If a curve $$y = f(x)$$, passing through the point $$(1, 2)$$, is the solution of the differential equation $$2x^2dy = (2xy + y^2)dx$$, then $$f\left(\frac{1}{2}\right)$$ is equal to:

We have the differential equation $$2x^{2}\,dy=(2xy+y^{2})\,dx$$ and the required curve passes through the point $$(1,2).$$ Our objective is to determine the value of $$y$$ when $$x=\dfrac12,$$ that is, to find $$f\!\left(\dfrac12\right).$$

First we rewrite the differential equation in the standard form $$\dfrac{dy}{dx}=\,\ldots$$ by dividing both sides by $$2x^{2}\,dx:$$

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

So we have

$$\dfrac{dy}{dx}-\dfrac{y}{x}= \dfrac{y^{2}}{2x^{2}}.$$

This is a Bernoulli differential equation of the form

$$\dfrac{dy}{dx}+P(x)\,y = Q(x)\,y^{n},$$

with $$P(x)=-\dfrac1x,\; Q(x)=\dfrac1{2x^{2}},\; n=2.$$

For a Bernoulli equation we set $$z=y^{\,1-n}=y^{-1},$$ i.e. $$z=\dfrac1y.$$ Then $$y=\dfrac1z$$ and, by differentiation,

$$\dfrac{dy}{dx}= -\dfrac{1}{z^{2}}\dfrac{dz}{dx}.$$

Substituting these in the differential equation we get

$$-\dfrac1{z^{2}}\dfrac{dz}{dx}-\dfrac1x\cdot\dfrac1z=\dfrac1{2x^{2}}\cdot\dfrac1{z^{2}}.$$

Multiplying throughout by $$z^{2}$$ to remove denominators,

$$-\dfrac{dz}{dx}-\dfrac{z}{x}=\dfrac1{2x^{2}}.$$

Rearranging gives a linear first-order equation in $$z$$:

$$\dfrac{dz}{dx}+\dfrac{1}{x}\,z=-\dfrac1{2x^{2}}.$$

The integrating factor (I.F.) for the linear equation $$\dfrac{dz}{dx}+P(x)z=R(x)$$ is $$\mu(x)=e^{\int P(x)\,dx}.$$ Here $$P(x)=\dfrac1x,$$ so

$$\mu(x)=e^{\int\frac{1}{x}\,dx}=e^{\ln|x|}=x.$$

Multiplying the entire linear equation by the integrating factor $$x,$$ we obtain

$$x\dfrac{dz}{dx}+z=-\dfrac1{2x}.$$

The left-hand side is the derivative of the product $$xz$$ because

$$\dfrac{d}{dx}(xz)=x\dfrac{dz}{dx}+z.$$

Thus

$$\dfrac{d}{dx}(xz)=-\dfrac1{2x}.$$

Integrating both sides with respect to $$x$$ gives

$$xz=\int -\dfrac1{2x}\,dx = -\dfrac12\int\dfrac1x\,dx=-\dfrac12\ln|x|+C,$$

where $$C$$ is the constant of integration. Hence

$$xz = C-\dfrac12\ln|x|.$$

Substituting back $$z=\dfrac1y,$$ we get

$$x\left(\dfrac1y\right)=C-\dfrac12\ln|x|,$$

or

$$\dfrac1y=\dfrac{C-\dfrac12\ln|x|}{x}.$$

Taking the reciprocal gives the explicit solution for $$y$$:

$$y=\dfrac{x}{\,C-\dfrac12\ln|x|\,}.$$

Now we use the given point $$(1,2)$$ to determine $$C.$$ Putting $$x=1,\;y=2$$ in the above expression, we have

$$2=\dfrac{1}{C-\dfrac12\ln 1}=\dfrac1C,$$

because $$\ln1=0.$$ Hence

$$C=\dfrac12.$$

Therefore the particular solution is

$$y=\dfrac{x}{\dfrac12-\dfrac12\ln|x|}\;=\;\dfrac{2x}{\,1-\ln|x|\,}.$$

We now evaluate $$y$$ at $$x=\dfrac12.$$ First compute the denominator:

$$1-\ln\!\bigl|\tfrac12\bigr|=1-\ln\!\left(\dfrac12\right)=1-\bigl(-\ln 2\bigr)=1+\ln2.$$

Substituting $$x=\dfrac12$$ in $$y=\dfrac{2x}{1-\ln|x|},$$ we obtain

$$y=\dfrac{2\left(\dfrac12\right)}{\,1+\ln2\,} =\dfrac1{1+\ln2}.$$

Hence $$f\!\left(\dfrac12\right)=\dfrac1{1+\ln_e2}.$$

Hence, the correct answer is Option A.