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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Differentiation

NDA 2018 · Mathematics · Differentiation

If y= ( x) , then which one of the following is correct?

If y=sin⁡(ln⁡x)y=\sin (\ln x), then which one of the following is correct?

Options

  1. Ad2ydx2+y=0\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}+\mathrm{y}=0
  2. Bd2ydx2=0\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=0
  3. Cx2d2ydx2+xdydx+y=0x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0
  4. Dx2d2ydx2−xdydx+y=0x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+y=0

Correct answer

C. x2d2ydx2+xdydx+y=0x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0

Step-by-step solution

 (c) y=sin⁡(log⁡x)⇒dydx=cos⁡(log⁡x)x⇒x(dydx)=cos⁡(log⁡x)\begin{aligned} & \text{ (c) } y=\sin (\log x) \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{\cos (\log x)}{x} \\ & \Rightarrow \quad x\left(\frac{d y}{d x}\right)=\cos (\log x) \end{aligned} Again differentiating, xd2ydx2+dydx=−sin⁡log⁡xx=−yx⇒x2d2ydx2+xdydx+y=0\begin{aligned} & x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=-\frac{\sin \log x}{x}=\frac{-y}{x} \\ \Rightarrow \quad & x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0 \end{aligned}
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