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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If y=2 (2 x)+3 (2 x) , then x^2 y^ +x y^ +2 y=

If y=2cos⁡(2log⁡x)+3sin⁡(2log⁡x)y=2 \cos (2 \log x)+3 \sin (2 \log x), then x2y′′+xy′+2y=x^2 y^{\prime \prime}+x y^{\prime}+2 y=

Options

  1. A−2y-2 y
  2. B2y2 y
  3. C0
  4. D4

Correct answer

A. −2y-2 y

Step-by-step solution

We have, y=2cos⁡(2log⁡x)+3sin⁡(2log⁡x)∴y′=−2sin⁡(2log⁡x)⋅2x+3cos⁡(2log⁡x)⋅2x⇒xy′=2[−2sin⁡(2log⁡x)+3cos⁡(2log⁡x)]⇒xy′′+y′′=2[−2cos⁡(2log⁡x)⋅2x−3sin⁡(2log⁡x)⋅2x]⇒x2y′′+xy′=−4[2cos⁡(2log⁡x)+3sin⁡(2log⁡x)]⇒x2y′′+xy′=−4y⇒x2y′′+xy′+2y=−2y\begin{aligned}& y=2 \cos (2 \log x)+3 \sin (2 \log x) \\ \therefore \quad & y^{\prime}=-2 \sin (2 \log x) \cdot \frac{2}{x}+3 \cos (2 \log x) \cdot \frac{2}{x}\\ \Rightarrow \quad & x y^{\prime}=2[-2 \sin (2 \log x)+3 \cos (2 \log x)] \\ \Rightarrow & x y^{\prime \prime}+y^{\prime \prime}=2 \left[-2 \cos (2 \log x) \cdot \frac{2}{x}-3 \sin (2 \log x) \cdot \frac{2}{x}\right]\\ \Rightarrow & x² y^{\prime \prime}+x y^{\prime}=-4[2 \cos (2 \log x)+3 \sin (2 \log x)] \\ \Rightarrow & x² y^{\prime \prime}+x y^{\prime}=-4 y \\ \Rightarrow & x² y^{\prime \prime}+x y^{\prime}+2 y=-2 y \end{aligned}
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