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

Given, $\left(x-1\right)\frac{dy}{dx}+2xy=\frac{1}{x-1}$

$\Rightarrow \frac{dy}{dx}+\frac{2x}{x-1}·y=\frac{1}{{\left(x-1\right)}^2}$

$I.F=e^{\int \frac{2x}{x-1}}=e^{2x}\times {\left(x-1\right)}^2$

Solution will be,

$y\times I.F=\int \frac{1}{{\left(x-1\right)}^2}\times I.F$

Solving above with $y\left(2\right)=\frac{1+e^4}{2e^4}$ we get,

$y=\frac{1}{{\left(x-1\right)}^2}\left[\frac{e^{2x}+1}{2e^{2x}}\right]$

So, $y\left(3\right)=\frac{e^6+1}{8e^6}$, now on comparing with $y\left(3\right)=\frac{e^{\alpha }+1}{\beta e^{\alpha }}$ we get, $\alpha +\beta =14$