A) \[{y}''=-{{\omega }^{2}}y\]
B) \[{y}''+y=0\]
C) \[{y}''+{y}'=0\]
D) \[{y}''-{{\omega }^{2}}y=0\]
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 \[{y}''=-{{\omega }^{2}}y\].
You need to login to perform this action.
You will be redirected in
3 sec
