Question

If y = a cos (log_e x) + b sin (log_e x), then x² y₂ + xy₁ =

विकल्प

• 0

• y

• −y

• none of these

Answer 1

(c) −y

Here,

\[y = a \cos\left( \log_e x \right) + b \sin\left( \log_e x \right)\]

\[ \Rightarrow y_1 = - a\sin\left( \log_e x \right)\frac{1}{x} + b \cos\left( \log_e x \right)\frac{1}{x}\]

\[ \Rightarrow y_2 = \frac{- a\sin\left( \log_e x \right) + b \cos\left( \log_e x \right)}{x}\]

\[ \Rightarrow y_2 = \frac{- a\cos\left( \log_e x \right) - b \sin\left( \log_e x \right) - \left\{ - a\sin\left( \log_e x \right) + b \cos\left( \log_e x \right) \right\}}{x^2}\]

\[ \Rightarrow x^2 y_2 = - \left\{ a\cos\left( \log_e x \right) + b \sin\left( \log_e x \right) \right\} - \left\{ - a\sin\left( \log_e x \right) + b \cos\left( \log_e x \right) \right\} \]

\[ \Rightarrow x^2 y_2 = - y - x y_1 \]

\[ \Rightarrow x^2 y_2 + x y_1 = - y\]