Question

If $y=a\cos (\log x)+b\sin (\log x)$where $a,b$are parameters then $x^{2}y\text{'}\text{'}+xy\text{'}$ equals to ?

• A. $y$
• B. $-y$
• C. $-2y$
• D. $2y$

Answer 1

The correct option is B

$-y$

Step 1: Given function

The Given function $y=a\cos (\log x)+b\sin (\log x)$

Step 2: Differentiate the function with respect to $x$

$$\begin{gathered} y\text{'}=-a\sin (\log x)\times (1/x)+b\cos (\log x)\times (1/x) \\ y\text{'}=(1/x)[-a\sin (\log x)+b\cos (\log x\left)\right] \\ xy\text{'}=[-a\sin (\log x)+b\cos (\log x\left)\right] \end{gathered}$$

Differentiate again with respect to $x$

$$\begin{gathered} xy\text{'}\text{'}+y\text{'}=-a\cos (\log x)(1/x)–b\sin (\log x)(1/x) \\ xy\text{'}\text{'}+y\text{'}=(-1/x)[a\cos (\log x)+b\sin (\log x\left)\right] \\ xy\text{'}\text{'}+y\text{'}=(-1/x)y \\ xy\text{'}\text{'}+y\text{'}=-y/x \end{gathered}$$

Multiply with $x$

$x^{2}y\text{'}\text{'}+xy\text{'}=-y$

Hence option (B) is the correct answer