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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If y = acos ( log x ) + bsin ( log x ) , where a and b are&...

If y=acos(logx)+bsin(logx)y = acos ( log x ) + bsin ( log x ), where a and ba \text{ and } b are& parameters, then x2y"+xy′x^{2}y^{"} + xy^{'}is equal to

Options

  1. Ayy
  2. B−y- y
  3. C2y2y
  4. D−2y- 2y

Correct answer

B. −y- y

Step-by-step solution

y=acos(logx)+bsin(logx)y = acos ( log x ) + bsin ( log x )

On differentiating w.r.t. xx, we get

y′=−asin(log⁡x)x+bcos(log⁡x)xy^{′} = \dfrac{- asin ( \log x )}{x} + \dfrac{bcos ( \log x )}{x}

⇒xy′=−asin(log⁡x)+bcos(log⁡x)⇒ xy^{′} = - asin ( \log x ) + bcos ( \log x )

Again, on differentiating w.r.t. xx, we get

xy″+y′=−acos(log⁡x)1x−bsin(log⁡x)1xxy^{″} + y^{′} = - acos ( \log x ) \dfrac{1}{x} - bsin ( \log x ) \dfrac{1}{x}

⇒x2y″+y′x=−[acos(log⁡x)+bsin(log⁡x)]⇒ x^{2} y^{″} + y^{′} x = - \left[acos ( \log x ) + bsin ( \log x\right) ]

⇒x2y″+xy′=−y⇒ x^{2} y^{″} + xy^{′} = - y
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