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Question

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$ Differentiating with respect to $t$: $y' = -A \omega \sin \omega t + B \omega \cos \om

Vedclass · Accepted Answer

$y'' = - \omega^2 y$

Vedclass · Answer

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$ Differentiating with respect to $t$: $y' = -A \omega \sin \omega t + B \omega \cos \omega t$ Differentiating again with respect to $t$: $y'' = -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t$ Factoring out $-\omega^2$: $y'' = -\omega^2 (A \cos \omega t + B \sin \omega t)$ Since $y = A \cos \omega t + B \sin \omega t$,we substitute $y$ into the equation: $y'' = -\omega^2 y$ Thus,the differential equation is $y'' + \omega^2 y = 0$.

The differential equation obtained by eliminating $A$ and $B$ from the equation $y = A \cos \omega t + B \sin \omega t$ is

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