The D.E whose solution is y = (a + bx)cos x + (c + dx) sin x is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$y_{4} = y$

$y_{4} - 2 y_{2} + 4 y = 0$

$y_{4} + 2 y_{2} + y = 0$

$y_{4} + 2 y_{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} y = (a + b x) \cos x + (c + d x) \sin x \\ d i f f w . r . t o x \\ y_{1} = - (a + b x) \sin x + b \cos x + (c + d x) \cos x + d \sin x \\ d i f f w . r . t o' x' \\ y_{2} = - (a + b x) \cos x - b \sin x - b \sin x + d \cos x - (c + d x) \sin x + d \cos x \\ = - ((a + b x) \cos x + (c + d x) \sin x) - 2 b \sin x + 2 d \cos x \\ = - y - 2 b \sin x + 2 d \cos x \\ y_{2} + y = - 2 b \sin x + 2 d \cos x - - (1) \end{array}$

$\begin{array}{l} d i f f w . r . t o x \\ y_{3} + y_{1} = - 2 b \cos x - 2 d \sin x \\ d i f f w . r . t o x \\ \Rightarrow y_{4} + y_{2} = 2 b \sin x - 2 d \cos x \\ = - (y_{2} + y) f r o m (1) \\ \Rightarrow y_{4} + 2 y_{2} + y = 0 \end{array}$