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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Differentiation

JEE Main 2017 · 9 April · Online · Mathematics · Differentiation

If 2x = y ^ (1)/(5) + y ^ - (1)/(5) and ( x ^ 2 - 1) d ^ 2 y d x ^...

If 2x=y15+y−15and(x2−1)d2ydx2+\lambdaxdydx+ky=02x = {y}^{\dfrac{1}{5}} + {y}^{- \dfrac{1}{5}} and \left({x}^{2} - 1\right)\dfrac{{d}^{2}y}{d{x}^{2}} + \lambdax\dfrac{dy}{dx} + ky = 0, then λ+k\lambda + k is equal to

Options

  1. A2626
  2. B−24- 24
  3. C−23- 23
  4. D−26- 26

Correct answer

B. −24- 24

Step-by-step solution

Given y15+y−15=2x{y}^{\dfrac{1}{5}} + {y}^{- \dfrac{1}{5}} = 2x

⇒(15y−45−15y−65)dydx=2⇒ \left(\dfrac{1}{5}{y}^{- \dfrac{4}{5}} - \dfrac{1}{5}{y}^{- \dfrac{6}{5}}\right)\dfrac{dy}{dx} = 2

⇒y′(y15−y−15)=10y⇒ {y}^{′}\left({y}^{\dfrac{1}{5}} - {y}^{- \dfrac{1}{5}}\right) = 10y

⇒y′(2x2−1)=10y⇒ {y}^{′}\left(2\sqrt{{x}^{2} - 1}\right) = 10y

⇒y′x2−1=5y⇒ {y}^{′}\sqrt{{x}^{2} - 1} = 5y

Differentiating again

y′′(2x2−1)+y′22x2x2−1=10y′{y}^{′′}\left(2\sqrt{{x}^{2} - 1}\right) + {y}^{′}2\dfrac{2x}{2\sqrt{{x}^{2} - 1}} =10y'

⇒y′′(x2−1)+xy′=5x2−1(y′)⇒ {y}^{′′}\left({x}^{2} - 1\right) + x{y}^{′} = 5\sqrt{{x}^{2} - 1}\left({y}^{′}\right)

⇒y′′(x2−1)+xy′−25y=0⇒ {y}^{′′}\left({x}^{2} - 1\right) + x{y}^{′} - 25y = 0

⇒λ=1,k=−25⇒ \lambda = 1, k = - 25

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