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Question

# If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0.

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Solution

## Here, $y=3\mathrm{cos}\left(\mathrm{log}x\right)+4\mathrm{sin}\left(\mathrm{log}x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{y}_{1}=-3\mathrm{sin}\left(\mathrm{log}x\right)×\frac{1}{x}+4\mathrm{cos}\left(\mathrm{log}x\right)×\frac{1}{x}\phantom{\rule{0ex}{0ex}}=\frac{-3\mathrm{sin}\left(\mathrm{log}x\right)+4\mathrm{cos}\left(\mathrm{log}x\right)}{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{y}_{2}=\frac{\left(\frac{-3\mathrm{cos}\left(\mathrm{log}x\right)}{x}-\frac{4\mathrm{sin}\left(\mathrm{log}x\right)}{x}\right)×x-\left\{-3\mathrm{sin}\left(\mathrm{log}x\right)+4\mathrm{cos}\left(\mathrm{log}x\right)\right\}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{-3\mathrm{cos}\left(\mathrm{log}x\right)-4\mathrm{sin}\left(\mathrm{log}x\right)-\left\{-3\mathrm{sin}\left(\mathrm{log}x\right)+4\mathrm{cos}\left(\mathrm{log}x\right)\right\}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{-3\mathrm{cos}\left(\mathrm{log}x\right)-4\mathrm{sin}\left(\mathrm{log}x\right)-\left\{-3\mathrm{sin}\left(\mathrm{log}x\right)+4\mathrm{cos}\left(\mathrm{log}x\right)\right\}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{-3\mathrm{cos}\left(\mathrm{log}x\right)-4\mathrm{sin}\left(\mathrm{log}x\right)}{{x}^{2}}-\frac{\left\{-3\mathrm{sin}\left(\mathrm{log}x\right)+4\mathrm{cos}\left(\mathrm{log}x\right)\right\}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{-\left\{3\mathrm{cos}\left(\mathrm{log}x\right)+4\mathrm{sin}\left(\mathrm{log}x\right)\right\}}{{x}^{2}}-\frac{\left\{-3\mathrm{sin}\left(\mathrm{log}x\right)+4\mathrm{cos}\left(\mathrm{log}x\right)\right\}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{-y}{{x}^{2}}-\frac{{y}_{1}}{x}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}{y}_{2}=-y-x{y}_{1}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}{y}_{2}+y+x{y}_{1}=0$ Hence proved.

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