Let $$x=x(y)$$ be the solution of the differential equation$$ y^{2}dx+\left( x-\frac{1}{y}\right)dy=0 $$ . If $$x(1)=1$$, then $$\frac{1}{2}$$ is :

The given differential equation is:

$$y^{2}dx + \left( x - \frac{1}{y} \right)dy = 0$$

Rearranging terms to express in standard linear form:

$$y^{2} \frac{dx}{dy} = -x + \frac{1}{y}$$

$$y^{2} \frac{dx}{dy} + x = \frac{1}{y}$$

Dividing both sides by $$y^{2}$$:

$$\frac{dx}{dy} + \frac{1}{y^{2}} x = \frac{1}{y^{3}}$$

This is a linear differential equation of the form:

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

where $$P(y) = \frac{1}{y^{2}}$$ and $$Q(y) = \frac{1}{y^{3}}$$.

To solve, use the integrating factor method. The integrating factor (IF) is:

$$IF = e^{\int P(y) dy} = e^{\int \frac{1}{y^{2}} dy}$$

Compute the integral:

$$\int \frac{1}{y^{2}} dy = \int y^{-2} dy = -y^{-1} = -\frac{1}{y}$$

Thus,

$$IF = e^{-\frac{1}{y}}$$

The solution is given by:

$$x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$$

Substituting:

$$x \cdot e^{-\frac{1}{y}} = \int \frac{1}{y^{3}} e^{-\frac{1}{y}} dy + C$$

Evaluate the integral:

Let $$I = \int \frac{1}{y^{3}} e^{-\frac{1}{y}} dy$$

Use substitution: let $$u = -\frac{1}{y}$$, then $$du = \frac{1}{y^{2}} dy$$.

Now, $$\frac{1}{y} = -u$$, so:

$$\frac{1}{y^{3}} dy = \frac{1}{y} \cdot \frac{1}{y^{2}} dy = (-u) du$$

Thus,

$$I = \int (-u) e^{u} du = -\int u e^{u} du$$

Integrate by parts: let $$v = u$$, $$dw = e^{u} du$$, so $$dv = du$$, $$w = e^{u}$$.

Then,

$$\int u e^{u} du = u e^{u} - \int e^{u} du = u e^{u} - e^{u} + C_1 = e^{u} (u - 1) + C_1$$

Therefore,

$$I = - \left[ e^{u} (u - 1) \right] + C = -e^{u} (u - 1) + C$$

Substitute back $$u = -\frac{1}{y}$$:

$$I = -e^{-\frac{1}{y}} \left( -\frac{1}{y} - 1 \right) + C = -e^{-\frac{1}{y}} \left( -\left( \frac{1}{y} + 1 \right) \right) + C = e^{-\frac{1}{y}} \left(1 + \frac{1}{y}\right) + C$$

So,

$$x \cdot e^{-\frac{1}{y}} = e^{-\frac{1}{y}} \left(1 + \frac{1}{y}\right) + C$$

Dividing both sides by $$e^{-\frac{1}{y}}$$:

$$x = 1 + \frac{1}{y} + C e^{\frac{1}{y}}$$

This is the general solution.

Apply the initial condition: when $$y = 1$$, $$x = 1$$.

Substitute:

$$1 = 1 + \frac{1}{1} + C e^{\frac{1}{1}}$$

$$1 = 1 + 1 + C e$$

$$1 = 2 + C e$$

$$C e = -1$$

$$C = -\frac{1}{e}$$

Thus, the particular solution is:

$$x = 1 + \frac{1}{y} - \frac{1}{e} e^{\frac{1}{y}} = 1 + \frac{1}{y} - e^{\frac{1}{y} - 1}$$

Now, find $$x\left(\frac{1}{2}\right)$$ by substituting $$y = \frac{1}{2}$$:

$$x\left(\frac{1}{2}\right) = 1 + \frac{1}{\frac{1}{2}} - e^{\frac{1}{\frac{1}{2}} - 1}$$

$$= 1 + 2 - e^{2 - 1}$$

$$= 3 - e^{1}$$

$$= 3 - e$$

The value of $$x\left(\frac{1}{2}\right)$$ is $$3 - e$$, which corresponds to option C.