Differential Equations — Maths STD 12 Science — Question

y = A cos (log x) + B sin (log x)    ...(1)
Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=-A \sin (\log x) \cdot \frac{d}{d x}(\log x)+B \cos (\log x) \cdot \frac{d}{d x}(\log x) \\ & =\frac{-A \sin (\log x)}{x}+\frac{B \cos (\log x)}{x} \\ & \therefore x \frac{d y}{d x}=-A \sin (\log x)+B \cos (\log x)\end{aligned}$
Differentiating again w.r.t. x, we get
$\begin{aligned} & x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{-A \cos (\log x)}{x}+\frac{B \sin (\log x)}{x} \\ & \therefore x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=-[A \cos (\log x)+B \sin (\log x)]=-y \quad \ldots .[\text { By (1)] } \\ & \therefore x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0 \text { is the required D.E. }\end{aligned}$