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(B) Given the equation: $y=x[a \cos (\log x)+b \sin (\log x)]$. First, differentiate with respect to $x$: $\frac{d y}{d x} = [a \cos (\log x) + b \sin

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$x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+2 y=0$

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(B) Given the equation: $y=x[a \cos (\log x)+b \sin (\log x)]$. First, differentiate with respect to $x$: $\frac{d y}{d x} = [a \cos (\log x) + b \sin (\log x)] + x [-a \sin (\log x) \cdot \frac{1}{x} + b \cos (\log x) \cdot \frac{1}{x}]$ $\frac{d y}{d x} = \frac{y}{x} - a \sin (\log x) + b \cos (\log x)$. Multiply by $x$: $x \frac{d y}{d x} = y - ax \sin (\log x) + bx \cos (\log x)$. Differentiate again with respect to $x$: $x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{d y}{d x} - [a \sin (\log x) + ax \cos (\log x) \cdot \frac{1}{x}] + [b \cos (\log x) - bx \sin (\log x) \cdot \frac{1}{x}]$ $x \frac{d^2 y}{d x^2} = -a \sin (\log x) - a \cos (\log x) + b \cos (\log x) - b \sin (\log x)$. Multiply by $x$: $x^2 \frac{d^2 y}{d x^2} = -ax \sin (\log x) - ax \cos (\log x) + bx \cos (\log x) - bx \sin (\log x)$. Rearranging terms: $x^2 \frac{d^2 y}{d x^2} = -[ax \cos (\log x) + bx \sin (\log x)] - [ax \sin (\log x) - bx \cos (\log x)]$. Using $y = ax \cos (\log x) + bx \sin (\log x)$ and $x \frac{d y}{d x} - y = -ax \sin (\log x) + bx \cos (\log x)$, we get: $x^2 \frac{d^2 y}{d x^2} = -y - (x \frac{d y}{d x} - y) = -x \frac{d y}{d x}$. Wait, simplifying the original expression $y = x[a \cos(\log x) + b \sin(\log x)]$: Let $t = \log x$, then $x = e^t$. $y = e^t [a \cos t + b \sin t]$. This is a linear differential equation with constant coefficients in $t$. The characteristic equation is $(m-1)^2 + 1 = 0 \Rightarrow m^2 - 2m + 2 = 0$. Converting back to $x$: $x^2 \frac{d^2 y}{d x^2} - x \frac{d y}{d x} + 2y = 0$.

If $a$ and $b$ are arbitrary constants, then the differential equation corresponding to the family of curves given by $y=x[a \cos (\log x)+b \sin (\log x)]$ is

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