Question

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

(a) 0
(b) y
(c) −y
(d) none of these

Answer 1

(c) −y

Here,
$$\begin{gathered} y=a\cos \left({\log }_{e}x\right)+b\sin \left({\log }_{e}x\right) \\ \Rightarrow y_1=-a\sin \left({\log }_{e}x\right)\frac{1}{x}+b\cos \left({\log }_{e}x\right)\frac{1}{x} \\ \Rightarrow y_2=\frac{-a\sin \left({\log }_{e}x\right)+b\cos \left({\log }_{e}x\right)}{x} \\ \Rightarrow y_2=\frac{-a\cos \left({\log }_{e}x\right)-b\sin \left({\log }_{e}x\right)-\left\{-a\sin \left({\log }_{e}x\right)+b\cos \left({\log }_{e}x\right)\right\}}{x^2} \\ \Rightarrow x^2{y}_2=-\left\{a\cos \left({\log }_{e}x\right)+b\sin \left({\log }_{e}x\right)\right\}-\left\{-a\sin \left({\log }_{e}x\right)+b\cos \left({\log }_{e}x\right)\right\} \\ \Rightarrow x^2{y}_2=-y-x{y}_1 \\ \Rightarrow x^2{y}_2+x{y}_1=-y \end{gathered}$$