If $$y = y(x)$$ is the solution of the differential equation, $$x\frac{dy}{dx} + 2y = x^2$$ satisfying $$y(1) = 1$$, then $$y\left(\frac{1}{2}\right)$$ is equal to:

We start with the given differential equation $$x\dfrac{dy}{dx}+2y = x^{2}.$$

To bring it to the standard linear form $$\dfrac{dy}{dx}+P(x)\,y = Q(x),$$ we divide every term by $$x$$ (note that we are looking around $$x>0$$, so division is legitimate):

$$\dfrac{dy}{dx} + \dfrac{2}{x}\,y = x.$$

Here we can clearly read $$P(x)=\dfrac{2}{x}$$ and $$Q(x)=x.$$

For a first-order linear differential equation, the integrating factor (I.F.) is defined by the formula

$$\text{I.F.}=e^{\displaystyle\int P(x)\,dx}.$$

Substituting $$P(x)=\dfrac{2}{x},$$ we obtain

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

(we use $$x>0$$ so $$|x|=x$$).

Now we multiply the whole differential equation in its linear form by this integrating factor:

$$x^{2}\left(\dfrac{dy}{dx}\right)+x^{2}\left(\dfrac{2}{x}\right)y = x^{2}\cdot x.$$

Performing the obvious simplifications, we get

$$x^{2}\dfrac{dy}{dx}+2x\,y = x^{3}.$$

The left-hand side is the exact derivative of the product $$y\,(x^{2})$$ because of the rule $$\dfrac{d}{dx}\bigl(y\cdot x^{2}\bigr)=x^{2}\dfrac{dy}{dx}+2x\,y.$$ Hence we can rewrite the equation compactly as

$$\dfrac{d}{dx}\Bigl(y\,x^{2}\Bigr)=x^{3}.$$

We integrate both sides with respect to $$x$$:

$$\int\dfrac{d}{dx}\Bigl(y\,x^{2}\Bigr)\,dx=\int x^{3}\,dx.$$

On integration, the left side simply gives back the function inside the derivative, while the right side yields $$\dfrac{x^{4}}{4}+C,$$ where $$C$$ is the constant of integration. Thus,

$$y\,x^{2}= \dfrac{x^{4}}{4}+C.$$

We now solve for $$y$$ by dividing by $$x^{2}$$:

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

To determine the constant $$C,$$ we use the initial condition $$y(1)=1.$$ Substituting $$x=1,\,y=1$$ in the general solution, we obtain

$$1=\dfrac{(1)^{2}}{4}+\dfrac{C}{(1)^{2}}=\dfrac{1}{4}+C.$$

Hence, $$C=1-\dfrac{1}{4}=\dfrac{3}{4}.$$

Substituting this value of $$C$$ back into the expression for $$y,$$ we get the particular solution

$$y(x)=\dfrac{x^{2}}{4}+\dfrac{\tfrac{3}{4}}{x^{2}}=\dfrac{x^{2}}{4}+\dfrac{3}{4x^{2}}.$$

Now we evaluate $$y\!\left(\dfrac{1}{2}\right).$$ First compute $$x^{2}$$ when $$x=\dfrac{1}{2}:$$

$$x^{2}=\left(\dfrac{1}{2}\right)^{2}=\dfrac{1}{4}.$$

Substituting into the formula for $$y$$, we have

$$y\!\left(\dfrac{1}{2}\right)=\dfrac{\dfrac{1}{4}}{4}+\dfrac{3}{4\left(\dfrac{1}{2}\right)^{2}}=\dfrac{1}{16}+\dfrac{3}{4\cdot\dfrac{1}{4}}.$$

Because $$4\left(\dfrac{1}{2}\right)^{2}=4\cdot\dfrac{1}{4}=1,$$ the denominator of the second term is $$1,$$ so that term simplifies straightaway to $$3.$$

Thus

$$y\!\left(\dfrac{1}{2}\right)=\dfrac{1}{16}+3=\dfrac{1}{16}+\dfrac{48}{16}=\dfrac{49}{16}.$$

Hence, the correct answer is Option D.