Question

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

Answer 1

$\frac{ dy }{ dx }+\frac{2 x }{ x -1} \cdot y =\frac{1}{( x -1)^{2}}$
$y =\frac{1}{( x -1)^{2}}\left[\frac{ e ^{2 x }+1}{2 e ^{2 x }}\right]$
$y (3)=\frac{ e ^{6}+1}{8 e ^{6}}$
$\alpha+\beta=14$