If the solution curve of the differential equation $$(y-2\log_e x)dx + (x\log_e x^2)dy = 0$$, $$x \gt 1$$ passes through the points $$(e, \dfrac{4}{3})$$ and $$(e^4, \alpha)$$, then $$\alpha$$ is equal to ______.

We consider the differential equation $$(y - 2\log_e x)\, dx + (x \log_e x^2)\, dy = 0$$ for $$x \gt 1$$ and seek a solution passing through the points $$(e, \frac{4}{3})$$ and $$(e^4, \alpha)$$.

Using the identity $$\log_e x^2 = 2\log_e x$$ transforms the equation into $$(y - 2\ln x)\, dx + 2x\ln x\, dy = 0$$ which can be rearranged into $$\frac{dy}{dx} = \frac{2\ln x - y}{2x\ln x}\,. $$

Introducing the substitution $$v = \ln x$$ leads to $$\frac{dv}{dx} = \frac{1}{x}$$ and hence $$\frac{dy}{dv} = x\frac{dy}{dx}\,. $$ In terms of $$v$$, the equation becomes $$\frac{dy}{dv} = \frac{2v - y}{2v}\,, $$ or equivalently $$\frac{dy}{dv} + \frac{y}{2v} = 1\,. $$

The integrating factor for this linear equation is $$\text{I.F.} = e^{\int \frac{1}{2v}\, dv} = e^{\frac{1}{2}\ln v} = \sqrt{v}\,. $$ Multiplying through by $$\sqrt{v}$$ yields $$\frac{d}{dv}\bigl(y\sqrt{v}\bigr) = \sqrt{v}\,, $$ so that integration gives $$y\sqrt{v} = \int \sqrt{v}\, dv = \frac{2}{3}v^{3/2} + C\,. $$ Solving for $$y$$ produces $$y = \frac{2}{3}v + \frac{C}{\sqrt{v}} = \frac{2}{3}\ln x + \frac{C}{\sqrt{\ln x}}\,. $$

Imposing the condition at $$x = e$$, where $$\ln e = 1$$, we have $$\frac{4}{3} = \frac{2}{3}(1) + \frac{C}{\sqrt{1}} = \frac{2}{3} + C$$ which implies $$C = \frac{2}{3}\,. $$ Therefore the general solution becomes $$y = \frac{2}{3}\ln x + \frac{2/3}{\sqrt{\ln x}}\,. $$

Finally, evaluating at $$x = e^4$$ gives $$\ln(e^4) = 4$$ and $$\sqrt{\ln x} = 2$$, so $$\alpha = \frac{2}{3}\cdot 4 + \frac{2/3}{2} = \frac{8}{3} + \frac{1}{3} = 3\,. $$

The answer is $$\alpha = 3$$.