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

Answer 2

Step 1: Arrange the differential equation
Given: \( (x^2 + 4)^2\, dy + (2x^3y + 8xy - 2) dx = 0 \)
Rewrite: \( (x^2 + 4)^2 \frac{dy}{dx} = 2 - 2x^3y - 8xy \)
So, \( \frac{dy}{dx} + \frac{2x^3 + 8x}{(x^2+4)^2} y = \frac{2}{(x^2+4)^2} \)

Step 2: Use integrating factor method
Integrating factor (IF):
\( \int \frac{2x^3 + 8x}{(x^2+4)^2} dx = \int \frac{2x(x^2+4)}{(x^2+4)^2} dx = \int \frac{2x}{x^2+4} dx \)
Let \( u = x^2 + 4 \), \( du = 2x dx \):
So, IF = \( \int \frac{du}{u} = \ln|x^2+4| \)
Thus, IF is \( x^2+4 \)

Multiply equation by IF:
\[ (x^2 + 4) \frac{dy}{dx} + (2x^3 + 8x) y / (x^2+4) = 2 / (x^2+4) \] or simply work with integrating factor and standard form.
The solution is:
\[ y(x) (x^2 + 4) = \int \frac{2}{x^2+4} dx + C \] \[ y(x) (x^2 + 4) = 2 \int \frac{1}{x^2 + 4} dx + C \] \[ = 2 \cdot \frac{1}{2} \tan^{-1}\frac{x}{2} + C \] \[ = \tan^{-1}\frac{x}{2} + C \] So, \( y(x) = \frac{\tan^{-1}\frac{x}{2} + C}{x^2+4} \)

Step 3: Use initial condition
Given: \( y(0) = 0 \)
\[ 0 = \frac{\tan^{-1} 0 + C}{0^2 + 4} \implies C = 0 \] So, \( y(x) = \frac{\tan^{-1}\frac{x}{2}}{x^2+4} \)

Step 4: Find \( y(2) \)
\[ y(2) = \frac{\tan^{-1} 1}{2^2 + 4} = \frac{\frac{\pi}{4}}{8} = \frac{\pi}{32} \]
Final Answer: \( y(2) = \frac{\pi}{32} \)