A) \[y\]
B) \[-y\]
C) \[2y\]
D) \[-2y\]
Correct Answer: B
Solution :
\[y=a\cos (\log x)+b\sin (\log x)\] Þ \[y'=\frac{-\,a\sin (\log x)}{x}+\frac{b\cos (\log x)}{x}\] Þ \[xy'=-\,a\sin (\log x)+b\cos (\log x)\] Þ \[x{y}''+{y}'=\frac{-a\cos (\log x)}{x}-\frac{b\sin (\log x)}{x}\] Þ \[{{x}^{2}}{y}''+x{y}'=-[a\cos (\log x)+b\sin (\log x)]\] Þ \[{{x}^{2}}y''\,+\,xy'=-y.\]You need to login to perform this action.
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