If cos−1(yb)=2log(x2), x>0⇒x2d2ydx2+xdydx=

y=bcos(logx2b2)→(1) dydx=−2bxsinx(logx2b2)→(2)x2d2ydx2=2bsin[log(x2b2)]−4bcos[log(x2b2)]→(3) from (1), (2) & (3)x2d2ydx2+xdydx=−4y

If cos−1(yb)=2log(x2), x>0⇒x2d2ydx2+xdydx=

y=bcos(logx2b2)→(1) dydx=−2bxsinx(logx2b2)→(2)x2d2ydx2=2bsin[log(x2b2)]−4bcos[log(x2b2)]→(3) from (1), (2) & (3)x2d2ydx2+xdydx=−4y

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If $\cos^{- 1} (\frac{y}{b}) = 2 \log (\frac{x}{2}), x > 0 \Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} =$

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– 4y

– 8y

answer is B.

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Detailed Solution

$y = b \cos (\log \frac{x^{2}}{b^{2}}) \to (1) \begin{array}{l} \frac{d y}{d x} = \frac{- 2 b}{x} \sin x (\log \frac{x^{2}}{b^{2}}) \to (2) \\ x^{2} \frac{d^{2} y}{d x^{2}} = 2 b \sin [\log (\frac{x^{2}}{b^{2}})] - 4 b \cos [\log (\frac{x^{2}}{b^{2}})] \to (3) \end{array} \begin{array}{l} f r o m (1), (2) & (3) \\ \frac{x^{2} d^{2} y}{d x^{2}} + \frac{x d y}{d x} = - 4 y \end{array}$