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


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

Differentiating with respect to x, we get


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

Again differentiating with respect to x, we get


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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