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
$\frac{dy}{dx}=y_1=-3\sin (\log x)\frac{d}{dx}(\log x)+4\cos (\log x)\frac{d}{dx}\log x$
$=-3\sin (\log x)\frac{1}{x}+4\cos (\log x)\frac{1}{x}\left(\because \frac{dy}{dx}=y_1\right)$
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
$x{y}_2+y_1\cdot 1=-3\cos (\log x)\frac{d}{dx}(\log x)-4\sin (\log x)\frac{d}{dx}\log x$
$=-3\cos (\log x)\frac{1}{x}-4\sin (\log x)\frac{1}{x}$
Multiplying throughout by $x$, we have
$x^2{y}_2+x{y}_1=-(3\cos (\log x)+4\sin (\log x))[\text{ from Eq. (i)] }$
$\Rightarrow x^2{y}_2+x{y}_1=-y\Rightarrow x^2{y}_2+x{y}_1+y=0$