मराठी

The Differential Equation Obtained on Eliminating a and B from Y = a Cos ωT + B Sin ωT, is

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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

MCQ
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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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पाठ 21: Differential Equations - MCQ [पृष्ठ १४०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
MCQ | Q 7 | पृष्ठ १४०

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