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Question

If y=3cos(logx)+4sin(logx), then show that x2d2ydx2+xdydx+y=0

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Solution

Given, y=3cos(logx)+4sin(logx)

Differentiate both sides with respect to x, we get

dydx=3sin(logx)1x1+4cos(logx)1x1

dydx=1x[3sin(logx)+4cos(logx)]

xdydx=3sin(logx)+4cos(logx)

Differentiate both sides with respect to x, we get

xd2ydx2+dydx=3cos(logx)1x4sin(logx)1x

xd2ydx2+dydx=1x[3cos(logx)+4sin(logx)]

xd2ydx2+dydx=y/x

x2d2ydx2+xdydx+y=0
Hence, proved.

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