If f(x) = asin (log x),then {x^2}f''(x) + xf' | vedclass

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(B) Given $f(x) = a\sin (\log x)$.Differentiating with respect to $x$,we get:$f'(x) = a\cos (\log x) \cdot \frac{1}{x}$$x f'(x) = a\cos (\log x)$Diffe

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$-f(x)$

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(B) Given $f(x) = a\sin (\log x)$.Differentiating with respect to $x$,we get:$f'(x) = a\cos (\log x) \cdot \frac{1}{x}$$x f'(x) = a\cos (\log x)$Differentiating again with respect to $x$ using the product rule on the left side:$\frac{d}{dx}(x f'(x)) = \frac{d}{dx}(a\cos (\log x))$$x f''(x) + f'(x) = -a\sin (\log x) \cdot \frac{1}{x}$Multiply the entire equation by $x$:$x^2 f''(x) + x f'(x) = -a\sin (\log x)$Since $f(x) = a\sin (\log x)$,we substitute this into the equation:$x^2 f''(x) + x f'(x) = -f(x)$.

If $f(x) = a\sin (\log x)$,then ${x^2}f''(x) + xf'(x) = . . . $

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