If $$y\frac{dy}{dx} = x\frac{y^2}{x^2} + \frac{\phi\frac{y^2}{x^2}}{\phi'\frac{y^2}{x^2}}$$, $$x > 0$$, $$\phi > 0$$, and $$y(1) = -1$$, then $$\phi\frac{y^2}{4}$$ is equal to:

The given equation is:

$$y \frac{dy}{dx} = x \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi'\left(\frac{y^2}{x^2}\right)}$$

Let $$v = \frac{y^2}{x^2}$$. This implies $$y^2 = vx^2$$.

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

$$2y \frac{dy}{dx} = v(2x) + x^2 \frac{dv}{dx}$$

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

Substitute $$y \frac{dy}{dx}$$ and $$\frac{y^2}{x^2} = v$$ into the original equation:

$$vx + \frac{x^2}{2} \frac{dv}{dx} = xv + \frac{\phi(v)}{\phi'(v)}$$

Subtract $$xv$$ from both sides:

$$\frac{x^2}{2} \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$$

Rearrange the terms to separate $$v$$ and $$x$$:

$$\frac{\phi'(v)}{\phi(v)} dv = \frac{2}{x} dx$$

Integrating both sides:

$$\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{2}{x} dx$$

$$\ln|\phi(v)| = 2 \ln|x| + \ln|C|$$

$$\ln|\phi(v)| = \ln|Cx^2|$$

$$\phi(v) = Cx^2$$

Substituting $$v = \frac{y^2}{x^2}$$ back:

$$\phi\left(\frac{y^2}{x^2}\right) = Cx^2$$

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

$$\phi\left(\frac{(-1)^2}{1^2}\right) = C(1)^2$$

$$\phi(1) = C$$

So, the general solution is:

$$\phi\left(\frac{y^2}{x^2}\right) = \phi(1) \cdot x^2$$

We need to find the value of $$\phi\left(\frac{y^2}{4}\right)$$. In the context of the question's structure, this refers to the value of the function when $$x = 2$$:

$$\phi\left(\frac{y^2}{2^2}\right) = \phi(1) \cdot 2^2$$

$$\phi\left(\frac{y^2}{4}\right) = 4\phi(1)$$

Correct Option: D ($$4\phi(1)$$)