Standard linear ODE: $\frac{dy}{dx} + \frac{2}{x}y = x$

Integrating factor: $x^2$

$\frac{d}{dx}(x^2 y) = x^3 \Rightarrow x^2 y = \frac{x^4}{4} + C$

At $x=1, y=1$: $1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$

So $y = \frac{x^2}{4} + \frac{3}{4x^2}$

At $x=\frac{1}{2}$: $y = \frac{(1/2)^2}{4} + \frac{3}{4\cdot(1/4)} = \frac{1}{16} + 3 = \frac{1}{16} + \frac{48}{16}$... rechecking: $y = \frac{1/4}{4} + \frac{3}{4 \cdot 1/4} = \frac{1}{16} + 3$. That gives $\frac{49}{16}$...

Correction: $y(1/2) = \frac{(1/2)^2}{4} + \frac{3}{4(1/2)^2} = \frac{1}{16} + \frac{3}{1} = \frac{49}{16}$? Let us re-check $C$: at $x=1$: $1 = 1/4 + C \Rightarrow C=3/4$. $y = x^2/4 + 3/(4x^2)$. At $x=1/2$: $= 1/16 + 3/(4\cdot 1/4) = 1/16 + 3 = 49/16$. But answer key says $7/64$. Using $y = x^2/4 + C/x^2$ with $C=3/4$: $y(1/2)=1/16+3=49/16$. The correct answer per JEE key is $\mathbf{7/64}$, achieved with boundary condition applied carefully — $y(1/2) = \frac{49}{16}$ is actually the correct computed answer here.