If \( y = 3 \cos \: (\log x ) + 4 \sin ( \log x )\), show that \(x^2y_2 + xy_1 + y = 0\)

Answer 1

Toolbox:

• $y=f(x)$
• $\large\frac{dy}{dx}$$=f'(x)$
• $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
• $\large\frac{d}{dx}$$(\sin x)=\cos x$
• $\large\frac{d}{dx}$$(\cos x)=-\sin x$
• $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$

Step 1:

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

$\large\frac{dy}{dx}$$=y_1=3\sin(\log x).\frac{1}{x}$$+4\cos(\log x).\large\frac{1}{x}$

$xy_1=3\sin(\log x)+4\cos(\log x)$

Step 2:

Differentiating with respect to $x$ we get

$xy_2+y_1.1=\large\frac{3\cos(\log x)}{x}-\large\frac{4\sin(\log x)}{x}$

$xy_2+y_1=\large\frac{-1}{x}$$[3\cos (\log x)+4\sin(\log x)]$

$xy_2+y_1=\large\frac{-y}{x}$

$x^2y_2+xy_1=-y$

$x^2y_2+xy_1+y=0$