\((x - 1) \frac{dy}{dx} + 2xy = \frac 1{x - 1}\)
⇒ \(\frac{dy}{dx} + \frac{2x}{x -1}y = \frac1{(x - 1)^2}\)
which is a linear differential equation
\(\therefore I.F. = e^{\int pdx} = e^{\int \frac{2x}{x-1}dx}\)
\(= e^{\int \frac{2x -2 +2}{x -1}dx}\)
\(= e^{\int (2 + \frac 2{x-1})dx}\)
\(= e^{2x + 2log\,(x -1)}\)
\(= e^{2x}.e^{log(x -1)^2}\) \((\because e^{x + y} = e^x.e^y)\)
\(= (x - 1)^2 e^{2x}\) \((\because e^{log(f(x))} = f(x))\)
\(\therefore \) Common solution is
\(y \times I.F. = \int Q \times (I.F.) dx + C\)
⇒ \(y(x - 1)^2 e^{2x} = \int \frac 1{(x - 1)^2} \times e^{2x} (x - 1)^2 dx +C\)
⇒ \(y (x -1 )^2 e^{2x} = \frac{e^{2x}}2 + C\) ......(i)
\(\because y(2) = \frac{1+e^4}{2e^4}\) (given)
\(\therefore \) By putting x = 2 in equation (i), we obtain
\(\frac{1 + e^4}{2e^4} \times 1 \times e^4 = \frac{e^4}2 + C\)
⇒ \(C = \frac{1 + e^4}2 - \frac{e^4}2 = \frac 12\)
\(\therefore \) From (i), we get
\(y(x - 1)^2 e^{2x} = \frac{e^{2x}}2 + \frac12\)
⇒ \(y = \frac{e^{2x} + 1}{2(x - 1)^2 e^{2x}}\) ......(ii)
⇒ \(y(3) = \frac{e^{6x}+ 1}{2(3 - 1)^2 e^{6x}}\) (By putting x = 3 in equation (ii))
⇒ \(y(3) = \frac{e^{6x} + 1}{8 \,e^{6x}}\)
\(\therefore \alpha = 6, \beta = 8\) \((\because y = \frac{e^\alpha+1}{\beta \,e^\alpha}(given))\)
\(\therefore \alpha + \beta = 6 + 8 = 14\)