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MISCELLANEOUS EXERCISE 1 (II) · Q250

Q.If y=Acos⁡(log⁡x)+Bsin⁡(log⁡x)y=A\cos(\log x)+B\sin(\log x), show that x2d2ydx2+xdydx+y=0x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}+y=0

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y′=−Asin⁡(log⁡x)⋅1x+Bcos⁡(log⁡x)⋅1x=−Asin⁡(log⁡x)+Bcos⁡(log⁡x)x.y'=-A\sin(\log x)\cdot\frac1x+B\cos(\log x)\cdot\frac1x=\frac{-A\sin(\log x)+B\cos(\log x)}{x}.

So xy′=−Asin⁡(log⁡x)+Bcos⁡(log⁡x)xy'=-A\sin(\log x)+B\cos(\log x). Differentiate this again w.r.t. xx:

y′+xy′′=−Acos⁡(log⁡x)⋅1x−Bsin⁡(log⁡x)⋅1x=−1x[Acos⁡(log⁡x)+Bsin⁡(log⁡x)]=−yx.y'+xy''=-A\cos(\log x)\cdot\frac1x-B\sin(\log x)\cdot\frac1x=-\frac{1}{x}\big[A\cos(\log x)+B\sin(\log x)\big]=-\frac{y}{x}. …

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