Question
यदि $f(x) = a\sin (\log x)$, तो ${x^2}f''(x) + xf'(x) = . . . $
$y $ का $x $ के सापेक्ष अवकलन करने पर, $f(x) = a\cos (\log x) \frac{1}{x} $
पुन: अवकलन करने पर,
$f''\,(x) = - \frac{1}{{{x^2}}}a\cos (\log x) - \frac{1}{{{x^2}}}a\sin (\log x)$
$ \Rightarrow {x^2}f''(x) = - [a\cos (\log x) + a\sin (\log x)]$
==> ${x^2}f''(x) + xf'(x) = - a\sin (\log x) = - f(x)$.
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