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प्रश्न
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
पर्याय
y" + y' = 0
y" − ω2 y = 0
y" = −ω2 y
y" + y = 0
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उत्तर
y" = −ω2 y
We have,
y = A cos ωt + B sin ωt .....(1)
Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dt} = - A\omega \sin \omega t + B \omega \cos \omega t\] .....(2)
Differentiating both sides of (2) again with respect to x, we get
\[\frac{d^2 y}{d t^2} = - A \omega^2 \cos \omega t - B \omega^2 \sin \omega t\]
\[ \Rightarrow \frac{d^2 y}{d t^2} = - \omega^2 \left( A \cos \omega t + B \sin \omega t \right)\]
\[ \Rightarrow \frac{d^2 y}{d t^2} = - \omega^2 y ..........\left[ \text{Using }\left( 1 \right) \right]\]
\[ \therefore y'' = - \omega^2 y\]
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