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

Answer: (D)

$\frac{\pi}{32}$

Answer 2

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