Given differential equation

$x\frac{dy}{dx}+2y=x^2, \qquad y(1)=1$

We first write it in standard linear form by dividing by $x$ (for $x\neq 0$):

$\frac{dy}{dx}+\frac{2}{x}y=x$

Identify integrating factor

For the linear differential equation

$\frac{dy}{dx}+P(x)y=Q(x),$

here

$P(x)=\frac{2}{x}$

So the integrating factor is

$\text{I.F.}=e^{\int \frac{2}{x}\,dx}=e^{2\ln x}=x^2$

(taking $x>0$ since the condition is given at $x=1$ and we need $x=\tfrac12$).

Multiply throughout by the integrating factor

$x^2\frac{dy}{dx}+2xy=x^3$

The left-hand side is the derivative of $x^2y$:

$\frac{d}{dx}(x^2y)=x^3$

Integrate both sides

$x^2y=\int x^3\,dx=\frac{x^4}{4}+C$

Hence,

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

Use the initial condition $y(1)=1$

$1=\frac{1}{4}+C$

So,

$C=\frac{3}{4}$

Therefore,

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

Find $y\left(\frac12\right)$

$y\left(\frac12\right)=\frac{(1/2)^2}{4}+\frac{3}{4(1/2)^2}$

$=\frac{1/4}{4}+\frac{3}{4\cdot 1/4}$

$=\frac{1}{16}+3$

$=\frac{1}{16}+\frac{48}{16}=\frac{49}{16}$

Check options

$\frac{49}{16}$

matches Option B.