A) \[y''=-{{\omega }^{2}}y\]
B) \[y''+y=0\]
C) \[y''+y'=0\]
D) none of these
Correct Answer: A
Solution :
\[y=A\cos \omega t+Bsin\omega t\] \[y'=-A\omega \sin \omega t+B\omega cos\omega t\] \[y''=-A{{\omega }^{2}}\cos \omega t-B{{\omega }^{2}}\sin \omega t\] \[y''=-{{\omega }^{2}}(cos\omega t+sin\omega t)\] \[y''=-{{\omega }^{2}}y\]
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