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Question
If y = a cos (loge x) + b sin (loge x), then x2 y2 + xy1 =
Options
0
y
−y
none of these
Solution
(c) −y
Here,
\[y = a \cos\left( \log_e x \right) + b \sin\left( \log_e x \right)\]
\[ \Rightarrow y_1 = - a\sin\left( \log_e x \right)\frac{1}{x} + b \cos\left( \log_e x \right)\frac{1}{x}\]
\[ \Rightarrow y_2 = \frac{- a\sin\left( \log_e x \right) + b \cos\left( \log_e x \right)}{x}\]
\[ \Rightarrow y_2 = \frac{- a\cos\left( \log_e x \right) - b \sin\left( \log_e x \right) - \left\{ - a\sin\left( \log_e x \right) + b \cos\left( \log_e x \right) \right\}}{x^2}\]
\[ \Rightarrow x^2 y_2 = - \left\{ a\cos\left( \log_e x \right) + b \sin\left( \log_e x \right) \right\} - \left\{ - a\sin\left( \log_e x \right) + b \cos\left( \log_e x \right) \right\} \]
\[ \Rightarrow x^2 y_2 = - y - x y_1 \]
\[ \Rightarrow x^2 y_2 + x y_1 = - y\]
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