# If, show that

Question:

If $y=3 \cos (\log x)+4 \sin (\log x)$, show that $x^{2} y_{2}+x y_{1}+y=0$

Solution:

It is given that, $y=3 \cos (\log x)+4 \sin (\log x)$

Then,

$y_{1}=3 \cdot \frac{d}{d x}[\cos (\log x)]+4 \cdot \frac{d}{d x}[\sin (\log x)]$

$=3 \cdot\left[-\sin (\log x) \cdot \frac{d}{d x}(\log x)\right]+4 \cdot\left[\cos (\log x) \cdot \frac{d}{d x}(\log x)\right]$

$\therefore y_{1}=\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}=\frac{4 \cos (\log x)-3 \sin (\log x)}{x}$

$\therefore y_{2}=\frac{d}{d x}\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}\right)$

$=\frac{x\{4 \cos (\log x)-3 \sin (\log x)\}^{\prime}-\{4 \cos (\log x)-3 \sin (\log x)\}(x)^{\prime}}{x^{2}}$

$=\frac{x\left[4\{\cos (\log x)\}^{\prime}-3\{\sin (\log x)\}^{\prime}\right]-\{4 \cos (\log x)-3 \sin (\log x)\} \cdot 1}{x^{2}}$

$=\frac{x\left[-4 \sin (\log x) \cdot(\log x)^{\prime}-3 \cos (\log x) \cdot(\log x)^{\prime}\right]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}}$

$=\frac{x\left[-4 \sin (\log x) \cdot \frac{1}{x}-3 \cos (\log x) \cdot \frac{1}{x}\right]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}}$

$=\frac{-4 \sin (\log x)-3 \cos (\log x)-4 \cos (\log x)+3 \sin (\log x)}{x^{2}}$

$=\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}}$

$\therefore x^{2} y_{2}+x y_{1}+y$

$=x^{2}\left(\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}}\right)+x\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}\right)+3 \cos (\log x)+4 \sin (\log x)$

$=-\sin (\log x)-7 \cos (\log x)+4 \cos (\log x)-3 \sin (\log x)+3 \cos (\log x)+4 \sin (\log x)$

$=0$

Hence, proved.