Let y = y(x), x > 1 , be the solution of the differential equation (x - 1)(dy)/(dx) + 2xy = (1)/(x - 1) , with y(2) = 1 + e ^ 4 2 e ^ 4 . If y(3) = e ^ α + 1 β e ^ α . then the value of α + β is equal to ______.

Correct answer: 0 Given, (x - 1)(dy)/(dx) + 2xy = (1)/(x - 1) ⇒ (dy)/(dx) + (2x)/(x - 1) · y = 1 (x - 1) ^ 2 I . F = e ^ ∫ (2x)/(x - 1) = e ^ 2x × (x - 1) ^ 2 Solution will be, y × I . F = ∫ 1 (x - 1) ^ 2 × I . F Solving above with y(2) = 1 + e ^ 4 2 e ^ 4 we get, y = 1 (x - 1) ^ 2 [ e ^ 2x + 1 2 e ^ 2x ] So, y(3) = e ^ 6 + 1 8 e ^ 6 , now on comparing with y(3) = e ^ α + 1 β e ^ α we get, α + β = 14

JEE Main 202229 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2022 · 29 June · Evening Shift · Mathematics · Differential Equations

Let y = y(x), x > 1 , be the solution of the differential equation...

Let

y = y\left(x\right), x > 1

, be the solution of the differential equation \left(x - 1\right)\dfrac{dy}{dx} + 2xy = \dfrac{1}{x - 1}

, with

y\left(2\right) = \dfrac{1 + {e}^{4}}{2{e}^{4}}

. If

y\left(3\right) = \dfrac{{e}^{\alpha} + 1}{\beta{e}^{\alpha}}

. then the value of

\alpha + \beta is equal to ______.

Correct answer

Step-by-step solution

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

$⇒ \dfrac{dy}{dx} + \dfrac{2x}{x - 1} \cdot y = \dfrac{1}{{\left(x - 1\right)}^{2}}$

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

Solution will be,

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

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

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

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

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