Question

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

Answer 1

Here,
$$\begin{gathered} y=3\cos \left(\log x\right)+4\sin \left(\log x\right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_1=-3\sin \left(\log x\right)\times \frac{1}{x}+4\cos \left(\log x\right)\times \frac{1}{x} \\ =\frac{-3\sin \left(\log x\right)+4\cos \left(\log x\right)}{x} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_2=\frac{\left({\displaystyle \frac{-3\cos \left(\log x\right)}{x}}-{\displaystyle \frac{4\sin \left(\log x\right)}{x}}\right)\times x-\left\{-3\sin \left(\log x\right)+4\cos \left(\log x\right)\right\}}{x^2} \\ \Rightarrow y_2=\frac{-3\cos \left(\log x\right)-{\displaystyle 4\sin \left(\log x\right)}-\left\{-3\sin \left(\log x\right)+4\cos \left(\log x\right)\right\}}{x^2} \\ \Rightarrow y_2=\frac{-3\cos \left(\log x\right)-{\displaystyle 4\sin \left(\log x\right)-}\left\{-3\sin \left(\log x\right)+4\cos \left(\log x\right)\right\}}{x^2} \\ \Rightarrow y_2=\frac{-3\cos \left(\log x\right)-{\displaystyle 4\sin \left(\log x\right)}}{x^2}-\frac{\left\{-3\sin \left(\log x\right)+4\cos \left(\log x\right)\right\}}{x^2} \\ \Rightarrow y_2=\frac{-\left\{3\cos \left(\log x\right)+{\displaystyle 4\sin \left(\log x\right)}\right\}}{x^2}-\frac{\left\{-3\sin \left(\log x\right)+4\cos \left(\log x\right)\right\}}{x^2} \\ \Rightarrow y_2=\frac{-y}{x^2}-\frac{y_1}{x} \\ \Rightarrow x^2{y}_2=-y-x{y}_1 \\ \Rightarrow x^2{y}_2+y+x{y}_1=0 \end{gathered}$$

Hence proved.