Question

If $y=3\cos \left(\log \left(x\right)\right)+4\sin \left(\log \left(x\right)\right)$, show that $x^2{y}_2+x{y}_1+y=0$.

Answer 1

Prove the given condition:

$$\begin{gathered} y=3\cos \left(\log \left(x\right)\right)+4\sin \left(\log \left(x\right)\right) \\ \Rightarrow \frac{dy}{dx}=-\frac{3}{x}\sin \left(\log \left(x\right)\right)+\frac{4}{x}\cos \left(\log \left(x\right)\right) \\ \Rightarrow x\frac{dy}{dx}=-3\sin \left(\log \left(x\right)\right)+4\cos \left(\log \left(x\right)\right) \end{gathered}$$

Differentiating both sides,

$$\begin{gathered} \Rightarrow x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-\frac{3}{x}\cos \left(\log \left(x\right)\right)-\frac{4}{x}\sin \left(\log \left(x\right)\right) \\ \Rightarrow x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=-\{3\cos \left(\log \left(x\right)\right)+4\sin \left(\log \left(x\right)\right) \\ \Rightarrow x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=-y \\ \Rightarrow x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0 \\ \Rightarrow x^2{y}_2+x{y}_1+y=0 \end{gathered}$$

Hence, proved