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(D) The given differential equation is $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. Rearranging the terms,we get $(x^2+4)^2 \frac{dy}{dx} + 2x(x^2+4)y = 2$.

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$\frac{\pi}{32}$

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(D) The given differential equation is $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. Rearranging the terms,we get $(x^2+4)^2 \frac{dy}{dx} + 2x(x^2+4)y = 2$. Dividing by $(x^2+4)^2$,we obtain the linear differential equation: $\frac{dy}{dx} + \frac{2x}{x^2+4}y = \frac{2}{(x^2+4)^2}$. The integrating factor $(IF)$ is $e^{\int \frac{2x}{x^2+4} dx} = e^{\ln(x^2+4)} = x^2+4$. Multiplying both sides by the $IF$,we get $\frac{d}{dx} [y(x^2+4)] = \frac{2}{x^2+4}$. Integrating both sides with respect to $x$,we get $y(x^2+4) = \int \frac{2}{x^2+2^2} dx = 2 \cdot \frac{1}{2} 	an^{-1}(\frac{x}{2}) + C = 	an^{-1}(\frac{x}{2}) + C$. Given $y(0)=0$,we have $0(0+4) = 	an^{-1}(0) + C$,which implies $C=0$. Thus,$y(x^2+4) = 	an^{-1}(\frac{x}{2})$. At $x=2$,$y(2^2+4) = 	an^{-1}(\frac{2}{2}) = 	an^{-1}(1) = \frac{\pi}{4}$. Therefore,$y(8) = \frac{\pi}{4}$,so $y(2) = \frac{\pi}{32}$.

List $I$ (Differential Equation)	List $II$ (Integrating Factor)
$(P)$ $(x^3+1)\frac{dy}{dx}+x^2y=3x^2$	$(1)$ $x^3$
$(Q)$ $x^2\frac{dy}{dx}+3xy=x^6$	$(2)$ $(x^3+1)^2$
$(R)$ $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2$	$(3)$ $(x^2+1)^2$
$(S)$ $(x^2+1)\frac{dy}{dx}+4xy=\ln x$	$(4)$ $x^2+1$
	$(5)$ $(x^3+1)^{1/3}$
	$(6)$ $(x^3+1)^{1/2}$

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

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