If $$\frac{dx}{dy} = \frac{1 + x - y^2}{y}$$, $$x(1) = 1$$, then $$5x(2)$$ is equal to:

We have $$\frac{dx}{dy} = \frac{1 + x - y^2}{y}$$, with $$x(1) = 1$$.

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

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

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

This is a linear first-order ODE in $$x$$ as a function of $$y$$.

Integrating factor: $$e^{\int -\frac{1}{y}dy} = e^{-\ln y} = \frac{1}{y}$$

Multiplying by IF:

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

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

Integrating:

$$\frac{x}{y} = -\frac{1}{y} - y + C$$

$$x = -1 - y^2 + Cy$$

Using $$x(1) = 1$$: $$1 = -1 - 1 + C$$, so $$C = 3$$.

Therefore $$x = -1 - y^2 + 3y$$.

$$x(2) = -1 - 4 + 6 = 1$$

$$5x(2) = 5 \times 1 = 5$$

The answer is $$\boxed{5}$$.