Question

If $y=3 \cos (\log x)+4 \sin (\log x)$, then

• A. $x y_2+y_1+y=0$
• B. $x y_2+y_1-y=0$
• C. $x^2 y_2+x y_1+y=0$
• D. None of these

Answer 1

Answer: (C)

Given, $y=3 \cos (\log x)+4 \sin (\log x) ....$(i)
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=y_1 =-3 \sin (\log x) \frac{d}{d x}(\log x)+4 \cos (\log x) \frac{d}{d x} \log x $
$ =-3 \sin (\log x) \frac{1}{x}+4 \cos (\log x) \frac{1}{x} \left(\because \frac{d y}{d x}=y_1\right)$
Multiplying by $x$, we get
$x y_1=-3 \sin (\log x)+4 \cos (\log x) .....$(ii)
Again, differentiating w.r.t. $x$, we obtain
$x y_2+y_1 \cdot 1 =-3 \cos (\log x) \frac{d}{d x}(\log x)-4 \sin (\log x) \frac{d}{d x} \log x $
$ =-3 \cos (\log x) \frac{1}{x}-4 \sin (\log x) \frac{1}{x}$
Multiplying throughout by $x$, we have
$x^2 y_2+x y_1=-(3 \cos (\log x)+4 \sin (\log x))[\text { from Eq. (i)] } $
$ \Rightarrow x^2 y_2+x y_1=-y \Rightarrow x^2 y_2+x y_1+y=0$