JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Differential Equations

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 ^{α} + 1}{β e ^{α}}

, then the value of

α + β

is equal to ______.

Answer: 14

Solution

In standard form the equation is

\frac{d y}{d x} + \frac{2 x}{x - 1} y = \frac{1}{( x - 1 ) ^{2}}

. Since

\frac{2 x}{x - 1} = 2 + \frac{2}{x - 1}

, the integrating factor is

e^{2 x} (x - 1)^{2}

, and the equation becomes

\frac{d}{d x} [y e^{2 x} (x - 1)^{2}] = e^{2 x}

. Integrating gives

y e^{2 x} (x - 1)^{2} = \frac{e ^{2 x}}{2} + C

, so

y = \frac{1 + 2 C e ^{- 2 x}}{2 ( x - 1 ) ^{2}}

. The condition

y (2) = \frac{1 + e ^{4}}{2 e ^{4}} = \frac{1}{2} + \frac{1}{2 e ^{4}}

(where

(x - 1)^{2} = 1

) forces

C = \frac{1}{2}

, hence

y = \frac{1 + e ^{- 2 x}}{2 ( x - 1 ) ^{2}}

. At

x = 3

y (3) = \frac{1 + e ^{- 6}}{8} = \frac{e ^{6} + 1}{8 e ^{6}}

, so

α = 6

β = 8

and

α + β = 14

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer