Let $$y(x)$$ be the solution of the differential equation $$2x^2 dy + (e^y - 2x)dx = 0$$, $$x > 0$$. If $$y(e) = 1$$, then $$y(1)$$ is equal to:

To solve for $$y(1)$$, we first need to find the general solution to the given differential equation by transforming it into a linear form.

The given equation is:

$$2x^2 dy + (e^y - 2x) dx = 0$$

Divide throughout by $$dx$$ and rearrange:

$$2x^2 \frac{dy}{dx} + e^y - 2x = 0$$

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

Divide the entire equation by

$$-e^y \cdot x^2$$

to start the substitution process:

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

Let $$v = e^{-y}$$. Then, differentiating with respect to $$x$$ gives:

$$\frac{dv}{dx} = -e^{-y} \frac{dy}{dx}$$

Substitute these into our equation:

$$2\frac{dv}{dx} + \frac{2}{x}v = \frac{1}{x^2}$$

Divide by $$2$$ to get the standard linear form

$$\frac{dv}{dx} + P(x)v = Q(x)$$ :

$$\frac{dv}{dx} + \frac{1}{x}v = \frac{1}{2x^2}$$

The integrating factor is calculated as:

$$\text{I.F.} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$$

Multiply the linear equation by the I.F.:

$$x \left( \frac{dv}{dx} + \frac{1}{x}v \right) = x \left( \frac{1}{2x^2} \right)$$

$$\frac{d}{dx}(v \cdot x) = \frac{1}{2x}$$

Integrate both sides with respect to $$x$$ :

$$v \cdot x = \int \frac{1}{2x} dx$$

$$v \cdot x = \frac{1}{2} \ln x + C$$

Substitute back $$v = e^{-y}$$ :

$$x e^{-y} = \frac{1}{2} \ln x + C$$

We are given that $$y(e) = 1$$ . Substitute $$x = e$$ and $$y = 1$$:

$$e \cdot e^{-1} = \frac{1}{2} \ln e + C$$

$$1 = \frac{1}{2}(1) + C$$

$$C = \frac{1}{2}$$

So, the specific solution is:

$$x e^{-y} = \frac{1}{2} \ln x + \frac{1}{2}$$

$$y(1)$$ Substitute $$x = 1$$

into our specific solution:

$$(1) e^{-y} = \frac{1}{2} \ln(1) + \frac{1}{2}$$

$$e^{-y} = 0 + \frac{1}{2}$$

$$e^{-y} = \frac{1}{2}$$

Taking the natural logarithm of both sides:

$$-y = \ln\left(\frac{1}{2}\right)$$

$$-y = -\ln 2$$

$$y = \ln 2 \quad (\text{or } \log_e 2)$$

The value of $$y(1)$$ is $$\log_e 2$$ , which corresponds to Option B.