 \right) dx = 0. \] If \( y(0) = 0 \), then \( y(2) \) is equal to:

• A. \( \frac{\pi}{8} \)
• B. \( \frac{\pi}{16} \)
• C. \( 2\pi \)
• D. \( \frac{\pi}{32} \)

Answer 1

The Correct Option is D

Rewriting the given equation:

\[ \frac{dy}{dx} + y \frac{2x^3 + 8x}{(x^2 + 4)^2} = \frac{2}{(x^2 + 4)^2}. \]

This is a linear differential equation in the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \( P(x) = \frac{2x^3 + 8x}{(x^2 + 4)^2} \) and \( Q(x) = \frac{2}{(x^2 + 4)^2} \).

The integrating factor (IF) is: \[ \text{IF} = e^{\int P(x) \, dx} = e^{\int \frac{2x}{x^2 + 4} \, dx}. \]

Simplify: \[ \int \frac{2x}{x^2 + 4} \, dx = \ln(x^2 + 4). \] Thus: \[ \text{IF} = e^{\ln(x^2 + 4)} = x^2 + 4. \]

Multiply through by the integrating factor:

\[ y(x^2 + 4) = \int \frac{2}{(x^2 + 4)^2} \cdot (x^2 + 4) \, dx. \]

Simplify the integral: \[ \int \frac{2}{x^2 + 4} \, dx = \int \frac{2}{x^2 + 2^2} \, dx = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right). \]

Thus: \[ y(x^2 + 4) = \tan^{-1} \left( \frac{x}{2} \right) + c. \]

Using the initial condition \( y(0) = 0 \): \[ 0 \cdot (0^2 + 4) = \tan^{-1} \left( \frac{0}{2} \right) + c \implies c = 0. \]

Therefore: \[ y(x^2 + 4) = \tan^{-1} \left( \frac{x}{2} \right). \]

At \( x = 2 \):

\[ y(4 + 4) = \tan^{-1}(1) = \frac{\pi}{4}. \]

Thus: \[ y(2) = \frac{\pi}{32}. \]