Question

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 $y(3) = \frac{e^α + 1}{βe^α}$ ,
then the value of α + β is equal to ____.

Answer 1

The correct answer is 14
$\frac{dy}{dx} + y(\frac{2x}{x-1}) = \frac{1}{(x-1)^2}$
$\text{I.F.} = e^{\int \frac{2x}{x-1} \,dx}$
$= e^{2\int \left(\frac{x-1}{x-1} + \frac{1}{x-1}\right) \,dx}$
$= e^{2x + 2\ln(x-1)}$
$= e^{2x} ( x-1)²$
$⇒$ $\int d(y \cdot e^{2x}(x-1)^2) = \int e^{2x} \, dx$
⇒ $y \cdot e^{2x}(x-1)^2 = \frac{e^{2x}}{2} + c$
↓$$y(2) = \frac{1+e^4}{2e^4}
$\frac{1+e^4}{2e^4} \cdot e^4 = \frac{e^4}{2} + c$
$⇒$ $c = \frac{e^4}{2} \left( \frac{1+e^4 - e^4}{e^4} \right) = \frac{1}{2}$
$⇒$ $y \cdot e^{2x}(x-1)^2 = \frac{e^{2x+1}}{2}$
$↓y(3) = \frac{e^α+1}{βe^α}$
$⇒ \frac{e^α+1}{βe^α}.e^6.4 = \frac{e^6+1}{2}$
⇒ α=6 and β=8
⇒ α+β = 14