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If y = 3 cos (log x) + 4 sin (log x), show that

\(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0.\)

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Given,

y = 3 cos (log x) + 4 sin (log x)

Differentiating with respect to x, we get

\(\frac{dy}{dx}=-\frac{3sin(log\,x)}{x}+\)\(\frac{4cos(log\,x)}{x}\)

⇒ y1\(\frac{1}{x}\)[- 3 sin(log x) +4 cos (log x)]

Again differentiating with respect to x, we get

Now,

LHS = x2y2 + xy1 +y 

= (\(x^2(\frac{-sin(logx)-7\,cos(logx)}{x^2})\)\(x\times\frac{1}x\)[-3 sin(log x) + 4 cos(log x)] +3 cos (log x) + 4 sin(log x)

=  - sin (log x) - 7 cos (log x) - 3 sin (log x) + 4 sin (log x)

= 0 = RHS

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