Let $$y = y(x)$$, $$x > 1$$, be the solution of the differential equation $$(x-1)\frac{dy}{dx} + 2xy = \frac{1}{x-1}$$, with $$y(2) = \frac{1+e^4}{2e^4}$$. If $$y(3) = \frac{e^{\alpha}+1}{\beta e^{\alpha}}$$, then the value of $$\alpha + \beta$$ is equal to ______.

We have the differential equation $$(x-1)\frac{dy}{dx} + 2xy = \frac{1}{x-1}$$ with the initial condition $$y(2) = \frac{1+e^4}{2e^4}$$. To put it into standard linear form, we divide by $$x-1$$ and obtain $$\frac{dy}{dx} + \frac{2x}{x-1}y = \frac{1}{(x-1)^2}$$.

The integrating factor is given by $$\mathrm{IF} = e^{\int \frac{2x}{x-1}\,dx}$$. Since $$\int \frac{2x}{x-1}\,dx = \int \frac{2(x-1)+2}{x-1}\,dx = 2x + 2\ln\lvert x-1\rvert\,, $$ it follows that $$\mathrm{IF} = e^{2x + 2\ln\lvert x-1\rvert} = e^{2x}(x-1)^2\,. $$

Multiplying the differential equation by this integrating factor yields $$\frac{d}{dx}\bigl[y\cdot e^{2x}(x-1)^2\bigr] = \frac{e^{2x}(x-1)^2}{(x-1)^2} = e^{2x}\,. $$ Integrating both sides gives $$y\cdot e^{2x}(x-1)^2 = \int e^{2x}\,dx = \frac{e^{2x}}{2} + C$$ and hence the general solution $$y = \frac{1}{2(x-1)^2} + \frac{C}{e^{2x}(x-1)^2}\,.$$

We now apply the initial condition $$y(2) = \frac{1+e^4}{2e^4}$$. Substituting $$x=2$$ into the general solution gives $$\frac{1+e^4}{2e^4} = \frac{1}{2(1)^2} + \frac{C}{e^4(1)^2}\,, $$ which simplifies to $$\frac{1+e^4}{2e^4} = \frac{1}{2} + \frac{C}{e^4}\,, $$ so $$\frac{1+e^4}{2e^4} - \frac{1}{2} = \frac{C}{e^4}\,, \quad \frac{1}{2e^4} = \frac{C}{e^4}\,, $$ giving $$C = \frac12\,. $$

With $$C=\tfrac12$$, it follows that $$y(3) = \frac{1}{2(2)^2} + \frac{1/2}{e^6(2)^2} = \frac{1}{8} + \frac{1}{8e^6} = \frac{e^6 + 1}{8e^6}\,. $$ Comparing this with the form $$y(3) = \frac{e^\alpha + 1}{\beta e^\alpha}$$ shows that $$\alpha = 6$$ and $$\beta = 8$$, so $$\alpha + \beta = 14$$. 14