Differential Equations Question 129
Question: The differential equation obtained on eliminating A and B from the equation $ y=A\cos \omega t+B\sin \omega t $ is
[Karnataka CET 2000; Pb. CET 2001]
Options:
A) $ {y}’’=-{{\omega }^{2}}y $
B) $ {y}’’+y=0 $
C) $ {y}’’+{y}’=0 $
D) $ {y}’’-{{\omega }^{2}}y=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {y}’=-A\omega \sin \omega t+B\omega \cos \omega t $
Again, $ {y}’’=-A{{\omega }^{2}}\cos \omega t-B{{\omega }^{2}}\sin \omega t $
$ =-{{\omega }^{2}}(A\cos \omega t+B\sin \omega t) $ . Therefore $ {y}’’=-{{\omega }^{2}}y $ .
BETA
